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CHEMICAL KINETICS
The rate of a reaction means the speed with which the reaction takes place. This isexpressed either in terms of decrease in the concentration of a reactant per unit time orincrease in the concentration of a product per unit time. Hence the rate of a reaction may bedefined as follows :The rate of reaction is the change in the concentration of any one of the reactants or productsper unit time i.e.,
Rate of reaction =intervalTime
reactantaofionconcentratin theDecrease
or =intervalTime
productaofionconcentratin theIncrease
For example, consider the reaction ,PCl
5 PCl
3 + Cl
2
Suppose in a time interval t, decrease in concentration of PCl5 is [PCl
5] and increase in
the concentration of PCl3 and Cl
2 are [PCl
3] and [Cl
2] respectively, where square
brackets indicate molar concentrations (moles/litre) of the substances involved. Then we have,
Rate of reaction = 5 3 2[PCl ] [PCl ] [Cl ]
t t t
It may be emphasized that the rate of reaction is always positive. The minus sign alongwiththe first term is used simply to show that the concentration of the reactant (PCl
5) is
decreasing while plus sign along with the other two terms is used to show that theconcentration of the products (PCl
3) and (Cl
2) is increasing.
In general, for any reaction of the typeA + B C + D
Rate of reaction =t
]D[
t
]C[
t
]B[
t
]A[
Units of the Rate of Reaction :Since concentration is usually expressed in moles/litre and the time is taken in seconds orminutes, the unit of the rate of reaction is moles litre–1 sec–1 (mol L–1 s–1) or moles litre–1 min–1
(mol L–1 min–1).In case of gaseous reactions, pressures are used in place of molar concentrations. Aspressures are expressed in atmospheres, therefore the units of the rate of reaction are atmmin–1 or atm s–1 etc. The relationship between partial pressure of a gas in a reaction mixture
and its molar concentration follows from the relationship PV = nRT i.e.n P
V RT
Where partial pressure , P = pressureTotalmolesofno.Total
gas theofmolesofNo.
CHEMICAL KINETICS
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Average Rate and Instantaneous Rate of Reaction :A difficulty arises in stating the rate of reaction as above. This is because according to theLaw of Mass Action , the rate of reaction depends upon the molar concentrations ofreactants which keep on decreasing with the passage of time (while those of the productskeep on increasing) . Therefore, the rate of reaction does not remain constant throughout.Thus the rate of reaction as defined above is the ‘average rate of reaction’ during the timeinterval chosen.To know the rate of reaction at any instant of time during the course of a reaction, weintroduce the term ‘instantaneous rate of reaction’ which may be defined as follows :The rate of reaction at any instant of time is the rate of change of concentration (i.e. change ofconcentration per unit time) of any one of the reactants or products at that particular instant oftime.To express the instantaneous rate of reaction , as small interval of time (dt) is chosen at thatparticular instant of time during which the rate of reaction is supposed to be almostconstant. Suppose the small change in concentration is dx in the small interval of time dt. Then
the rate of reaction at that instant is given bydt
dx.
Measurement of the Rate of Reaction :In order to measure the rate of a reaction, the progress of the reaction is followed by studyingthe concentration of one of the reactants or products at different intervals of time. the mostcommon practice to do so is to withdraw small amount of the reaction mixture (2 cm3 or 5 cm3)at different intervals of time, cool it down immediately to nearly 0° C to arrest the reaction(called freezing the reaction) and then find out the concentration of the reactant or the productby suitable method usually by titration against a suitable reagent. However, this method is notpreferred when some observable property like volume, pressure, optical rotation etc. changeswith time and can be observed directly at different intervals of time without stopping the reaction.it is important to mention that except concentration, all other factors (like temperature etc.)which affect the rate of the reaction are kept constant during the kinetic study of the reaction.
0.00
0.01
0.02
0.03
0.04
0.05
5 10 15 20 25
O
A
P
B
x = 0.04MOLE
TIME (IN MINUTES)
CO
NC
EN
TR
AT
ION
OF
AR
EA
CT
AN
T(I
NM
OL
ES
/LIT
RE
)
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CHEMICAL KINETICS
If concentration of one of the reactants is studied at different intervals of time and these valuesare plotted against the corresponding times, a graph of the type shown in figure is obtained.
Calculation of instantaneous rate of reaction :To know the rate of the reaction at any time t, a tangent is drawn to the curve at the pointcorresponding to that time figure and it is extended on either side so as to cut the axes, sayat the points A and B. then
Rate of reaction =Time
ionconcentratin theChange
=t
x
=OB
OA Slope of the tangent
Thus the slope of the tangent gives the rate of reaction.For example , from the (figure) in the present case, at time t = 10 minutes, x = 0.04 moleand t = 20 minutes = 20 60 = 1200 seconds , therefore, rate of reaction at the end of10 minutes will be 0.04/1200 = 3.33 10–5 mol L–1 s–1.
Calculation of the average rate of reaction :To calculate the average rate of reaction between any two instants of time say t
1 and t
2, the
corresponding concentrations x1 and x
2 are noted from the graph. Then
Average rate of reaction =12
12
tt
xx
For example, from the (figure) between the time interval 5 to 15 minutes,
Average rate 0018.010
018.0
515
012.003.0
and mol L–1 min–1
If concentration of one of the products is plotted against time, the type of curve obtained andthe rate of reaction at any instant of time are calculated as shown in the figure.
0.00
0.01
0.02
0.03
0.04
0.05
5 10 15 20 25
O
x1
TIME (IN MINUTES)
CO
NC
EN
TR
AT
ION
OF
RE
AC
TAN
T(M
OL
L)
-1
x2
t1 t2TIME
B
P
A
O
CO
NC
EN
TR
AT
ION
OF
APR
OD
UC
T
Illustration 1 :In a reaction the concentration of a reactant (A) changes from 0.200 mol litre –1 to0.150 mol litre–1 in 10 minutes. What is the average rate of reaction during this interval?
Solution :[A] = [A]
final – [A]
initial
= [0.150 – 0.200]= – 0.050 mol litre–1
t = 10 minutes
Average rate of reaction =t
]A[
=10
50.0 = 0.005 mol litre–1 min–1
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CHEMICAL KINETICS
Illustration 2 :Decomposition of N2O5 is expressed by the equation ,
N2O5 2NO2 + ½O2
If in a certaion time interval, rate of decomposition of N2O5 is 1.8× 10–3 mol litre–1
min–1, what will be the rates of formation of NO2 and O2 during the same interval ?Solution :. The rate expression for the decomposition of N
2O
5 is
–t
]ON[ 52
=2
1
t
]NO[ 2
= 2 ×t
]O[ 2
Sot
]NO[ 2
= 2t
]ON[ 52
= 2 × 1.8 × 10–3
= 3.6 × 10–3 mol litre–1 min–1
andt
]O[ 2
=2
1
t
]ON[ 52
=2
1 × 1.8 × 10–3
= 0.9 ×10–3 mol litre–1 min–1
[Rate is always positive and hence =t
]ON[ 52
is taken positive.
Illustration 3 :If the decomposition of nitrogen (V) oxide 2N2O5 4NO2 + O2, following a firstorder kinetics.(i) Calculate the rate constant for a 0.04 M solution, if the instantaneous rate is
1.4 10–6 mol L s–1.(ii) Also calculate the rate of reaction when the concentration of N2O5 is 1.20 M.(iii) What concentration of N2O5 would give a rate of 2.45 105 mol L–1s–1?
Solution :(i) As the given reaction is of first order, therefore,
Rate = k[N2O
5]
or6
2 5
rate 1.4 10k
[N O ] 0.04
[M = conc. in mol L–1]
= 3 . 5 10–5s–1
(ii) Now if the concentration of N2O
5 is 1.20 M, then
rate = k[N2O
5]
= 3.5 10–5 1.20 = 4.2 10–5 mol L–1 s–1
(iii) To obtain concentration of N2O
5when the rate is 2.45 10–5 mol –1 s–1, we note that
[N2O
5] =
5
5
Rate 2.45 10
k 3.5 10
= 0.7 mol L–1 or 0.7 M
Factors Affecting the Reaction Rate :The rate of any particular reaction depends upon the following factors:
(i) Concentration of the reactants : Greater the concentrations of the reactants, fasteris the reaction. Conversely, as the concentrations of the reactants decrease, the rate ofreactions also decreases.
(ii) Temperature : The rate of reaction increases with increase of temperature. In most of thecases, the rate of reaction becomes nearly double for 10 K rise of temperature. In some cases,reactions do not take place at room temperature but take place at higher temperature.
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(iii) Presence of Catalyst : A catalyst generally increases the speed of a reaction withoutitself being consumed in the reaction. In case of reversible reactions, a catalyst helps to attainthe equilibrium quickly without disturbing the state of equilibrium.
(iv) Surface area of the reactants : For a reaction involving a solid reactant or catalyst,the smaller is the particle size, i.e., greater is the surface area, the faster is the reaction.
(v) Presence of light : Some reactions do not take place in the dark but take place in thepresence of light e.g., H
2 + Cl
2 2HCl. Such reactions are called “photochemical reactions”.
Now we shall discuss the quantitative effect of concentration and temperature on therate of reaction. The study of quantitative effect of concentration on the rate of reactionleads to the introduction of a number of new terms such as order of reaction, specificreaction rate (rate constant) and molecularity of reaction. It also leads to the study ofmechanism of the reactions. These different aspects are discussed in the following sections.
Single step reaction is called elementary reaction while a reaction occurring in two or moresteps is called complicated or consecutive reaction. Each step of a complicated reaction is anelementary reaction.The minimum number of molecules, atoms or ions of reactants required for an elementaryreaction to occur is indicated by the sum of the stoichiometric coefficients of the reactant(s) inthe chemical equation, is known molecularity of the reaction. Thus for an elementary reactionrepresented by the general chemical equation .
aA + bB productsmolecularity = a + bExamples: Reactions Molecularity
PCl5
PCl3 + Cl
21
H2 + I
2 2HI 2
A complicated reaction has no molecularity of its own but molecularity of each of the stepsinvolved in its mechanism.For example :
The reaction : 2NO + 2H2
N2 + 2H
2O, takes place in the sequence of following
three steps . I. NO + NO N2O
2(fast and reversible)
II. N2O
2 + H
2 N
2O + H
2O (slow)
III. N2O + H
2 N
2 + H
2O (fast)
The molecularity of each step involved in mechanism is 2 i.e., each step is bimolecular. Soas regards the molecularity of the reaction under consideration we simply say that thereaction has mechanism and each step involved in it is bimolecularHowever, there is another view also , according to which molecularity of a complicatedreaction is taken to be equal to the number of molecules, atoms or ions of reactant(s) and/orintermediates coming into contact and colliding simultaneously in the slowest step i.e. the rate-determining step (RDS) of the reaction.
For example the reaction ROH + HCl ZnCl2 RCl + H2O, which actually is a nucleophilic
substitution reaction
I + H (fast)
II R + (slow)
III R + R – (fast)
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Here the step II is the slowest step and hence the R.D.S. the molecularity of which is one.Therefore, the mechanism is called unimolecular nucleophilic substitution (S
N1) and the
reaction is said to follow SN1 mechanism.
The mathematical equation showing the relation of rate of reaction with concentration ofreactant(s) is called rate-law of the reaction. For example, the kinetic experiment carried onthe reaction.
2 NO + 2 H2(g) N
2(g) + 2H
2O (g)
has shown that :(i) rate increases 4 - fold when conc. of NO is doubled keeping the conc. of H
2 constant
(ii) rate gets just doubled when conc. of H2 is doubled keeping that of NO constant, and
(iii) rate increases 8 - fold when concentrations of both NO and H2 are doubled
simultaneously. These experimental rate data fit into following equation.Rate [NO]2[H
2]
This equation is called experimentally observed “Rate law” of the reaction.Order of reaction is defined as the sum of the powers of the concentration termsappearing in experimentally observed rate law.Thus for the reaction (reduction of nitric oxide by hydrogen) considered above .
Order w.r.t. NO = 2 Order w.r.t. H2= 1
Overall order = 2 + 1 = 3In general , let a reaction represented by the chemical equation:
aA + bB Productsobeys the following rate law.
Rate [A]m [B]n, or Rate = k[A]m [B]n
Where k = rate constant of reaction, a constant at constant temp and is actually the rate ofreaction when conc. of each reactant is equal to unity k is also called velocity constant orspecific reaction rate.Note that m and n are experimental quantities which may or may not be equal to therespective stoichiometric coefficients.Also note that if either of A or B is taken in large excess as compared to another,the order w.r.t this reactant will be zero so order = m (when B is in large excess), andorder = n (when A is large excess) , i.e.,
Rate = k[A]m, k = k[B]n another constant, when B is in large excess.Order w.r.t. A = mOrder w.r.t. B = 0
Rate = k[B]n, k = k[A]m = another constant, when A is in large excess.Order w.r.t. A = 0Order w.r.t. B = n
k or k is actually known as pseudo rate constantThus, a bimolecular reaction conforms to the first - order when one of the reactants istaken in large excess and the reaction is said to be pseudo unimolecular or pseudofirst order.An example of this is the hydrolysis of ester by dil. acid i.e.
CH3COOC
2H
5 + H
3O+ H CH
3COOH + C
2H
5OH
The reaction is originally of second - order obeying the kinetics.Rate = k [CH
3COOC
2H
5] [H
2O]
But the reaction is usually carried out taking dilute aqueous solution of ester and acid (HCl)such that in the reaction mixture water exists in large excess as compared to ester or betterto say, water has almost its maximum concentration of 55.5 mole L–1. So there is noappreciable change in the concentration of water and the same remains practically constant.
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k[H2O] becomes another constant (k) called pseudo rate constant and reactions obeys the
following 1st order kineticsRate = k [CH
3COOC
2H
5]
Thus , the molecularity of the above reaction is ‘two’ but its order is ‘one;.In the light of the above concept one may define order of reaction as “ the number ofmolecules of the reactant(s) whose concentration alters during the chemical change isthe order of reaction”.
Distinction between Molecularity and Order(i) Molecularity is a theoretical property while order is an experimental property.(ii) Molecularity concerns to mechanism, order concerns to kinetics(iii) Molecularity is always a whole number and never zero while order may be any number
zero, fractional and integral.(iv) Molecularity is invariant but order may vary with change in experimental condition. As
for example, the isomerisation of cyclo - propane represented by the chemical equationgiven below is a first order reaction at high cyclo propane pressure and is a secondorder reaction at low cyclo propane pressure.
isomerisation CH3 CH3
(v) Reactions of higher molecularity is rare since the chance of coming into contact andcolliding simultaneously decreases as the number of molecules involved in collisionincreases.Unit of rate constant
nA Product
ndxk[conc.]
dt
n[conc.]k[conc.]
sec
k = [conc.]1–n sec–1
= [mole/litre]1–n sec–1
Illustration 4 :State the order w.r.t. each reactant, order of overall reaction and units of rateconstant in each of the following reactions(i) H2O2 + 3I– + 2H+ 2H2O + I3
–(aq)Rate = k [H2O2] [I
–](ii) CO + Cl2 COCl2
Rate = k[CO]2[Cl2]1/2
Solution :(i) Rate = k[H
2O
2] [I–]
Order w.r.t. H2O
2 =1; Order w.rt. I– = 1
Overall order = 2Units of k = (mol L–1)1–2 s–1 = mol–1 Ls–1
(ii) Rate = k[CO]2 [Cl2]1/2
Order w.r.t. to CO = 2; Order w.r.t. Cl2 = ½
Overall order = 2.5
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Illustration 5 :For the reaction , 2 NO + Cl2 2 NOClat 300 K following data are obtained
expt. No.][Cl[NO]
ionConcentratInitial
2 Initial rate
1. 0.010 0.010 1.2 ×10–4
2. 0.010 0.020 2.4 × 10–4
3. 0.020 0.020 9.6 ×10–4
Write rate law for the reaction. What is the order of the reaction? Also calculate thespecific rate constant.
Solution :Let the rate law for the reaction be
Rate = k[NO]x[Cl2]y
From Expt. (1) , 1.2 × 10–4 = k[0.010]x[0.010]y ...... (i)From Expt. (2), 2.4 × 10–4 = k[0.010]x[0.020]y ...... (ii)Dividing Eq. (ii) by Eq. (i),
4
4
102.1
102.4
= y
y
100.0
020.0
2 = 2y
y = 1From Expt. (2), 2.4 × 10–4 = k[0.010]x[0.020]y ...... (ii)From Expt. (3), 9.6 × 10–4 = k[0.020]x[0.020]y ...... (iii)Dividing Eq. (iii) by Eq. (ii),
4
4
104.2
106.9
= y
x
100.0
020.0
4 = 2x
x = 2Order of reaction = x + y = 2 + 1 = 3Rate law for the reation is , Rate = k [NO]2[Cl
2]
Considering Eq. (i) (again) , 1.2 × 10–4 = k[0.010]2[0.010]
k = 3
4
]010.0[
102.1 = 1.2 × 102 mol–2 litre2 sec–1
Illustration 6 :For the hypothetical reaction , 2 A + B productsThe following data are obtained :Expt No. Initial conc. of (A) initial conc. of (B) initial rate
(mol L–1) (mol L–1) (mol L–1s–1)1. 0.10 0.20 3 × 102
2. 0.30 0.40 3.6 × 103
3. 0.30 0.80 1.44 × 104
4. 0.10 0.40 ........5. 0.20 0.60 ........6. 0.30 1.20 ........Find out how the rate of the reaction depends upon the concentration of A and Band fill in the blanks.
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Solution :From Expt (2) and (3), it is clear that when concentration of A is kept constant and that of Bis doubled, the rate increases four times. This show that the reaction is of second order withrespect to B.Similarly , from Expt. (1) and (2), it is observed that when concentration of A is increasedthree times and that of B two times, the rate becomes twelve times. Hence, the reaction is firstorder with respect to A.Thus the rate law for the reaction is :
Rate = k [A] [B]2
Fill in the blanks : Substituting the values of Expt. (1) in the rate equation,3 × 102 = k[0.10][0.20]2
or k = 2
2
]20.0][10.0[
103 = 7.5 ×104L2mol–2s–1
Expt.(4): Rate = k[0.10][0.40]2
= 7.5 ×104 ×0.10 ×0.40 × 0.40 = 1.2 × 103 mol L–1s–1
Expt (5) : Rate = k[0.20] [.60]2
= 7.5 ×104 ×0.20 ×0.60 × 0.60= 5.4 × 103 mol L–1s–1
Expt (6) : Rate = k[0.30] [1.20]2
= 7.5 ×104 ×0.30 ×1.20 × 1.20= 3.24 × 104 mol L–1s–1
As already mentioned in the reaction:2NO + 2H
2 N
2 + 2H
2O
Obeys the following third order kineticsRate [NO]2 [H
2]
The kinetics is not in tune with chemical equation. As the law of mass action suggests eachconcentration term should be raised to power 2. Neither experimental facts nor those comingfrom Law of mass action can be unacceptable to us. Under this condition we are led to believethat the reaction does not occur according to the chemical equation as written. That is thereaction is not elementary but is complicated. In order to explain the observed rate law followingmechanism has been proposed.
I NO + NO N2O
2(fast and reversible)
II N2O
2 + H
2 N
2O + H
2O (slow)
III N2O + H
2 N
2 + H
2O (fast)
The step II being the slowest i.e., the rate - determining step (R.D.S.). Note that the rate offormation of N
2 cannot be faster than the rate of formation of N
2O. So, the rate of overall
reaction or rate of formation of N2 will be equal to the rate of step II which according to law of
mass action may be given as.Rate of overall reaction = Rate of step II = k[N
2O
2] [H
2]
Where k = rate constant of step IIThe conc. of the intermediate (N
2O
2) may be evaluated by applying law of mass action upon
the equilibrium existing in step I as mentioned below.
Kc = 2 2
2
[N O ]
[NO] or [N
2O
2] = K
c [NO]2
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Putting this in the rate expression, we getRate = k.k
c[NO]2[H
2] = k
obs [NO]2[H
2]
Where kobs
= k.kc = observed rate constant of the reaction
Rate [NO]2 [H2]
This is the same rate - law as observed experimentally.
Illustration 7 :The possible mechanism for the reaction
2 NO + Br2 2 NOBris NO + Br2 NOBr2 (fast)
NOBr2 + NO 2 NOBr (slow).Establish the rate law.
Solution :As the slowest step is the rate determining step, therefore the rate law isR = k[NOBr
2] [NO]
Now since NOBr2 is an intermediate its concentration can be calculated from step 1 as
follows :
2
2
[NOBr ]K
[NO][Br ] [K = eq. constant]
or [NOBr2] = K [NO] [Br
2]
Substituting this value in above equationr = k. K[NO]2[Br
2]
or rate = k [NO]2 [Br2] [ k K = constant k ]Q
Rate = k [NO]2 [Br2]
Illustration 8 :Rate law for ozone layer depletion is
3d[O ]
dt = ][O
]K[O
2
23
Give the probable mechanism of reaction?Solution : O
3 O
2 + O (fast reaction)
O3 + O 2O
2(slow reaction, rate constant k)
Rate = k [O3][O] ....[1]
Kc=
]O[
]O][O[
3
2 or [O] = Kc ]O[
]O[
2
3 , (equilibrium constant Kc)
putting the value in (1)
From (1) Rate = k . [O3] . kc ]O[
]O[
2
3 = k . K
c.
]O[
]O[
2
23 = K .
]O[
]O[
2
23
k = k × Kc.
Illustratrion 9 :In hypothetical reaciton A2 + B2 2AB follows the mechanism as given below:
A2 A + A (fast reaction)A + B2 AB + B (slow reaction) A + B AB (fast reaction)
Give the rate law and order of reaction.
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Soluton :Slowest step is rate determining.
Rate = k[A][B2] ......(1)
Here [A] should be eleminated.
Kc =
]A[
]A][A[
2 =
]A[
]A[
2
2
[A] = 2/1cK [A
2]1/2
From (1). Rate = k 2/1cK [A
2]1/2[B
2]
= K[A2]1/2[B2] ; [K = k . K
c]
Order = 1 + 1/2 = 3/2
Illustration 10 :For the formation of phosgene from CO(g) and chlorine,
CO(g) + Cl2(g) COCl2(g)The experimentally determined rate equation is,
2d[COCl ]dt
= k[CO][Cl2]3/2
Is the following mechanism consistent with the rate equation?(i) Cl2 2Cl (fast)(ii) Cl + CO COCl (fast)(iii) COCl + Cl2 COCl2 + Cl (slow)
Solution :Multiplying equation (ii) by 2 and adding (i) , we get:
Cl2 + 2CO 2COCl
K =2
2
2
]CO][Cl[
]COCl[
[COCl] = (K)1/2 [Cl2]1/2 [CO] ...... (1)
Slowest step is rate determining, henceRate = k[COCl][Cl
2] ...... (2)
From (1) and (2), we getRate = kK1/2 [Cl
2]1/2[Cl
2][CO]
Rate = k'[Cl2]3/2[CO] [k' = k.k1/2]
Thus, rate law is in accordance with the mecahnism.
A reaction is said to be of the first order if the rate of the reaction depends upon only onconcentration term only. Thus we may haveFor the reaction : A Products
Rate of reaction [A].For the reaction : 2 A Products
Rate of reaction [A] only..For the reaction : A + B Products
Rate of reaction [A] or [B] only..
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Let us consider the simplest case viz.A Products
Suppose we start with ‘a’ moles per litre of the reactant A. After time t, suppose x moles perlitre of it, have decomposed. Therefore, the concentration of A after time t = (a – x) moles perlitre. Then according to Law of Mass Action.
Rate of reaction (a – x)
i.e., dt
dx(a – x)
or kdt
dx (a – x) ..… (1)
where k is called the rate constant or the specific reaction rate for the reaction of the first order.The expression for the rate constant k may be derived as follows :Equation (1) may be rewritten in the form
xa
dx
= kdt ..…(2)
Integrating equation (2), we get kdtx–a
dx
orxa
aln
t
1k
..…(3)
orxa
alog
t
303.2k
..…(4)
Equation (4) is sometimes written in another form which is obtained as follows :If the initial concentration is [A]
0 and the concentration after time t is [A], then putting a = [A]
0
and (a – x) = [A] equation (iv) becomes
]A[
]A[log
t
303.2k 0 ..… (5)
Further, putting a = [A]0 and (a – x) = [A] in eqn. (3), we get
kt = ln ]A[
]A[ 0
which can be written in the exponential form as
kt0 e]A[
]A[ or
kt
0
e]A[
]A[
or [A] = [A]0 e–kt ..… (6)
Illustration 11 :It was found that cane sugar solution in water was hydrolysed to the extent of25 per cent in one hour. Calculate the time that will be taken for the sugar to behydrolysed to the extent of 50 % , assuming that the reaction is of the first order.
Solution :
2.303 at log
k a x
Wherex
a = f
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f1 1
2.303 1 2.303t log log(1 f )
k (1 f ) k
..…(1)
Putting f = 0.25 at t0.25
= 1 hour from the data
t0.25
= 1 hr =1
2.303log(1 0.25)
k =
1
2.303log 4
k/3 ..…(2)
Similarly t0.5
=1
2.303 2.303log(1 0.5) log 2
k k ..…(3)
Dividing (iii) by (ii)0.5t log 2
1hr log 4 / 3 = 2.4 hours
Hence time required for 50% hydrolysis = 2.4 hours.
Illustration 12 :The half time of first order decomposition of nitramide is 2.1 hour at 15°C.
NH2NO2(s) N2O(g) + H2O(l)If 6.2 g of NH2NO2 is allowed to decompose, calculate(a) time taken for NH2NO2 to decompose 99% and(b) volume of dry N2O produced at this point measured at STP.
Solution :
For a first-order reaction, rate constant expression is k = 0
t
[A]2.303log
t [A]
Initial moles of nitramide =6.2
0.162
2.303 2.1 0.1t log 13.95 hours
0.693 0.001
Since, the decomposition is 99 %, so 99 % of the initial moles of NH2NO
2 would be
converted to N2O.
Moles of N2O =
0.1 99
100
Volume of N2O at STP =
0.1 99 22.42.217 litre
100
.
Illustration 13 :A drug becomes ineffective after 30 % decomposition. The original concentration ofa sample was 5mg/mL which becomes 4.2 mg/mL during 20 months. Assuming thedecomposition of first order , calculate the expiry time of the drug in months. What isthe half life of the product?
Solution : k =t
303.2log
xa
a
=20
303.2 log
10
2.4
5
= 0.00872 min–1
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Expiry time 't' may be calculated as
k =t
303.2 log
10 xa
a
0.00872 =
t
303.2 log
10 70
100
t = 40.9 41 months t1/2
=00872.0
693.0 = 79.4 months
Illustration 14 :A first order reaction is 20 % completed in 10 minutes. Calculate the time taken forthe reaction to go to 80 % completion.
Solution :Applying first order equation,
k =t
303.2 log
10 )20100(
100
=
10
303.2 log
10 80
100 = 0.0223 min–1
Again applying first order equation,
t =2.303
k log
10 )80100(
100
=0223.0
303.2 log
10 20
100 = 72.18
Some Important Characteristics of First Order Reactions :(i) Any reaction of the first order must obey equation (5) or (6). This may be tested in any one
of the following ways :(a) Substitution method : Starting with a known concentration ‘a’ or [A]
0, the
concentration of the reactant (a – x) or [A] at different intervals of time may be noted.For every value of t, the corresponding value of (a – x) i.e. [A] may be substituted inequation (vi) or (vii). If the values of k thus obtained are nearly constant (within theexperimental error), the reaction is of the first order.
(b) Graphical method : Equation (6) may be written as
]Alog[]Alog[]A[
]A[logt
303.2
k0
0
or 0]A[logt303.2
k]A[log ..... (7)
This is the equation of a straight line (y = mx + c). Thus if log [A] or log (a – x) values areplotted against time ‘t’, the graph obtained should be a straight line if the reaction is of thefirst order.The intercept made on the y-axis would be ‘log [A]
0’ and the slope of the line would be
equal to
303.2
ki.e., slope = –
303.2
k
From this, the value of k can be calculated.
log A
t
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(ii) Half-life Period : The time taken for any fraction of the reaction to be completed isindependent of the initial concentration. For example, let us test the truth of this statement forhalf of the reaction to be completed. Equation (5) may be written as
xa
alog
k
303.2t
..… (8)
When half of the reaction is completed, x = a/2. Representing the time taken for half of thereaction to be completed by t
1/2 equation (8) becomes
t1/2
=k
693.02log
k
303.2
2
aa
alog
k
303.2
i.e,
k
693.0t 2/1
Thus ‘a’ does not appear in this equation so that t1/2
is independent of a. Similarly, it can beseen that t
1/2, t
2/3 etc. will also be independent of the initial concentration.
The time taken for half of the reaction to be completed i.e., the time in which the concentrationof a reactant is reduced to half of its original value is called Half-life period of the reaction.The above result can also be used to test a reaction of the first order. The method is calledfractional life method.
(iii) The units of k are independent of the units in which the concentrations are expressed. This isobvious from equation (5) because in this equation k depends upon the ratio of two
concentrations viz.xa
a
and the time t. Thus so long as both a and (a – x) are expressed in the
same concentration units, the value of k is not affected. The units of k, therefore, depend onlyupon the units of time ‘t’. Depending upon whether t is expressed in second, minutes or hours,k would be in sec–1, min–1 or hr–1 respectively.
Examples of the Reactions of First Order1. Decomposition of Nitrogen Pentoxide : The compound, nitrogen pentoxide, is a
volatile solid which decomposes in the gaseous state as well as in the form of its solution in aninert solvent like carbon tetrachloride, chloroform etc. according to the equation.
224252 O2
1NO2ONON
2NO2
(i) When the reaction is carried out in the solution, N2O
4 and NO
2 remain in the
solution and the volume of oxygen gas collected is noted at different intervals of time. Itis obvious that Volume of oxygen gas collected at any time (V
t) x VV
t.Amount of
N2O
5 decomposed (x) i.e.,
(ii)
(a)takeninitiallyONofAmount
essel)reaction v theheating
bydoneis(which)(V timeinfinite
atcollectedgasoxygenofVolume
52
Va
Vx t
i.e., Substituting these values in the first order equation viz.
xa
alog
t
303.2k
We get
tVV
Vlog
t
303.2k
The constancy in the value of k proves the reaction to be of the first order.
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Illustration 15 :For decomposition of N2O5 in CCl4 solution at 320 K
2N2O5 4NO2 + O2
show that the reaction is of first order and also calculate the rate constant:Time in minutes 10 15 20 25 Volume of O2 evolved (in ml) 6.30 8.95 11.40 13.50 34.75
Solution :If the reaction is of first order, it must obey the equation
2.303 ak log
t (a x)
In the above reaction, NO
2 remains in solution and oxygen is liberated and collected at
different intervals of time.Therefore , V
t x V a
Substituting these values in the first order equation
t
V2.303 a 2.303k log log
t a x t V V
Time Vt
V – Vt
t
V2.303k log
t V V
10 6.30 34.75 – 6.30 = 28.452.303 34.75
k log 0.019810 28.45
15 8.95 34.75 – 8.95 = 25.802.303 34.75
k log 0.019815 25.80
20 11.40 34.75 – 11.40 = 23.352.303 34.75
k log 0.0198200 23.35
25 13.50 34.75 – 13.50 = 21.252.303 34.75
k log 0.019825 21.25
Since the value of k comes out to be constant the reaction, therefore, is of first order. Theaverage value of rate constant is 0.0198 min–1.
Decomposition of Hydrogen Peroxide : The decomposition of hydrogen peroxidein aqueous solution (catalysed by the presence of finely divided platinum) takes placeaccording to the equation
H2O
2 H
2O + 2
1 O2
The kinetics of this reaction may be studied either by the same method as done earlier (i.e.collecting the oxygen gas produced and noting its volume at different intervals of time) or bymaking use of the fact that H
2O
2 solution can be titrated against KMnO
4 solution. Thus by
withdrawing equal amounts of the solution (usually 5 cc) at regular intervals of time and titratingagainst the same KMnO
4 solution, the amount of H
2O
2 present can be found every time. It is
obvious that for the same volume of the reaction solution withdrawn,
(a)OHof
ionconcentartInitial
)(V timezeroati.e.reaction
theofntcommenceme thebefore
usedsolutionKMnOofVolume
220
4
i.e., 0Va
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x)-(ai.e.instant
at thatpresent
OHofAmount
)(V t timeofinstantanyat
usedsolutionKMnOofVolume22
V) x-(at
4
t
Substituting these values in the first order equation, we get
xa
alog
t
303.2k
or
t
0
V
Vlog
t
303.2k
The decomposition of hydrogen peroxide, as tested by this equation, is found to be of the firstorder.
Illustration 16 :From the following data show that the decomposition of an aqueous solution ofhydrogen peroxide is of first order :Time (minutes) 0 10 20 30V (ml) 46.1 29.8 19.6 12.3where V is the volume of potassium permanganate solution in ml required todecompose a definite volume of the peroxide solution.
Solution :It is evident from the given data that at zero time, titre value is proportional to the originalconcentration of hydrogen peroxide, i.e., a. The titre value at any time t corresponds toundecomposed hydrogen peroxide, i.e., (a – x).Substituting the value in the equation
1
2.303 ak log , we get
t a x
(i)2
1
2.303 46.1k log 4.364 10
10 29.8
(ii)2
1
2.303 46.1k log 4.276 10
20 19.6
(iii)2
1
2.303 46.1k log 4.404 10
30 12.3
Since k1 comes out to be constant in the two cases, the reaction is a first order one with the
average value of three.
Consider the following acid-catalysed reactions:(i) Hydrolysis of ethyl acetate
alcoholEthyl52
acidAcetic3
H2
acetateEthyl523 OHHCCOOHCHOHHCOOCCH
(ii) Inversion of cane-sugar
Fructose6126
ecosGlu6126
H2
Sucrose112212 OHCOHCOHOHC
Both the above reactions are bimolecular but are found to be of the first order, asexperimentally it is observed that
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For the first reaction , Rate of reaction [CH3COOC
2H
5] only
and for the second reaction, Rate of reaction [C12
H22
O111
] only.The reason for such a behaviour is obvious from the fact that water is present in such alarge excess that is concentration remains almost constant during the reaction.Such reactions which are not truly of the first order but under certain conditions becomereactions of the first order are called pseudo-unimolecular reactions.The kinetics of the above reactions have been studied as follows :
(a) Hydrolysis of Ethyl acetate
CH3COOC
2H
5 + H
2O
H CH3COOH + C
2H
5OH
In this reaction acetic acid is one of the products, the amount of which can be found by titrationagainst standard NaOH solution. But being an acid-catalysed reaction, the acid present originallyas catalyst, also reacts with NaOH solution. Hence a little careful though reveals that for thesame volume of reaction mixture withdrawn at different times.
0)at tproducedisCOOHCHno(as
catalystasonlypresentacidofAmount
)(V timezeroati.e.beginningthe
inusedsolutionNaOHofVolume
30..… (i)
producedCOOHCHofamount
catalystaspresentacidofAmount
)(V t timeofinstantanyat
usedsolutionNaOHofVolume
3t..… (ii)
Combining results (i) and (ii), we find that
)VV( timeofinstantanyat
producedCOOHCHofAmount0t
3
..… (iii)
3 3 2 5But amount of CH COOH produced Amount of CH COOC H
at any instant of time that has reacted (x)
Hence )VV(x 0t ..… (iv)Further
3
Volume of NaOH solution used after the reaction Amount of acid present as
has taken place for a long time, say 24 hours or so, catalyst Max. amount
called infinite time (V ) of CH COOHproduced
..…(v)
Combining results (i) and (v), we find that
Max. amount of CH3COOH produced )VV( 0
But Max. amount of CH3COOH produced
Initial concentration of CH3COOC
2H
5..… (vi)
Hence , a )VV( 0
From equations (iv) and vi), we have
(a – x) )VV( 0 – (Vt – V
0)
or (a – x) )VV( t ..… (vii)Substituting the values of a and (a – x) from equations (vi) and (vii) in the first order equation,we get
xa
alog
t
303.2k
ort
0
VV
VVlog
t
303.2k
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Illustration 17 :In an experiment to study hydrolysis of an ester 0.5 M HCl at 300 K was used. 5 cm 3
of the reaction mixture was withdrawn after definite intervals and titrated against0.2 M NaOH solution. Calculate the rate constant at 300K from the following data.t (sec) 0 600 1200 1800 v (cm3 of NaOH used) 11.5 12.0 12.5 13.0 25.5
Solution :From the given data, V
0 = 11.5 cm3; V = 25.5 cm3
a V – V0 = 25.5 – 11.5 = 14.0 and (a – x) V – V
t
Putting the values of Vt = 12.0 at 600 sec.; 12.5 at 1200 sec. and 13.0 at 1800 seconds in the
equation
01
t
V V2.303k log , we get
t V V
(i) 5 11
2.303 14.0 2.303 14.0k log log 6.061 10 s
600 (25.5 12.0) 600 13.5
(ii) 5 11
2.303 14.0 2.303 14.0k log log 6.176 10 s
1200 (25.5 12.0) 1200 13.0
(iii) 5 11
2.303 14.0 2.303 14.0k log log 6.296 10 s
1800 (25.5 12.0) 1800 12.5
The average value of rate constant k
1= 6.177 10–5 s–1
(b) Optical Rotation MethodInversion of Cane Sugar (Sucrose)
H12 22 11 2 6 12 6 6 12 6(( );66.5 ) glu cose fructose
(( );52.5 ) (( ); 92 )
C H O H O C H O C H O
laevo-rotatorydextro-rotatory
In this reaction by the hydrolysis of dextro-rotatory sucrose produces a mixture of glucose(dextro-rotatory) and fructose (laevo rotatory). As laevo rotation of fructose is more thereforethe resulting mixture is laevo rotatory.The kinetics of above reaction is studied by noting the angle of rotation at different intervals oftime using polarimeter.Say angle of rotation at the start of experiment = r
0
Angle of rotation at any time t = rt
Angle of rotation at time = rNow , Angle of rotation at any instant of time = (r
0 – r
t) amount of sucrose hydrolysed
or x (r0 – r
t)
Similarly, angle of rotation at time
0
90 90
180
rt
r0
r¥
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= (r0 – r) initial conc. of sucrose (a)
or a (r0 – r
t)
Thus, (a – x) (r0 – r) – (r
0 – r
t)
(rt – r)
Hence , 0
t
r rk 2.303log
r r
Illustration 18 :The optical rotation of sucrose in 0.5 N-hydrochloric acid at 308 K and at varioustime intervals are given below. Find out the rate constant for the first order hydrolysisof sucrose.Time (minutes) 0 10 20 30 60 Rotation (degrees) +32 25.5 20.0 15.5 5.0 -10.50
Solution :Here a , the initial concentration (r
0 – r)
x, the change in time t (r0 – r
t)
(a –x) (rt – r)
Substituting these values for different (a – x) values corresponding to time t from the data in theequation for first order reaction
2.303 ak log
t a x
=
0
t
r r2.303log , we get
t r r
(i) -11
2.303 32 ( 10.5) 2.303 42.5k log log 0.0166 min
10 25.5 ( 10.5) 10 36.0
(ii) -11
2.303 32 ( 10.5) 2.303 42.5k log log 0.0166 min
20 20.0 ( 10.5) 20 30.0
(iii) -11
2.303 32 ( 10.5) 2.303 42.5k log log 0.0164 min
30 15.5 ( 10.5) 30 26.0
(iv) -11
2.303 32 ( 10.5) 2.303 42.5k log log 0.0168 min
60 5.0 ( 10.5) 60 15.5
Average value of k
1 = 0.0166 min–1
A Productt = 0 a 0t = t a – x x
n)xa(dt
dx
n)xa(Kdt
dx
1/ 2ta / 2
n0 0
dxKdt
(a x)
Let a – x = z
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– dx = dz , when x = 0 , z = aWhen x = a/2, z = a/2
1/ 2ta / 2
na 0
dzKdt
z
> 2/1
2/a
a
1n
kt1n
Z
1 n1 n
1/ 2
1 aa kt
n 1 2
2/11n
1nkt]12[
)1n(a
1
,
)12(
)1n(K
1t).a( 1n
2/11n
= k
Therefore for a nth order reaction, half life period in inversely related to initial amount
1n
21at
= constant
Berzelius used the term catalyst for the first time for the substances which accelerate the rateof chemical reaction. Now the term catalyst has been used for the foreign substances whichinfluence the rate of a reaction and the phenomenon is known as catalysis. Usually two termsare used for catalysis :Positive Catalysis : The phenomenon in which presence of catalyst accelerates therate of reaction.Negative catalysis : The phenomenon in which presence of catalyst retards the rate ofreaction. Such substances are also known as inhibitors or negative catalyst.
Characteristics Of Catalyst(i) A catalyst remains unchanged chemically at the end of reaction, however its physical state
may change. e.g. MnO2 used as catalyst in granular form for the decomposition of KClO
3, is
left in powder form at the end of reaction.(ii) A catalyst never initiate a chemical reaction . It simply influences the rate of reaction.
Exception : combination of H2 and Cl2 takes place onlywhen moisture (catalyst) is present.(iii) A small quantity of catalyst is sufficient to influence the rate of reaction e.g. 1 g atom of
Platinum is sufficient to catalyse 108 litre of H2O
2 decomposition.
(iv) A catalyst does not influence the equilibrium constant of reaction. It simply helps in attainingequilibrium earlier. It alters the rate of forward & backward reactions equally.
(v) A catalyst normally does not alter the nature of products in a reaction. However some exceptionsare found to this characteristic.
(a) HCOOH 3AlCl H
2O + CO, HCOOH Cu
H2 + CO
2
(b) CO + 3H2Ni CH
4 + H
2O, CO + 2H
2
ZnOCr2O3
CH3OH
CO + H2Cu HCHO
It may therefore be concluded that in some cases nature of products formed depends upon thepresence and nature of catalyst used.
(vi) A catalyst does not make reaction more exothermic(vii) Catalyst’s activity is more or less specific : A catalyst for one reaction is not necessary to
catalyse the another reaction.
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1. Homogeneous catalysis : The reaction in which catalyst and the reactants have samephase. It includes two types.(a) Gaseous phase catalysis : When reactants and catalyst are in gaseous phase
e.g. chamber process for H2SO
4.
2SO2(g) + O
2(g)
catalystNO 2SO
3(g)
(b) Solution phase catalysis : The reactions in which catalyst and reactants arein solution phase or liquid phase of miscible nature.
OHRRCOOHOHRRCOO H)l(2
ester)l(
2. Heterogeneous catalysis : The reactions in which catalyst and reactants formphase 2.
(a) Solid-liquid catalysis :Reactant : Liquid
ImmiscibleCatalyst : Solid
2H2O
2(l)2(s)MnO
Catalyst 2H
2O + O
2
(b) Liquid-Liquid catalysis ::Reactant : Liquid
ImmiscibleCatalyst : Liquid
2H2O
2(l)
Hg( )catalystl 2H
2O + O
2
(c) Solid-gas catalysis :Reactant : gases
ImmiscibleCatalyst : solid
Two aspects of solid catalyst are more significant, one is activity and the other is selectivity.Activity is the ability of catalyst to accelerate chemical reactions; the degree of accelerationcan be as high as 1010 times in certain reactions. Selectivity is the ability of catalyst to directreaction to yield particular reaction (excluding others). For example , n-heptane selectivitygives toluene in presence of Pt catalyst . Similarly,CH
3– CH = CH
2 + O
2 give CH
2 = CHCHO selectivity over Bismuth molybdate catalyst .
In 1889 Arrhenius recognised the temperature dependence of rates or rate constant. He hasgiven an emperical relation which can be written as
k = RT/EaAe ....… (1)k is the rate constant (of any order other than zero order), A is the pre-exponential factor,E
a is the activation energy, R is the universal gas constant and T is the absolute temperature.
Activation energy (Ea) is the minimum energy required by a reactant at a certain temperature
to undergo transformation into product . Arrhenius clearly assumed that reactions occurbecause of collisions between atoms and molecules of the reactant.He assumed the activation energy to be the least value of energy which the collidingmolecules must possess for the collision to yield a product. If we plot graph betweenactivation energy and progress of a reaction (expressed as reaction co-ordinate),we get agraph as shown in figure .
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(E )a f
R
T.S.
E
Reaction Co-ordinate
(E )a b
PH
The difference between the energies of the reactant and the transition state (TS) is called (Ea)
f.
H = (Ea)
f– (E
a)
b
A, the pre-exponential factor or the frequency factor or the Arrhenius constant is the numberof effective collisions occurring per unit time. Effective collision is the number of collisions
occurring per unit time in which orientation of the colliding molecules is proper. RTEae / givesthe fraction of the effective collision that have the sufficient activation energy. Therefore the
product of RT/EaeandA gives the number of collisions per unit time that forms the productand is called its number of productive collisions which is the rate constant, k.For all practical calculations , we shall assume that E
a and A are temperature independent.
Both A and Ea are characteristics of the reaction.
Determination of A and EaFirst Method :
Taking log of both sides of equation (1)
ln k = ln A – aE
RT
Converting (natural log) to common log , log k = log A –RT303.2
E a
If log k is plotted against 1/T, a straight line is obtained which is shown as below:
The slope of this line is given by slope = aE
2.303 R
Thus, knowing the slope, the Ea can be easily calculated. The intercept of the line will give the
value of log A.
Second Method :The logarithmic form of Arrhenius equation is rearranged as
ln k = –RT
E a + ln AA
Differentiating with respect to temperature, we get :
1/T
log k
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dT
klnd = 2
a
RT
E
Integrating with in the limits of temperature T1 and T
2, we get
dTRT
Eklnd
2
1
2
1
k
k
T
T 2a
21
a
1
2
T
1
T
1
R
E
k
kln
or log
21
a
1
2
T
1
T
1
R303.2
E
k
k..… (2)
Where k1and k
2are rate constants at temperatures T
1 & T
2 respectively. Thus, knowing these
values Ea can be calculated. When the value of E
a is known, the value of A can be calculated by
substituting its value in equation k = RT/EaAe . In equation (2), the value of R has to be insertedin the same unit in which E
a is desired.
Temperature Coefficient : “Temperature coefficient of a chemical reaction is definedas the ratio of rate constants of a reaction at two temperatures differing by 10C”.
Temperature coefficient = T+10
T
k2 to 3
k»
where kT is the rate constant at temperature TK and k
T+10 is the rate constant at
temperature (T+10) K. This ratio generally falls between 2 and 3 which means for most of thechemical reactions, the rate becomes two or three folds for every 10C rise in temperature.
Illustration 19 :The rate of a reaction triple when temperature changes from 20ºC to 50ºC.Calculate energy of activation for the reaction (R = 8.314 JK–1 mol–1).
Solution :The Arrhenius equation is
log10
1
2
k
k =
303.2R
Ea
21
12
TT
TT
Given1
2
k
k = 3 ; R = 8.314 JK–1 mol–1; T
1 = 20 + 273 = 293 K
and T2 = 50 + 273 = 323 K
Substituting the given vlues in the Arrhenius equation,.
log10
3 =303.2314.8
Ea
323 293
323 293
Ea =
2.303 8.314 323 293 0.477
30
= 28811.8 J mol–1
= 28.8118 kJ mol–1
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Illustration 20 :In Arrhenius equation for a certain reaction, the value of A and E a (activationenergy) are 4 × 1013 sec–1 and 98.6 kJ mol–1 respectively. At what temperature, thereaction will have specific rate constant 1.1 × 10–3 sec–1 ?
Solution :According to Arrhenius equation
k = RT/EaAe
or logek = log
eA –
RT
Ea logee
2.303 log10
k = 2.303 log10
A–RT
Ea
or 2.303 log (1.1 × 10–3) = 2.303 log (4×1013)–T314.8
106.98 3
T =56.16303.2314.8
106.98 3
K = 311 K
Illustration 21 :The rate constant is giveny by Arrhenius equation
k = /RTEaAe
Calculate the ratio of the catalysed and uncatalysed rate constant at 25ºC if thenergy of activation of a catalysed reaxction is 162 kJ and for the uncatalysed reactionthe value is 350 kJ.
Solution :Let k
ca and k
un be the rate constants for catalysed and uncatalysed reactions.
2.303 log10 kca
= 2.303 log10
A –RT
10162 3......(i)
and 2.303 log 10un
= 2.303 log10
A–RT
10350 3......(ii)
Substracting Eq. (ii) from Eq.(i)
log10
un
ca
k
k =
RT303.2
103
(350 – 162)
=298314.8303.2
10188 3
= 32.95
un
ca
k
k = 8.88 × 1032
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Illustration 22 :At 407 K the rate constant of a chemical reaction is 9.5 × 10–5s–1 and at 420 K, the rateconstant is 1.9 × 10–4s–1. Calculate the frequency factor of the reaction.
Solution :The Arrhenius equation is,
log10 1
2
k
k =
R303.2
Ea
21
12
TT
TT
Given k1 = 9.5 × 10–5 s–1; k
2 = 1.9 × 10–3 s–1;
R = 8.314 J mol–1K–1;T
1 = 407 K and T
2 = 420 K
Substituting the values in Arrhenius equation.
log10 5
4
105.9
109.1
=314.8303.2
Ea
407420
407420
Ea = 75782.3 J mol–1
Applying now log k1 = log A –
1
a
RT303.2
E
log 9.5 ×10–5 = log A –407314.8303.2
3.75782
or log 5105.9
A
=407314.8303.2
3.75782
= 9.7246A = 5.04 × 105 s–1
Illustration 23 :The energy of activation for a reaction is 100 kJ mol–1. Presence of a catalyst lowersthe energy of activation by 75%. What will be effect on rate of reaction at 20ºC, otherthings being equal ?
Solution :The arrhenius equation is
k = RT/EaAe
In absence of catalyst, k1 = RT/100Ae
In presence of catalyst, k2 = Ae–25/RT
So1
2
k
k = RT/75e or 2.303 log
1
2
k
k =
RT
75
or 2.303 log1
2
k
k=
29310314.8
753
or log1
2
k
k =
303.229310314.8
753
or1
2
k
k = 2.34 × 1013
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As the things being equal in presence or absence of a catalyst,
1
2
k
k = catalystofabsenceinrate
catalystofpresenceinrate
i.e.,1
2
r
r =
1
2
k
k = 2.34 ×1013
These are reactions in which a reacting substance do not follow a particular path to give aparticular set of products. It follows one or more paths to give different products. For example,
A
k1
k2C
B
The reactant A follows two different paths to form B and C. (The change in concentration of A,B and C is given in figure below)
AC
B
tChange in conc. of A, B and C
Let initial conc. of A = a mol L–1 and say the amount transformed in time t = x mol L–1.If y and z be amounts of B and C formed at time t, then
dt
dz
dt
dy
dt
dx CBA
Now if the reaction is unimolecular and k1 and k
2 are rate constants for formation of B and C
from A then
rate of formation of y, );xa(kdt
dy1
B
and rate of formation of z, )xa(kdt
dz2
C
Thus, )xa)(kk()xa(k)xa(kdt
dx2121
A
If kkk 21 , Then )xa(kdt
dx A (Here CrBr ff )
or the rate of reaction will be k(a – x)
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Some examples of side reactions
(i) C6H
5N = NCl + CH
3CH
2OH
C H + N + HCl + CH CHO6 6 2 3
C H OC H + N HCl6 6 2 5 2
(ii) CH3CH
2Br + KOH
CH CH OH + KBr3 3
CH = CH + KBr + H O2 2 2
Reactions involving opposing or reversible reactions :Such reactions results in equilibrium. In other words the reactant changes to product andvice versa.Say we have an opposing reaction in which both forward and backward reactions are firstorder, viz.,
(k1 and k
2 are rate constant of forward and backward reaction)
Say initial conc. of A and B are a and b mol L–1 respectively.If after time t, x moles/L of A change into B, then conc. of A and B will be (a – x) and (b + x)respectively.The net rate of the reaction would be given as :Rate = k
1 (a – x) – k
2 (b + x) … (i) [ both processes occur simultaneously]
When equilibrium is reached, the net rate is zeroThus, k
1 (a – x
e) = k
2 (b + x
e) (e = equilibrium)
Hence, (b + xe)
2
1
k
k (a – x
e) or b =
2
1
k
k(a – x
e) –x
e
Substituting value of b in eqn. (i)
Rate of reaction,
xx)xa(
k
kk)xa(k
dt
dxee
2
121
on solving, we getRate = (k
1 + k
2) (x
e – x)
After rearranging and integrating the equation, we get an equation similar to first order reactionas shown below.
A Bk1
k2
or dt)kk(xx
dx21
e
x
x
t
0 21e0
dt)kk(xx
dx
or log t)kk(xx
xx21
e
0e
orxx
xxlog
t
1)kk(
e
0e21
The equation is similar to first order reaction except that the measured rate constant is now thesum of the forward and the reverse rate constants.
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A Typical Example : In many cases, the product formed in one of the elementaryreactions acts as the reactant for some other elementary reaction. One of the examples ofconsecutive reactions involves the following steps.
1 1k kA B C Let the initial concentration of A be [A]
0 and let after time t, the concentrations of A, B and C
be [A], [B] and [C], respectively. It is obvious that[A]
0 = [A] + [B] + [C] ....…(1)
Differential Rate Law : The differential rate expressions are
1
d[A]k [A]
dt ....…(2)
1 1
d[B]k [A] k [B]
dt ....…(3)
1
d[C]k [B]
dt ....…(4)
On integrating equation (2), we get1k t
0[A] [A] e ....…(5)Substituting [A] from equation (5) into equation (3), we get
1k t1 0 1
d[B]k [A] e k [B]
dt
1k t1 1 0
d[B]k [B] k [A] e
dt
Integrated Rate Law : Multiplying the above expression throughout by exp (k1t), we
get
1 1 1k t (k k )t1 1 0
d[B]k [B]e k [A] e
dt
The left side of the above expression is equal to 1k td([B]e ) / dt. Hence, the above expression
can written as , 1 1 1k t (k k )t1 0d{[B]e } k [A] e dt
Integrating the above expression with [B] = 0 at t = 0, we get
1 1(k k )tk t
1 01 1 1 1
e 1[B]e k [A]
(k k ) k k
1 1k t k t
1 01 1 1 1
e e[B] k [A]
(k k ) k k
1 1k t k t11 0
1 1
k[B] k [A] {e e }
k k
....…(6)
Substituting equations (5) and (6) in equation (1), we get
1 1 1k t k t k t10 0 0
1 1
k[A] [A] e [A] (e e ) [C]
k k
or 1 1k t k t0 1 1
1 1
1[C] [A] 1 (k e k e )
k k
....…(7)
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Figure (1) Illustrates the general appearance of the variations of concentrations of A, B and Cduring the progress of the reaction.
[C]
[B]
Time
Fig. 1: Typical variations of concentration of A, Band C during the progress of the reactionA ? B ? C. The actual variations on thevalues of k1 and k.
Con
cent
ratio
n
[A]
In general concentration of A decreases exponentially, the concentration of B initially increasesupto a maximum and then decreases thereafter, and the concentration of C increases steadilyuntil it reaches its final value [A]
0, when all A has changed into C.
Maximum Concentration of B : Equation (6) is
1 1k t k t10
1 1
k[B] [A] {e e }
k k
....…(6)
At the maximum concentration of B, we have
d[B]0
dt
Hence, differentiating equation (6) with respect to t, we get
1 1k t k t10 1 1
1 1
d[B] k[A] { k e k e }
dt k k
....…(8)
Equating equation (8) to zero, we get1 max 1 maxk t k t
1 1k e k e 0
or 1 1 max(k k )t1
1
ke
k
or 11 1 max
1
kln (k k )t
k
or 1max
1 1 1
k1t ln
k k k
....…(9)
substituting equation (9) in equation (6), we get
1 1 1k /(k k )
1max 0
1
k[B] [A]
k
....…(10)
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Pressure Change Method :This method is used for gaseous reactions.As reaction proceeds there is change in pressure.For a reaction , A
(g) B
(g) + C
(g)
Initial pressure at t = 0 P0
0 0Pressure at time (t) (P
0 – x) x x
(Here x is no. of moles of A which change to produce)Thus, total pressure (P
t) at time (t) = P
0 – x + x + x
or Pt = P
0 + x, x = P
t – P
0
a – x = P0 – (P
t – P
0) = 2P
0 – P
t
Thus , 0
0 t
P2.303k log
t (2P P )
Illustration 24 :The following rate data was obtained for the first order thermal decompositionof SO2Cl2(g) at a constant volume .SO2Cl2(g) SO2(g) + Cl2(g)
Exp. Time (sec–1) Total pressure (atm)1. 0 0.52 100 0.6Calculate the reaction rate when total pressure is 0.65 atmosphere.
Solution :Let us say that the pressure of SO
2Cl
2 decreases by x atm, then the increase of pressure of
SO2 and Cl
2 = x atm each. [Q 1 mole of SO
2Cl
2 decomposes to give 1 mole of SO
2 and 1
mole of Cl2].
SO2Cl
2(g) SO
2(g) + Cl
2(g)
Pressure at t = 0 0.5 atm 0 0Pressure at t, (0.5 – x) atm x atm x atm
Since total pressure PT
=2 2 2 2SO Cl SO ClP P P
= (0.5 – x) + x + xP
T= 0.5 + x or x = P
T – 0.5
Hence2 2SO ClP = 0.5 – (P
T – 0.5)
= 0.5 – PT + 0.5 = 1.0 – P
T
Since , at t = 100 sec, PT = 0.6 atm
2 2SO ClP = 1.0 – 0.6 =0.4 atm
(a) Evaluation of k
k =2.303 Initial pressure
logt Pressure at time t
=2.303 0.5 2.303
log log1.25100 0.4 100
=2.303
0.0969100
= 2.23 10–3 sec–1
(b) Rate at PT = 0.65 atm
2 2SO ClP at total pressure of 0.65 atm = 1.0 – 0.65 = 0.35 atmRate = k[N
2O
5] = 2.23 10–3 0.35 = 7.8 10–4 atm sec–1
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Illustration 25 :For a homogeneous gaseous phase reaction: 2A 3B + C, the initial pressure ofreactant was P° while pressure at time ‘t’ was P. Find the pressure after time 2t.Assume I order reaction.
Solution : 2A 3B + CInitial Pressure P° 0 0Pressure at t = t P° – 2a 3a aPressure at t = 2t P° – 2x 3x xGiven P° – 2a + 3a + a = P ....… (1) P° + 2a = PNow for I order reaction at time t
K =o o
o o
2.303 P 2.303 Plog log
t P 2a t 2P P
....… (2)
Let total pressure at t = 2t be A, then P° – 2x + 3x + x = A or P° + 2x = A
K =o o
o o
2.303 P 2.303 Plog log
2t P 2x 2t 2P A
....… (3)
K =
1/ 2o
o
2.303 Plog
t 2P A
....… (4)
By equations (2) and (4)2o o
o o
P P
2P P 2P A
[2P° – A] . P° = [2P° – P]2
P°. A = 2 (P°)2 – (2P° – P)2
A = 2P° –o 2
o
(2P P)
P
.
Illustration 26 :The decomposition of Cl2O7 at 400 K in the gas phase to Cl2 and O2 is of I order.After 55 sec at 400 K, the pressure of Cl2O7 falls from 0.062 to 0.044 atm. Calculate :(a) The rate constant.(b) Pressure of Cl2O7 after 100 sec.
Solution : Cl2O
7 Cl
2 + 7
22 O
Mole at t = 0 a 0 0Mole at t = 55 sec. (a – x) x 7x/2
(a) Since Pressure of Cl2O
7 is given and therefore,
a 0.062(a – x) 0.044
Q K = 10
2.303 0.062log
t 0.044K = 6.23 10–3 sec–1.
(b) Let at t = 100 sec, (a – x) P
6.23 10–3 = 10
2.303 0.062log
100 P P = 0.033 atm.
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RADIOACTIVITYRadioactivity may be defined as a process in which nuclei of certain elements spontaneouslydisintegrate (transformation into another element by the ejection of -or - particle)at a ratecharacteristic for each particular active isotope (Becquerel, 1896). All the heavy elements frombismuth (atomic number 83) through uranium and also a few of the lighter elements possessradioactive properties. However, the radioactive property of the different radioactive elementsdiffers widely, e.g. radium atoms have about three million times the activity of uranium atoms.Uranium in the form of potassium uranyl sulphate, KUO
2(SO
4)
2 was the first compound found
to be radioactive. Radioactive changes are spontaneous. These are not controlled bytemperature, pressure or nature of chemical combination.
Radioactive disintegration
The atomic nuclei of radioactive elements can disintegrate any moment. During disintegration, atomsof new elements having different physical and chemical properties are formed, called daughterelements.Disintegration occurs by the following processes :(i)-particle emission : When an-particle (4
2He) is emitted from the nucleus of parent element,
the new element formed called daughter element, possess atomic mass less by 4 unit & atomicnumber less by 2 units.
Parent element Daughter elementAtomic mass A A – 4Atomic number Z Z – 2
For e.g., U23892 Th234
90 + He42
(ii) -particle emission - When -particle is emitted from parent element thus formed daughterelements possesses same atomic mass but atomic no. is increased by 1 unit.
Parent element Daughter element
Atomic mass A AAtomic number Z Z + 1
* Elements having same mass number called isobars. (A - same, Z- different)daughter element formed by the-particle emission is an isobar of parent element.
e.g., Th23490 Pa234
91 + e01
If in a radioactive transformation & both are emitted then atomic mass & atomic numberchanges accordingly & produces an isotope of the parent element.
DCBA 4AZ
4A1Z
4A2Z
AZ
(A & D are isotopes)
* Elements having same no. of protons called isotopes. e.g., C126 , C14
6 and O168 , O17
8 etc.
(A - different, Z- same)
* Elements having same no. of neutrons are called isotones. e.g., C146 , O16
8 & F199 , O18
8 etc.
(A - different, Z- different)
* Elements having same value of (A – 2Z ) are called isodiaphers. e.g., F199 , K39
19
(A - different, Z- different, A – 2Z -same)* Elements having same number of electrons are called isoelectronic. e.g., SO
4–2, PO
4–3
* Compounds having same number of atoms as well as same number of electrons are called isosters.
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1. Law of Radioactive Disintegration :(i) Atoms of all radioactive elements undergo spontaneous disintegration and form new radioactive
elements. The disintegration is accompanied by the emission of , , or -rays.(ii) The disintegration is at random, i.e. every atom has equal chance for disintegration at
any time.(iii) The number of atoms that disintegrate per second is directly proportional to the number of
remaining unchanged radioactive atoms present at any time. The disintegration is independentof all physical and chemical conditions like temperature, pressure chemical combination etc.The two laws of radioactive disintegration can be summed up as below:
1. Group displacement law : The result of - and - particle changes can be summed upin the form of group displacement law. “In an -particle change the resulting element hasan atomic weight less by four units and atomic number less by two units and it falls in agroup of the periodic table two columns to the left of the original element, and in a -particle change the resulting element has same atomic weight but its atomic number isincreased by one than its parent and hence it lies one column to right”.e.g.,(i) Po214
84 Pb21082 + He4
2 (ii) C146 N14
7 + e01
(VI A) (IV A) (IV A) (V A)
2. Law of radioactive decay : According to the law of radioactive decay, the quantity ofa radioelement which disappears in unit time (rate of disintegration) is directlyproportional to the amount present.
Determination of the number of -and -particles emitted in anuclear reaction:
Consider the following general reaction.
01
42
mn
mn baYX
Then , m = m + 4a + 0b (ii) n = n + 2a - bSolve for a and b
Where a is the number of He42
emitted and b is the number of 0
1 emitted
Points to Remember1. Rate of decay (activity, A) is the number of atoms undergoing decay to unit time; it is
represented bydt
dN t .
2. Rate of decay of a nuclide is directly proportional to the number of atoms of that nuclidepresent at that moment , hence.
Ndt
dN t or bt N
dt
dN
(the negative sign shows that the number of radioactive atoms, Nt decreases as time t increases)
3. Rate of decay of nuclide is independent of temperature, so its energy of activation is zero.4. Since the rate of decay is directly proportional to the amount of the radioactive nuclide
present and as the number of undecomposed atoms decreases with increase in time, the rate ofdecay also decreases with the increase in time.Various forms of equation for radioactive decay are
tt 0N N e
0 tlog N log N 2.303 t
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303.2
t
N
Nlog
t
0
t
0
N
Nlog
t
303.2 ....…( I)
Note that the equation (I) is similar to that of first order reaction, hence we can say thatradioactive disintegration are examples of first order reactions.Where N
0 = Initial number of atoms of the given nuclide, i.e. at time 0
Nt = No. of atoms of that nuclide present after t
= Decay constantHowever , unlike first order rate constant (K), the decay constant () is independent oftemperature.Decay constant : The ratio between the number of atoms disintegrating in unit time tothe total number of atoms present at that time is called the decay constant of that nuclide.Characteristics of decay constant ()1. It is characteristic of a nuclide (not of an element)2. Its units are time–1
3. Its value is always less than one
Half-life Period (T1/2 or t1/2) : Rutherford in 1904 introduced a constant known ashalf-life period of the radio-element for evaluating its radioactivity or for comparing itsradioactivity with the activities of other radio-elements. The half-life period of a radio-element is defined as the time required by a given amount of the element of decay to one-halfof its initial value.
Mathematically, 1/ 2
0.693T
Average Life Period (T) : Since total decay period of any element is infinity, it ismeaningless to use the term total decay period (total life period) for radio elements. Thus theterm average life is used which is determined by the following relation.
nucleiofnumberTotal
nuclie theoflivesofSum(T)lifeAverage
Relation between average life and half-life . Average life (T) of an element is the inverse of its
decay constant i.e.,
1T
Substituting the value of in the above equation , 1/ 21/ 2
TT 1.44T
0.693
Thus, Average life (T) = 1.44 Half-life (T1/2
) = 2/1T2 Thus the average life period of a radioisotope is approximately under-root two times of its halflife period.Specific activity : It is the measure of radioactivity of a radioactive substance. It isdefined as ‘the number of radioactive nuclei which decay per second per gram ofradioactive isotope’. Mathematically, if ‘m’ is the mass of radioactive isotope, then
Specific activity = ginmassAtomic
numberAvogadro
m
N
m
decayofRate
Where N is the number of radioactive nuclei, which undergoes disintegration.
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2. Radioactivity :(a) The phenomenon of spontaneous disintegration of nuclei of unstable atoms is defined as
radioactivity.(b) Generally it is exhibited by atoms with A>192 and Z>82(c) It was discovered by Henry Becqurel(d) Lead isotope is the stable end product of any natural radioactive series(e) Radio activity is a nuclear process and not an atomic process(f) Radioactivity is not associated with the electron configuration of the atom.
3. Rutherford-Soddy Theory :(a) If N is the number of radioactive nuclei present in a sample at a given instant of time, then
the rate of decay at that instant is proportional to N i.e.,
dNN
dt
(b) If N0 is the number of radioactive nuclei at time t = 0, then the number of radioactive nuclei
at a later time t is given by , 0tN N e
(c) The nuclei of unstable atoms decay spontaneously emitting , particles and rays(d) Radioactivity remains unaffected due to the physical and chemical changes of the material.(e) Radioactivity obeys the law of probability i.e it is uncertain that when a particular atom will
decay.
4. Soddy-Fajaan's Laws :(a) During an -decay, mass number decreases by 4 units and atomic number by 2 units.
42
rayA AZ ZX Y E
VDaughter nucleus will occupy two positions before that of parent nucleus, in periodic table.
(b) During -decay mass number of the atom will not change and atomic number increases by1 unit
1rayA A
Z ZX Y
Daughter nucleus will occupy one position on the right of that of parent nucleus in periodictable
(c) During -decay, the mass number and atomic number of the nucleus remain unchangedrayA A
Z ZX Y E V4.(a) Emission of -particle means loss of two protons and two neutrons
(b) Emission of -particle means loss of an electron.
(c) Emission of a -ray means no change in charge and mass, but only energy changes
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(a) The number of atoms of any material decaying per second is defined as the activity of thatmaterial.
(b) Its value depends on the quantity and nature of that material.(c) Units of activity –
fundamental unit – disintegrations per second i.e., Bq1Bq =1disintegration/s
(d) Practical units: curie and rutherford.1 curie = 3.7 1010 disintegration/second:1 Rutherford = 106 disintegrations/second.
(e) Formulae of activity
(i) dNA ii A N
dt 0 0 0
tiii A N iv A A e 0.693 AN mv A
WT
where 0A = maximum initial activity; A= activity after time t, =decay constant,
NA = Avogadro number , m = mass of material ,
W = atomic weight of material , T = half life of material
(a) Decay constant is equal to the reciprocal of that time in which the activity of the material
reduces to e1 or 37 % of its initial activity..
(b) . sec ( sin )
.
dNNo of atoms decaying per ond Rate of di tegrationdt
N No of atoms remaining after decay
i.e. The rate of disintegration per atom is defined as decay constant(c) Decay constant does not depend on temperature, pressure and volume. It depends on the
nature of material.(d) Its unit per second.
(e) Decay constant
N
tdNd
represents the probability of decay per second.
(f)2.303
t 0 0 0
10 10 10
2.303 2.303log log log
N A M
N t A t M
(g) Its value is equal to the negative of the slope of N-t curve.(h) The decay constant of stable element is zero.
(a) The time, in which the number of atoms (N) reduces to half of its initial value (N0) , is defined
as the half-life of the element (i.e. half of the atoms decay). 01
2
,2
Nt T N
(b) The time in which the activity reduces to half of its initial value is defined as half life.
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At 01
2
,2
At T A
(c) Its unit is second(d) Formulae of half life
(i) 1
2
log 20.693 eT
(ii) 10 0 02
10 10 10
log 2 log 2 log 2
log log log
e e eTN A M
N A M
(iii) 1
2
tT
n where n=No. of half lives
(iv) Time of disintegration
0 010 10
10
log log
log 2 0.3010
N NT T
N Nt
(a) The time, for which a radioactive material remains active, is defined as mean life of thatmaterial :
(b)0
t dNSum of lives of all atoms
total number of atoms present N
(c) The average time taken in decaying by the atoms of an element is defined as its mean life .(d) = 1/(e) Its units are second, minute, hour day, month, year etc.(f) Mean life does not depend on the mass of material. It depends on the nature of the material.(g) The magnitude of slope of decay curve is equal to the mean life.(h) Relation between the mean life and half-life.
(i)1
2
0.693
T
(ii) 1
2
1.44T (iii) 1
2
T
(iv) The time, in which the number of radioactive atoms decays to 1/e or 37% of itsinitial value, is defined as the mean life of that material.
(a) 0tN N e (b) 0
tA A e (c) 0tM M e
(d)
0102.3027 log
N
Nt
(e)
0102.3027 log
A
At
(f)
0102.3027 log
M
Mt
(g)
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(h)
(When two particles decay simultaneously)
(i)
1/ 2
0 0
22
n T
T
N NN (j)
1/ 2
0
2
T
T
AA (k)
1/ 2
0
2
T
T
MM
(i) Percentage decreases in activity =0
1 100A
A
(ii) Number of atoms remaining after n half lives 0
2n
NN
(iii) Number of atoms decayed after time 0 t N N 0
11
2nN
(iv) The fraction of radioactive material at time T .=1/ 20
11 1
2
T
T
N
N
(v) Percentage of radioactive material decayed at time T =1/ 20
11 100 1 100
2
T
T
N
N
(vi) Percentage of radioactive material decayed in n halflives =1/ 2
0
1100 100
2
T
T
N
N
(vii) Fraction of radioactive material decayed in n half lives =0
11 1
2n
N
N
(viii) Percentage of radioactive material decayed in n half lives0
11 100 1 100
2n
N
N
(ix) Percentage of radioactive material remaining after n half-lives.0
1100 100
2n
N
N
(x) When decay process is too slow then 0 0 0 0N N N t or N N t N
(xi) N-t graph is a straight line with –ve slope, for slow decay process.
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11. Characteristics of , , and raysS.No Property -Particles -Particles -rays
1. Nature andvalue of charge
Positive and double of the chargeof the proton
Negative andequal to the
charge of electron191.6 10 C
Uncharged(Neutral)
2. Nature ofparticle
Doubly ionized helium atom (2protons and 2 neutrons)
Electron (or)positron
Electromagneticwaves
3. Mass Four times the mass of the proton 274 1.67 10 kg
Equal to the massof electron
319.1 10 kg
Mass less
4. Specific chargeq
m
19
27
3.2 10
4 1.67 10
= 4.79 × 10711 11.7 10 Ckg Uncharged and
mass less
5. Explained by Tunnel effect Neutrinohypothesis
Transitions ofnuclei into theground energy
level after and decay
6. Effect of electricand magnetic
fields
Deflected by electric and magneticfields
Deflected byelectric and
magnetic fields
Unaffected
7. Penetratingpower
1 100 10000
8. Ionizing power 100000 100 1
9. Velocity Less than the velocity oflight 7 7 11.4 10 / 2.2 10m s to ms
Approximatelyequal to the
velocity of light
83 10 /m s
10. Mutualinteraction with
matter
Produce heat Produce heat Produce thephenomenon of
Photoelectriceffect, Compton
12. -emission
(a) Characterstictics of –decay :(i) The spectrum of -particles is a discrete line spectrum.(ii) Spectrum of -particles has fine structure i.e. every spectral line consists of a number
of fine lines.(iii) The -emitting nuclei have discrete energy levels i.e energy levels in nuclei are analogous
to discrete energy levels in atoms..(iv) -decay is explained on the basis of tunnel effect.
(v) Geiger-Muller law - log loge eA B R For radioactive series B is same whereas A is
different
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(b) Range of -particles :(i) The maximum distance traversed by -particles in air before being finally stopped is
defined as the range of -particles.(ii) The maximum distance traversed by -particles before being finally absorbed after
ionizing gas molecules, is defined as the range of -particles.(iii) The range of -particles in air is from 2.6cm to 8.6cm.(iv) Relations between the range of -particles and their energy
(I) 3/ 20.318R E (II)3
log log 0.318 log2
R E (c) Size of the nucleus decreases by emission
(i) The energy spectrum of -particles is continuous i.e. -particles of all energies uptoa certain maximum are emitted
(ii) The number of such -particles is maximum whose energy is equal to the maximum
probable energy i.e. at ,mpE EBN =maximum.
(iii) There is a characteristic maximum value of energy in the spectrum of -particles which
is known as the end point energy 0E
(iv) In -decay process, a neutron is converted into proton or proton is converted intoneutron.
1 1 00 1 1n p e Particle 1 1 0
1 0 1p n e Particle
(v) The energy of -particles emitted by the same radioactive material may be same ordifferent.
(vi) The number of -particles with energy 0E E (end point energy) is zero.
(a) According to Pauli, whenever neutron is converted into proton or proton into neutron then thisprocess is accompanied with the emission of a new particle to which he named as neutrino.
1 1 01 0 1p n e v ; 1 1 0
0 1 1n p e v (b) Properties of neutrino :
(i) The charge on neutrino is zero(ii) The rest mass of neutrino is zero
(iii) Its spin angular momentum is2
h
(iv) Its speed is equal to that of light.(v) It has finite magnetic moment but the magnitude is very small(vi) Its antiparticle is anti-neutrino.
(vii) The linear momentum vector p and spin vector S are mutually in opposite directions.
(viii) Its energy is equal to endE E .
(ix) It does not interact with matter.(x) Neutrino was discovered by Pauli and its experimental verification is done by Reines
and Cowan.
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15. (a) Characterstics of -decay
(i) The spectrum of -rays is a discrete line spectrum.
(ii) Whenever or - particles is emitted by a nucleus then the daughter nucleus is leftin the excited state. It suddenly makes transition in the ground state thereby emitting -rays.
(iii) Knowledge about nuclear energy levels is obtained by -spectrum.(iv) -rays interact with matter as a consequence of which the phenomena of photoelectric
effect, Compton effect and pair production happen. (At low energy photoelectric effectand at high energy pair-production are effective).
15. (b) Intensity of rays in materials-
(i) When -rays penetrate matter, then their intensity (a) decreases exponentially withdepth (x) inside the matter. The intensity of -rays at depth x inside the matter is given
by 0xI I e .
(ii) The thickness of matter, at which the intensity of -rays (I) reduces to half its initial
maximum value 0I , is known as its half-value thickness.
693.
X 2/1
(iii) The reciprocal of the distance inside matter, at which the intensity (I) reduces to1
e or
37 % of its maximum value 0I , is defined as the coefficient of absorption of thatmaterial.
(iv) Coefficient of absorption
(I)/dI I
dx
(II) depends on the wavelength of -rays 3 and the nature of absorbing
material
If parent element is unstable then it will dissociate into daughter element & if this daughter element isstill unstable, then it will again dissociate into a new daughter element & process continuous till theformation of a stable element. Series of element obtained from parent element to the finally stablenon-radioactive element is known as radioactive disintegration series.(4n + 1) is artificial series & 4n, (4n + 2), (4n + 3) are natural series.
S.No Series Name of theseries
Initialelement
Finalelement
Nature ofseries
No of&
Particlesemittted
1. 4n+2 Uranium series 23892U
20682 Pb Natural 8 ,6
2. 4n+3 Actinium series 23592U
20782 Pb Natural 7 ,4
3. 4n Thorium series 23290Th 208
82 Pb Natural 6 , 4
4. 4n+1 Neptunium series 23793 Np 209
83 Bi Artificial 7 , 4
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17. To Calculate no of -particles and -Particles emitted 1
1A AZ Z
X Y x y
x: no of -particles emitted y: no of -particles emitted1
14 02 1
A AZ Z
X Y xHe ye
1 4A A x 1
4
A Ax
1 2Z Z x y 1 2y Z Z x 1
1
2
A Ay Z Z
eg : 238 206 4 092 82 2 1U Pb xHe ye
1 238 2068
4 4
A Ax
particles
1
1 238 20692 82 16 10 6
2 2
A Ay Z Z
particles
Units of radioactivity :The unit of radioactivity is curie (Ci). It is the quantity of any radioactivity substance which has decay
rate of 3.7 × 1010 disintegrations per second.
1 millicurie (mCi) = 3.7 × 107disintegrations per sec.
1 microcurie (Ci) = 3.7 × 104 disintegrations per sec.
There is another unit called rutherford (Rd) which is defined as the amount of a radioactive substance
which undergoes 106 disintegrations per second.
1 milli rutherford = 103 disintegration per sec.
1 micro rutherford = 1 disintegration per sec.
The SI unit radioactivity is proposed as Becquerel which refers to one dps.
1 curie = 3.7 × 104 Rutherford
1 curie = 3.7 GBq
Here, G stands for 109, i.e., giga.
S.No Isotopes Isobars Isotones1. The atoms of the same
elements whose chargenumber (Z) is same but massnumber is different are known
as isotopes.
The atoms with massnumber same and charge
number different are knownas isobars.
The atoms with sameneutron number but A andZ are different are known
as isotones
2. Chemical properties are same Chemical properties aredifferent
Chemical properties aredifferent
3. Number of electrons is same Number of electrons isdifferent
Number of electrons isdifferent
4. Occupy same place in periodictable
Occupy different places inperiodic table
Occupy different places inperiodic table.
5. Example 16 17 188 8 8, ,O O O
1 2 31 1 1, ,H H H
20 21 2210 10 10, ,Ne Ne Ne
3 31 2H and He
14 146 7C and N
17 178 9O and F
7 83 4Li and Be
2 31 2H and He
3 41 2H and He
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The isotopes of elements which spontaneously decay by emitting radioactive radiations aredefined as radioactive isotopes.They are two types .(a) Natural radioactive isotopes (b) Artificial radioactive isotopes
(a) Natural radioactive isotopes : Those radioactive isotopes which exist naturally are
known as natural radioactive isotopes. e.g. 232 240,Th Pu etc.
(b) Artificial radioactive isotopes : Those isotopes, which are prepared artificially by
bombarding fundamental particles like , , , p,n etc, no matter, are known asartificial isotopes.
(a) In Medicine :(i) For testing blood chromium-51(ii) For testing blood circulation-Sodium-24(iii) For detecting brain tumor-Radio mercury-203(iv) For detecting fault in thyroid gland-Ratio iodine-131(v) For cancer-Cobalt-60(vi) For blood-Gold-189(vii) For skin diseases-Phosphorous-31
(b) In Archaeology :(i) For determining age of archaeological sample (Carbon dating) - 14C(ii) For determining age of meteorites - 40K(iii) For determining age of earth-Land isotopes
(c) In Agriculture :(i) For protecting potato crop from earthworm – Cobalt-60(ii) For artificial rains AgI(iii) As fertilizers-Phosphorous-32
(d) As tracers :Very small quantity of radio isotopes present in a mixture is known as tracer . Tracertechnique is used for studying biochemical reactions in trees and animals.
(e) In industries :(i) For detecting leakage in oil or water pipe lines.(ii) For testing machine parts
(f) In research :(i) In the study of carbon-nitrogen cycle.(ii) For determining the age of planets
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1. In the reaction; A + 2B 3C + D which of the following expressions does not describechanges in the concentration of various species as a function of time:
(A)dt
]A[d3
dt
]C[d (B)
dt
]C[d
dt
]D[d3 (C)
dt
]C[d2
dt
]B[d3 (D)
dt
]A[d
dt
]B[d2
2. Which of the following statement is correct for a reaction X + 2Y prodcuts :(A) The rate of disappearance of X = twice the rate of disappearance of Y(B) The rate of disappearance of X = 1/2 rate of appearance of products(C) The rate of appearance of products = 1/2 the rae of disappearance of Y(D) The rat of appearance of products = 1/2 the rate of disappearance of X
3. For the reaction, 4A + B 2C + 2D, The statement not correct is :(A) The rate of disppearance of B is one fourth the rate of disappearance of A(B) The rate of appearance of C is half the rate of disappearance of B(C) The rate of formation of D is half the rate of consumption of A(D) The rates of formation of C and D are equal
4. For the reaction 2A + B C + D, –dt
]A[d = k[A]2 [B]. The expression for
dt
]B[d will
be:(A) K[A]2 [B] (B) 1/2K[A]2[B] (C) K[A]2[2B] (D) K[2A]2[B]
5. Which is correct relation in betweendt
dC,
dt
dn and
dt
dP where C, n, P, represents concentration,
mole and pressure terms for gaseous phase reactant A(g) product.
(A) –dt
dC = –
V
1
dt
dn = –
RT
1
dt
dP(B)
dt
dC =
dt
dn = –
dt
dP
(C)dt
dC =
V
RT
dt
dn = –
dt
dP(D) All
6. When ammonia is treated with O2 at elevated temperatures, the rate of disappearance of
ammonia is found to be 3.5 × 10–2 mol dm–3 s–1 during a measured time interval. Calculate therate of appearance of nitric oxide and water.
7. The following reaction was carried out at 44ºCN
2O
5 2NO
2 + ½ O
2
The concentration of NO2 is 6.0 × 10–3 M after 10 minutes of the start of the reaction.
Calculate the rate of production of NO2 over the first ten minutes of the reaction.
8. The oxidation of iodide ion by arsenic acid, H3AsO
4, is described by the balance equation :
3I– (aq) + H3AsO
4 (aq) + 2H+ (aq)I
3– (aq) + H
3AsO
3 (aq) + H
2O (l)
(a) If – [I–] / t = 4.8 × 10–4 M/s, what is the value of [I3
–] /t during the same timeinterval ?
(b) What is the average rate of consumption of H+ during that time interval ?
9. In a reaction, the rate expression is, rate = K[A] [B]2/3 [C]0, the order of reaciton is :(A) 1 (B) 2 (C) 5/3 (D) Zero
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10. For the reaction, H2(g + Br
2(g) = 2HBr(g), the reaction rate = K [H
2] [Br
2]1/2. Which statement
is true about this reaction :(A) The reaction is of second order (B) Molecularity of the reaction is 3/2(C) The unit of K is sec–1 (D) Molecularity of the reacion is 2
11. The dimensions of the rate constant of a sceond order reaction involves :(A) Neither time nor concentration (B) Time and concentration(C) Time and square of concentration (D) Only time
12. The rate constant for a reaction is 10.8 × 10–5 mol litre–1 sec–1. The rection obeys :(A) First order (B) Zero order (C) Second order (D) Half order
13. Which statement about the order of reaction is correct ?(A) The order of reaction must be a positive integer(B) A second order reaction is also bimolecular(C) The order of reaction increases with increasing temperature.(D) The order of reaction can only be determined by experiment
14. The rate of the reaction A + B + C P is given by :
r =dt
]A[d = k [A]1/2 [B]1/2 [C]1/4. The order of the reaction is:
The reaction is :(A) 1 (B) 2 (C) 1/2 (D) 5/4
15. Bromomethane is converted to methanol in an alkaline solution :CH
3Br(aq) + OH– (aq) CH
3OH(aq) + Br– (aq)
The reaction is first order in each reactant.(a) Write the rate law.(b) How does reaction rate change if the OH– concentration is decreased by factor of 5 ?(c) What is the change in rate if the concentrations of both reactants are doubled ?
16. The oxidation of Br – by BrO3– in acidic solution is described by the equation
5Br– (aq) + BrO3
– (aq) + 6H+ (aq) 3Br2(aq) + 3H
2O(l)
The reaction is first order in Br –, first order in BrO3
–, and second order in H+.(a) Write the rate law(b) What is the overall reaction order ?(c) How does the reaction rate change if the H+ concentration is tripled ?(d) What is the change in rate if the concentrations of both Br– and BrO
3– are halved ?
17. For the reaction : 2A + B2 + CA
2B + BC, the rate law expression has been determined
experimentally to be R = k [A]2 [C] with , k = 3.0 × 10–4 M–2 min–1.(i) Determine the initial rate of the reaction, started with concentration
[A] = 0.1 M, [B2] =0.35 M & [C] = 0.25 M
(ii) What is the effect on rate of reaction and rate constant on changing the volume to1/4th of initial value.
18. For the reaction 2NO + Cl2 2NOCl, it is found that doubling the concentration of both
reactants increases the rate by the factor of 8 , but doubling the Cl2 concentration alone,
only doubles the rate What is the order of the reaction with respect to NO and Cl2 ?
19. For the reaction , 2A + B + C A2B + C
the rate = k[A] [B]2 with k = 2.0 × 10–6 M–2 s–1. Calculate the initial rate of the reaction when[A] = 0.1 M, [B] = 0.2 M and [C] = 0.8 M.
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20. If the pressure of gaseous reactant is changed by 3 times the rate of reaction changes by5.2 times. Determine(i) order of reaction (ii) unit of rate constant
(iii) effect on rate of reaction if volume of container is reduced to 161 th of the original
1. The rate law for the single-step reaction , 2A + B 2C , is given by :(A) Rate = K[A] . [B] (B) Rate = K[A]2 . [B](C) Rate = K[2A] . [B] (D) Rate = K[A]2 . [B]0
2. The correct expression the rate of reaction of elementary reaction , A + B C is
(A)dt
]C[d = K [A] (B)
dt
]C[d = K[B] (C)
dt
]A[d = K[A][B] (D)
dt
]A[d = K[A]
3. Select the intermediate in the following reaction mechanism :O
3(g) O
2(g) + O(g), O
3 + O 2O
2
(A) O3(g) (B) O(g) (C) O
2(g) (D) None
4. The elementary step of the reaction , 2Na + Cl2 2NaCl is found to follow III order
kinetics, its molecularity is :(A) 1 (B) 2 (C) 3 (D) 4
5. For the reaction 2NO2 + F
2 2NO
2F, following mechanism has been provided :
NO2 + F
2 slow NO2F + F
NO2 + F fast
NO2F
Thus rate expression of the above reaction can be written as :(A) r = k [NO
2]2 [F
2] (B) r = k [NO
2][F
2]
(C) r = k [NO2] (D) r = k [F
2]
6. The reaction 2NO + Br2 2NOBr, Obey’s the following mechanism:
NO + Br2
fast NOBr
2; NOBr
2 + NO slow
2NOBrThe rate expression of the above reaction can be written as:(A) r = k [NO]2 [Br
2] (B) r = k [NO] [Br
2]
(C) r = k ][NO] [Br2]2 (D) r = k [NOBr
2]
7. The reaction 2A B + C follows zero order kinetics. The differential rate equation for thereaction is:
(A) tdxd
= K [A]0 (B) tdxd
= K [A]2 (C) tdxd
= K [B] [C] (D) tdxd
= K [A]
8. For the elementary step, (CH3)
3.CBr(aq) (CH
3)
3 C+(aq) + Br¯ (aq) the molecularity is:
(A) Zero (B) 1 (C) 2 (D) cannot ascertained
9. Following mechanism has been proposed for a reaction ,2 A + B D + EA + B C + D ........ (slow)A + C E ...... (fast)
The rate law expression for the reaction is:(A) r = K[A]2[B] (B) r = K [A] [B] (C) r = K[A]2 (D) r = K[A] [C]
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10. A reaction, A2 + B
2 2AB occurs by the following mechanism ;
A2 A + A ........ (slow)
A + B2 AB + B ..... (fast)
A + B AB ...... (fast)Its order would be:(A) 3/2 (B) 1 (C) zero (D) 2
11. The chemical reaction, 2O3 3O
2 proceeds as follows :
O3
O2 + O ...... (fast)
O + O3 2O
2...... (slow)
The rate law expression should be:(A) r = K [O
3]2 (B) r = K[O
3]2 [O
2]–1 (C) r = K [O
3][O
2] (D) Unpredictable
12. The following two step mechanism has been proposed for the gas-phase decomposition ofnitrous oxide (N
2O):
Step I : N2O(g) N
2(g) + O(g)
Step II : N2O(g) + O(g) N
2(g) + O
2(g)
(a) Write the chemical equation for overall reaction.(b) Identify any reaction intermediates.(c) What is the molecularity of each of the elementary reactions ?(d) What is the molecularity of the overall reaction ?
13. What is the order of reaction ,A
2 + B
2 2 AB ; having following mechanism
A2
A + AA ........ (fast)A + B
2 AB + B ........ (slow)
A + B AB ........ (fast)
14. The following reaction has a second-order rate law :H
2(g) + 2ICl (g)I
2(g) + 2HCl (g) Rate = k[H
2] [ICl]
Devise a possible reaction mechanism.
15. Initial rate data at 25ºC are listed in the table for the reactionNH
4+ (aq) + NO
2– (aq) N
2(g) + 2H
2O (l)
ExperimentNo.
Initial [NH4+] Initial [NO2
–] Initial rate of consumptionof NH4
+(M/s)1.2.3.
0.240.120.12
0.100.100.15
7.2 × 10–6
3.6 × 10–6
5.4 × 10–6
(a) What is the rate law ?(b) What is the value of the rate constant ?(c) What is the reaction rate when the concentrations are [NH
4+] = 0.39 M and
[NO2
–] = 0.052 M ?
16. Find :(i) rate law expression(ii) order of reaction with respect to each reactant and overall order of reaction(iii) value and unit of rate constant(iv) effect on rate of reaction on changing the volume to 1/8th of the original.For the reaction , A + B + Cproduct
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given :
ExperimentNo.
[A]0
10–4 (m/l)[B]0
10–2 (m/l)[C]0
10–1 (m/l)Initial rate
× 10–6 (m/l/s)1234
77714
3663
5525
912.712.74.51
17. Initial rate data at 25ºC are listed in the table for reaction , A2 + B
2 2AB
ExperimentNo.
[A2]0 (m/l) [B2]0 (m/l) Initial rate (m/l/s)
123
1.5 × 10–3
4.5 × 10–3
3 × 10–3
3.2 × 10–4
9.6 × 10–4
1.6 × 10–4
8.1 × 10–8
6.56 × 10–6
3.24 × 10–7
using that , determine(i) rate law(ii) order of reaction with respect to A
2 and B
2 and overall order of reaction
(iii) value and unit of rate constant(iv) effect on rate of reaction on doubling the volume of container.
18. The data given below are for the reacion of NO and Cl2 to form NOCl at 295 K.
[Cl2] [NO] Initial rate (mol litre–1 sec–1)0.05 0.05 1 × 10–3
0.15 0.05 3 × 10–3
0.05 0.15 9 × 10–3
(a) What is the order w.r.t. NO and Cl2 in the reaction
(b) Write the rate expression(c) Calculate the rate constant(d) Determine the reaction rate when conc. of Cl
2 and NO are 0.2 M and 0.4 M
respectively.
19. The decomposition of NH3 on tungsten surface follows zero order kinetics. The half life is
5 minutes for an initial pressure of 70 torr of NH3. If the initial pressure had been 0.25 atm ,
what would the total pressure after 3 minutes ? What is the half life ?
20. In a reaction , the decrease in reactant’s concentration is 20 % in 20 minute and 40 % in40 minute. Calculate order of reaction and rate constant .
1. Show that for a first order reaction , R P the concentration of product can berepresented as a function of time by , [P] = at + bt2 + ct3 + ------ and express a , b and c interms of [R]
0 and K.
2. (i) Substance A decomposes by the first order reaction. Starting initially with[A] = 2.00 M, after 200 minutes [A] = 0.25 M. For this reaction what is t
1/2 and k ?
(ii) A first order reaction is 40 % complete after 8 min. How long will it take before it is90 % complete ? What is the value of the rate constant ?
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(iii) For a certain reaction it takes 20 minutes for the initial concentrations of 34.8 mol L–1 tobecome 17.4 mol L–1 and another 20 minutes to become 8.7 mol L–1. Calculate the rateconstant of the reaction.
(iv) For a first order reaction in which k = 5.48 × 10–1 sec–1. Find : (a) t1/3
(b) t2/3
3. Show that for a first order reaction(i) t
25% = 0.415t
1/2(ii) t
87.5% = 3t
1/2(iii) t
90% = 3.33 t
1/2
(iv) t99%
= 6.66t1/2
(v) t99.9%
= 10t1/2
(vi) t99%
= 2t90%
(vii) t99.9%
= 3t90%
4. The solution of H2O
2 of normality 0.73 is catalytically decomposed. What will be the
concentration at the end of 45 minute, assuming the decomposition to follow first order ratelaw if half life is 15 minute ?
5. The change , A B shows I order :(a) How will the rate of reaction change when the concentration of A is tripled ?(b) What will be the change in half life period in doing so ?
6. The virus prepared in a chemical bath shows inactivation process obeying I order. Calculatethe rate constant for the viral inactivation if in the beginning 1.5% of the virus is inactivatedper minute. Also calculate the time required for its ,(a) 50 % inactivation (b) 80 % inactivation.
7. Show that for I order reaction , the time required for 99.9 % decomposition of the reactionis ten folds to the time required for half of the reaction .
8. The time required for 20 % completion for a reaction is 10 minute for I order reaction.Calculate :(a) specific reaction rate. (b) Time required for 75 % completion
9. The half life for a I order reaction A Product, is 10 minute. What % of A remains after(i) 1/2 hour (ii) 2/3 hour (iii) 1 hour
10. The reaction SO2Cl
2 SO
2 + Cl
2 obeys I order kinetics with rate const. 3.2 × 10–5 sec–1
at 320 ºC . What % of SO2Cl
2 will be decomposed on heating gas for 90 minute ?
11. For the first order reaction , AB, shown in the figure, what is the significance of thepoint at which the two curves cross each other. How is the slopes of the two curves berelated at this point ?
Con
cent
rati
on
Time
12. A first order gaseous reactions has K = 1.5 × 10–6 sec–1 at 200 ºC . If the reaction isallowed to run for 10 hour, what percentage of initial concentration would have changed intoproducts. What is the half life period of reaction ?
13. Thermal decomposition of a compound is of first order. If 50 % sample of the compound isdecomposed in 120 minute, how long will it take for 90 % of the compound to decompose ?
14. A substance is reduced to one third of its original concentration in 100 minute. Calculate thetime in which it will be reduced to one ninth of its original value. Assume I order .
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15. For the decomposition of N2O
5, the rate constant is 6.2 × 10–4 sec–1 at 45ºC. It begins with
one mole of N2O
5 in a litre flask, how long would it take for 20% N
2O
5 to decompose and
how long for 50 % ?
16. In the decomposition of oxalic acid following data were obtained:Time in minute 0 300 600Pressure in mm 22.0 17.0 13.4Determine the rate constant K and half life period, if reaction obeys I order kinetics.
17. Decomposition of N2O
5(g) into NO
2(g) and O
2(g) is a first order reaction. If the initial
concentration of N2O
5(g), i.e., [N
2O
5]
0 is 0.030 mol litre–1, what will be its concentration
after 30 minute? Rate constant of reaction is 1.35 × 10–4 sec–1.
18. For a homogeneous gaseous reaction , A B + C + D, the initial pressure was P0 while
pressure after time t was P. Derive an expression for rate constant K in terms of P0, P and
t. assume I order reaction.
19. Acetone on heating gives CO and other hydrocarbons at 600°C. The reaction obeys I orderkinetics w.r.t. acetone concentration. The half life period is 81 sec. Calculate the time in whichacetone taken in a container at 600 ºC reduces its pressure from 0.5 atm to 0.4 atm.
20. Dimethyl ether gaseous phase decomposition is :CH
3OCH
3 CH
4 + H
2 + CO at 750 K having rate constant 6.72 × 10–3 min–1. Calculate
the time in which initial pressure of 400 mm in closed container becomes 750 mm.
21. The decomposition of Cl2O
7 at 400 K in gaseous phase to Cl
2 and O
2 is of I order reaction.
After 55 sec at 400 K, the pressure of reaction mixture increase from 0.62 to 1.88 atm.Calculate the rate constant of reaction. Also calculate the pressure of reaction mixture after100 second.
22. In the vapour phase decomposition of ethylene oxide, C2H
4O CH
4 + CO at 414.5ºC,
the initial pressure and the pressure after 5 minute were 116.51 mm and 122.56 mm of Hgrespectively. If the reaction follows I order kinetics, what must be the pressure after 12 minutes?
23. Calculate the partial pressures of reactants and products , when azomethane decomposes atan initial pressure of 200 mm for 30 minute according to (CH
3)
2 N
2C
2H
6 + N
2. The rete
constant is 2.5 ×10–4 sec–1.
24. Rate constant of a first order reaction A B is 6.93 ×10–2 minute. If initial concentration ofreactant is 1 M . Then calculate :(i) Initial rate (ii) Rate after 600 seconds(iii) Rate after 75 % of the reaction is completed(iv) Rate after 30 minutes.
25. In a first order reaction concentration of reactant decreases from 10 M to 2.5 M in 500seconds . Calcualte :(i) t
1/2 of the reaction
(ii) rate of reaction when concentration is 5 M.
26. Rate constant of a first order reaction is 10–3 sec. Calculate time in seconds when concentrationis changing from :(i) 2 M to 0.25 M (ii) 16 M to 1 M(iii) 6.4 M to 1.6 M (iv) 4 M to 0.125 M
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1. The following data represent for the decomposition of NH4NO
2 in aqueous solution
Time in minute 10 15 20 25 Volume of N2 (in mL) 6.25 9.0 11.40 13.65 33.05
(a) Show that reaction is of I order. (b) Calculate velocity constant.
2. Derive the O. R. for decomposition of H2O
2 from the following data
Time in minute 10 15 20 25 Vol. of O2 given by H2O2 6.30 8.95 11.40 13.5 35.75
3. Decomposition of diazobenzene chloride was followed at constant temperature by measuringvolume of N
2 evolved at definite intervals of time. Calculate O. R. and rate constant
Time in minute 0 2.0 5.5 7.0 Volume of N2 in mL 0 10 25 35 163
4. The decomposition of N2O
5 in chloroform was followed by measuring the volume of
O2 gas evolved; 2N
2O
52N
2O
4 + O
2 (g). The maximum volume of O
2 gas obtainable
was 100cm3. In 500 minutes, 90cm3 of O2 were evolved. Calculate the first order rate
constant of the reaction.
5. The specific rate constant of the decomposition of N2O
5 is 0.008 min–1. The volume of O
2
collected after 20 minute is 16 mL. Find the volume that would be collected at the end ofreaction. NO
2 formed is dissolved in CCl
4.
6. Derive order of reaction , for the decomposition of H2O
2 from the following data .
Time in minute 0. 10 20 30Volume of KMnO4 needed for H2O2 25 16 10.5 7.09
7. The kinetics of hydrolysis of methyl acetate in excess dilute HCl at 25 ºC were followed bywithdrawing 2 mL of the reaction mixture at intervals of (t), adding 50 mL water and titratingwith baryta water. Determine the velocity constant of hydrolysis.
t (in minute) 0 75 119 259 Titre value (in mL) 19.24 24.20 26.60 32.23 42.03
8. The acid catalysed hydrolysis of an organic compound A at 30ºC has half life of 100 minutewhen carried out in a buffer solution of pH = 5 and 10 minute when carried out at pH = 4.Both the times the half life are independent of the initial concentration of A. If the rate ofreaction is given by : rate = k[A]m [H+]n, what are the values of m and n and also calculater therate of reaction ?
9. In the inversion of cane sugar in presence of an acid, the following polarimeter readings areobtained
Time in minute 0 30 90 230 Rotation in degree +46.75 +41.0 +30.75 +12.75 –18.75
Calculate rate constant.
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10. The inversion of cane sugar proceeds with constant half life of 500 minutes at pH = 5 for anyconcentration of sugar. However, if pH = 6, the half life changes to 50 minutes. Derive the ratelaw for inversion of cane sugar.
11. A solution of N2O
5 in CCl
4 at 45 ºC produces 5.02 mL of O
2 in 1198 sec and 9.58 mL O
2 after
a long time . Calculate rate constant assuming I order for the reaction.
N2O
5 N
2O
4 + 2
1 O2
1. An organic compound A decomposes following two parallel first order mechanisms :
1
2
K
K=
1
9 and k
1 = 1.3 × 10–5 sec–1
Calculate the concentration ratio of C to A , if an experiment is allowed to start with onlyA for one hour .
2. Trans-1,2-dideuterocyclopropane (A) undergoes a first order decomposition. The observedrate const. at certain temp., measured in terms of disappearance of ‘A’ was 1.52 × 10–4 sec–1.Analysis of products showed that the reaction followed two parallel paths, one leading todideuteropropane. (B) and the other to cis-1,2-dideuterocyclopropane (C). (B) was found toconstitute 11.2 % of the reaction product , independently of extent of reaction . What is theorder of reaction for each path and what is the value of the rate constant for the formation ofeach of the products ?
3. Bicyclohexane was found to undergo two parallel first order rearrangements. At 730 K , thefirst order rate constant for the formation of cyclohexane was measured as 1.26×10–4s–1, andfor the formation of methyl cyclopentene the rate constant was 3.8 × 10–5 s–1. What is thepercentage distribution of the rearrangement products ?
4. For the reaction ,
[Cr(H2O)
4Cl
2]+ (aq) 1K [Cr(H
2O)
5Cl]2+ (aq) 2K [Cr(H
2O)
6]3+ (aq)
k1 = 1.78 × 10–3 s–1 and k
2 = 5.8 × 10–5 s–1 for the initial concentration of [Cr(H
2O)
4Cl
2]+
is 0.0174 mol/litre at 0 ºC. Calculate the value of t at which the conc. of [Cr(H2O)
5Cl]2+ is
maximum .
5. Two first order reactions proceed at 25 ºC at the same rate. The temperature coefficient of therate of the first reaction is 2 and that of second reaction is 3 . Find the ratio of the rates of thesereactions at 75 ºC .
6. The half life of a substance in a first-order reaction is 100 minutes at 323.2 and 15 min at353.2 K . Calculate the temperature coefficient of the rate constant of this reaction .
7. The activation energy for the reaction , O3(g) + NO(g) NO
2(g) + O
2(g)
is 9.6 kJ/mole. Prepare an activation energy plot if Hº for this reaction is – 200 kJ/mole.What is the energy of activation for the reverse reaction ?
8. Which reaction will have the greater temperature dependence for the rate constant-one witha small value of energy of activation (E) or one with a large value of E ?
9. For a chemical reaction the energy of activation is 85 J mol–1. If the frequency factor is,4.0 × 109 L mol–1s–1 , what is the rate constant at 400 K ?
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10. For the displacement reaction[Co(NH
3)
5Cl]2+ + H
2O[Co(NH
3)
5(H
2O)]3+ + Cl–
the rate constant is given by , ln [k/(min–1)] = –11067
T + 31.33.
Evaluate k , E and A for the chemical reaction at 25 ºC .
11. For the reaction , 2 N2O
54 NO
2+ O
2
the rate constant is given by , ln [k(sec–1)] = –10500
T + 33.
Evaluate k , E and A for the chemical reaction at 27 ºC .
12. For a first order reaction, the rate constant is given by , ln [k(sec–1)] = –11400
T + 34.7
Evaluate k, E and A for the chemical reaction at 27 ºC.
13. The time required for 20 % completion of a first order reaction at 27 ºC is 1.5 times thatrequired for its 30 % completion at 37 ºC. If the pre exponential factor for the reaction is3 × 109 sec–1, calculate the time required for 40 % completion at 47 ºC and also the energyof activation .
14. The rate constant of a reaction increases by 7 % when its temperature is raised from 300 Kto 301 K while its equilibrium constant increases by 3 %. Calculate the activation energy of theforward and reverse reactions.
15. A bottle of milk stored at 300 K sours in 36 hours. When stored in a refrigerator at 275 K itsours in 360 hrs. Calculate the energy of activation of the reaction involved in the souringprocess.
16. Calculate the ratio of the catalysed and uncatalysed rate constant at 20 ºC if the energy ofactivation of a catalysed reaction is 20 kJ mol–1 and for the uncatalysed reaction is75 kJ mol–1.
17. A second order reaction where a = b is 20 % completed in 500 seconds. How long will thereaction take to be 60 % complete.
18. Two reactions of same order have equal pre-exponential factors but their activation energiesdiffers by 41.9 J/mole. Calculate the ratios between rate constants of these reactions at 600 K.
19. Rate constant of a reaction changes by 2 % by 0.1 ºC rise in temperature at 25 ºC. Thestandard heat of reaction is 121.6 kJ mol–1. Calculate E
a of reverse reaction.
20. The energy of activation and specific rate constant for a first-order reaction at 25ºC 2N
2O
5 2N
2O
4 + O
2
(in CCl4) (in CCl
4)
are 100 kJ/mole and 3.46 × 10–5 s–1 respectively. Determine the temperature at which the half-life of the reaction is 2 hours.
21. In Arrhenius’ equation for a certain reaction, the value of A and E (activation energy) are 4 ×1013 s–1 and 98.6 kJ mol–1 respectively. If the reaction is of first order, at what temperature willits half-life period be ten minutes?
22. Two reactions proceed at 25ºC at the same rate. The temperature coefficient of the rate ofthe first reaction is 2 and that of the second is 2.5. Find the ratio of the rates of these reactionsat 95 ºC .
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23. What is the energy of activation of a reaction if its rate doubles when the temperature is raisedfrom 290 K to 300 K ?
24. For the reaction A + B C + D; H = + 20 kJ/mole, the activation energy of the forwardreaction is 85 kJ/mole. Calculate the activation energy of the reverse reaction.
25. What is the value of the rate constant, predicted by the Arrhenius's equation if T?Is this value physically reasonable ?
26. The activation energy of a certain uncatalysed reaction at 300 K is 76 kJ mol –1. Theactivation energy is lowered by 19 kJ mol–1 by the use of catalyst. By what factor, the rate ofcatalysed reaction is increased ?
27. Given that K (sec–1) = 5 × 1014 e–124080/RT where activation energy is expressed in joule.Calculate the temperature at which reaction has t
1/2 equal to 25 minute. Assume I order reaction.
28. For the reaction A products , the time for half change is 5000 second at 300K and 1000second at 310 K. If the reaction obeys first order kinetic, calculate energy of activation .
29. Two reactions of same order have equal exponential factors but their activation energydiffer by 24.9 kJ mol–1. Calculate the ratio between the rate constant of these reactions at 27ºC
[ R = 8.314 JK–1 mol–1 ]
30. The energy of activation of a I order reaction is 104.5 kJ mol–1 and pre-exponential factor Ain Arrhenius equation is 5 × 1013 sec–1. At what temperature will the reaction have half life of1 minute ?
6. Daily Practice Problem Sheet
Q.1 Calculate the number of neutrons in the remaining atom after emission of an alpha particle from
U23892 atom.
Q.2 Radioactive disintegration of Ra22688 takes place in the following manner into RaC.
RaCRaBRaARnRa
Determine mass number and atomic number of RaC.
Q.3 A radioactive element A disintegrates in the following manner
DCBA
(i) Which one of the elements A, B, C, D are isotopes ?(ii) Which one of the elements A, B, C, D are isobars ?
Q.4 Write the particles emitted from each nucleides in the following reactions :
(a) Th23190
)i( Pa231
91
)ii( Ac22789 (b) At217
85
)i( Bi21383
)ii( Tl209
81
Q.5 An atom has atomic mass 232 and atomic number 90. During the course of disintegration, it emits2-particles and few-particles. The resultant atom has atomic mass 212 and atomic number 82.How many-particles are emitted during this process.
Q.6 In the sequence of the following nuclear reaction
MLZYX21890
n23898
what is the value of n(A) 3 (B) 4 (C) 5 (D) 6
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Q.7 The isotope U23592 decays in a number of steps to an isotope of Pb207
82 . The groups of particlesemitted in this process will be(A) 4, 7 (B) 6, 4 (C) 7, 4 (D) 10, 8
Q.8 The U23892 disintegrates to give 4-and 6-particles. The atomic number of the end product is
(A) 92 (B) 96 (C) 84 (D) 90
Q.9 Following are the atoms having the number of neutrons and protons as given below :Atoms Protons NeutronA 8 8B 8 9C 8 10D 7 8E 7 9
Select incorrect conclusion(s) :(A) A, B and C, D are isotopes (B) A and D are isotones(C) A and E are isobars (D) A and B are isodiaphers
Q.10 An alkaline earth element is radioactive. It and its daughter elements decay by emitting there-particles in sucession. The daughter element formed wil belong to group -(A) 8 (B) 16 (C) 14 (D) 12
Q.11 HePbPo 42
20682
21084
In above reaction, predict the position of Po in the periodic table when lead belongs to IVB group :(A) IIA (B) VI B (C) IV B (D) V B
Q.12 Radioactive substance of 1 curie is the amount that can produce ................. disintegrations persecond.
Q.13 The last member of 4n + 1 series is an isotope of .............. .
Q.14 The 4n series starts from Th-232 and ends at(A) Pb-208 (B) Bi-209 (C) Pb-206 (D) Pb-207
Q.15 Select the wrong statement(A) Nuclear isomers contain the same number of protons and neutrons(B) The decay constant is independent of the amount of the substance taken(C) One curie = 3.7 × 1010 dis/minute(D) Actinium series starts with U238
Q.16 The correct starting material and product of different disintegration series are :(A) 232Th, 208Pb (B) 235U, 206Pb (C) 238U, 207Pb (D) 237Np, 209Bi
Q.17 Which of the following is/are correct(A) 1 curie = 3.7 × 1010 d/s (B) 1 rutherford = 106 d/s(C) 1 becquerel = 1 d/s (D) 1 fermi = 103 d/s
Q.18 To which radioactive families do the following nucleides belong ?222Rn, 228Ra, 207Pb, 209Bi, 233Pa.
Q.19 Balance the following nuclear reactions.
(i) Be94 + He4
2 ........+ n10 (ii) Li6
3 + ........ Li73 + H1
1
(iii) Be94 + ...... nBe 1
084
(iv) nU 10
23592 Ba141
56 + ..........+ 3 n10
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(v) P3115 + H1
1 S3116 + ........ (vi) 75
33 As + H21 ......... + H1
1
(vii) Cu6329 + He4
2 3717 Cl + 14 H1
1 + ........
(viii) H21 + H3
1 ....... + n10
Q.20 Calculate and particles emitted during the process.
(i) X23090 Y210
85 (ii) X20882 Y196
82
(iii) Th25290 Pb208
82 (iv) U23892 U234
92
(v) U23892 Ra226
88 (vi) Ra22688 Bi214
83
(vii) Th23490 Po218
84 (viii) NP23793 Bi209
83
(ix) U23592 Pb207
92 (x) X22086 Y200
80
7. Daily Practice Problem Sheet
1. The triad of nuclei that represents isotopes is:(A)
6C14,
7N14,
9F19 (B)
6C12,
7N14,
9F19
(C)6C14,
6C13,
6C12 (D)
6C14,
7N14,
9F17
2. The triad of nuclei that represents isotones is:(A)
6C12,
7N14,
9F19 (B)
6C14,
7N15,
9F17
(C)6C14,
7N14,
9F17 (D)
6C14,
7N14,
9F19
3. The rate of radioactive disintegration........... with time:(A) Increases (B) Decreases (C) Is constant (D) May increase
4. When a radioactive element emits an electron the daughter element formed will have:(A) Mass number one unit less (B) Atomic number one unit less(C) Mass number one unit more (D) Atomic number one unit more
5. Decrease in atomic no. is observed during:(A) Alpha emission (B) Electron capture(C) Positron emission (D) all
6. Successive emission of an-particle and two -particles by an atom of an element results in theformation of its:(A) Isodiapher (B) Isomorph (C) Isotope (D) Isotherm
7. If N0 is the initial number of nuclei, number of nuclei remaining undecayed at the end of nth half life
is:(A) 2–n N
0(B) 2n N
0(C) n–2 N
0(D) n2 N
0
8. Which one of the following nuclear reaction is correct:(A)
6C13 +
1H1
7N13 + – + (B)
11Na23 +
1H1
10Ne20 +
2He4
(C)13
Al23 +0n1
11Na23 + –1e
0 (D) 12
Mg24 +2He4
13Al27 +
0n1
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9. The activity of a radionuclide (X100) is 6.023 curie. If the disintegration constant is 3.7 × 104 sec–1,the mass of radionuclide is:(A) 10–14 g (B) 10–6 g (C) 10–15 g (D) 10–3 g
10. The half life of I131 is 8 day. Given a sample of I131 at t = 0, we can assert that:(A) No nucleus will decay at t = 4 day(B) No nucleus will decay before t = 8 day(C) All nucleus will decay before t = 16 day(D) A given nucleus may decay at any time after t = 0
11. If 5g of a radioactive substance has t1/2
= 14 hr., 2 g of the same substance will have a t1/2
equal to:(A) 56 hr (B) 3.5 hr (C) 14 hr (D) 28 hr
12. The half life of a radioactive isotope is 2.5 hour. The mass of it that remains undecayed after 10 houris (If the initial mass of the isotope was 16 g):(A) 32 g (B) 16 g (C) 4 g (D) 1 g
13. The number of-and-particles emitted during the transformation of Th23290 to Pb208
82 is respec-
tively:(A) 2, 2 (B) 4, 2 (C) 6, 4 (D) 8, 6
14. If 75% quantity of a radioactive isotope disintegrates in 2 hour, its half life would be:(A) 1 hour (B) 45 minute (C) 30 minute (D)15 minute
15. A certain radioactive isotope has a half life of 50 day. Fraction of the material left behind after 100day will be:(A) 50% (B) 75% (C) 12.5% (D) 25%
16. The half life period of a radioactive elements is 140 day. After 560 day, 1 g of the element willreduce to:(A) 0.5 g (B) 0.25 g (C) 1/8 g (D) 1/16 g
17. 75% of a first order reaction was completed in 32 minute. When will be 50% of the reactioncomplete.(A) 24 minute (B) 16 minute (C) 8 minute (D) 4 minute
18. The half life of a radioactive isotope is 1.5 hour. The mass of it that decayed after 6 hour is (the initialmass of the isotope is 32 g):(A) 32 g (B) 16 g (C) 30 g (D) 2 g
19. Half life period of a substance is 1600 minute. What fraction of the substance will remain after 6400minute:(A) 1/16 (B) 1/4 (C) 1/8 (D) 1/2
20. The half life period of a radioactive nuclide is 3 hour. In 9 hour its activity will be reduced by(A) 1/9 (B) 7/8 (C) 1/27 (D) 1/6
21. A radioactive isotope having a half life of 3 day was received after 12 day. It was found that therewere 3 g of the isotope in the container. The initial weight of the isotope when packed was:(A) 12 g (B) 24 g (C) 36 g (D) 48 g
22. A sample of rock from moon contains equal number of atoms of uranium and lead (t1/2
for U = 4.5× 109 year). The age of the rock would be:(A) 4.5 × 109 year (B) 9 × 109 year (C) 13.5 × 109 year (D) 2.25 × 109 year
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23. Two radioactive elements X and Y have half lives of 50 and 100 minute respectively. Initial sampleof both the elements have same no. of atoms. The ratio of the remaining number of atoms of X andY after 200 minute is:(A) 2 (B) 1/2 (C) 4 (D) 1/4
24. The time in which activity of an element is reduced to 90% of its original value is, (given t1/2
= 1.4 ×1010 yr):(A) 1.128 × 109 yr (B) 2.128 × 109 yr (C) 3.128 × 109 yr (D) None
25. The number of- and - particles emitted in the nuclear reaction Th22890 Bi212
83 are:
(A) 4 and 1 (B) 3 and 7 (C) 8 and 1 (D) 4 and 7
26. Two radioisotopes P and Q of atomic weight 10 and 20 respectively are mixed in equal amount byweight. After 20 day, their weight ratio is found to be 1 : 4. Isotope P has a half life of 10 day. Thehalf life of isotope Q is:(A) Zero (B) 5 day (C) 20 day (D) infinite
27. The radioactivity due to the C14 isotope (t1/2 = 6000 year) of a sample of wood from an ancienttomb was found to be nearly half that of fresh wood; the tomb is, therefore, about:(A) 3000 year old (B) 6000 year old (C) 9000 year old (D) 12000 year old
28. The activity of a radioactive sample drops to 1/64th of its original value in 2hr. The decay constantfor the sample is:(A) 5.775 × 10–4 sec.–1 (B) 5.775 × 104 sec.–1
(C) 5.775 × 102 sec.–1 (D) None
29. A radioactive element has a half life of 4.5 × 109 year. If 80 g of this was taken, the time taken for itto decay to 40 g will be:(A) 2.25 × 109 year (B) 4.50 × 109 year (C) 6.75 × 109 year (D) 8.75 ×109 year
30. A certain nuclide has a half life period of 30 minute. If a sample containing 6000 atoms is allowed todecay for 90 minute, how many atoms will remain:(A) 200 atoms (B) 450 atoms (C) 750 atoms (D) 150 atoms
31. A substance is kept for 2 hour and three fourth disintegrates during this period. The half life of thesubstance is:(A) 2 hour (B) 1 hour (C) 30 minute (D) 4 hour
32. The binding energy of an element is 64 MeV. If BE/nucleon is 6.4, the number of nucleons are:(A) 10 (B) 64 (C) 16 (D) 6
33. The half life of 92U238 against-decay is 4.5 × 109 year. The time taken in year for the decay of 15/
16 part of this isotope is:(A) 9.0 × 109 (B) 1.8 × 1010 (C) 4.5 × 109 (D) 2.7 × 1010
34. A radioactive isotope has a half life of 10 day. If today there are 125g of it left, what was its originalweight 40 day earlier:(A) 600 g (B) 1000 g (C) 1250 g (D) 2000 g
35. Radium has atomic weight 226 and half life of 1600 year. The number of disintegration producedper sec. from 1 g are:(A) 4.8 ×1010 (B) 3.7 ×1010 (C) 9.2 ×106 (D) 3.7 ×108
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36. If the amount of radioactive substance is increased three times, the number of atoms disintegratedper unit time would:(A) Be double (B) Be triple (C) Remain one third (D) Not change
37. The half life of a radioactive element is 35 year. If there are 4 × 106 nuclei at the start, then after howmany year they will be left 0.5 × 106.(A) 35 (B) 70 (C) 105 (D) 140
38. Wooden article and freshly cut tree show activity 7.6 and 15.2 min–1 g–1 of carbon (t1/2 = 5760year) respectively. The age of the article is:
(A) 5760 year (B) 5760 ×6.7
2.15year (C) 5760 ×
2.15
6.7year (D) 5760 × (15.2 – 7.6) year
39. For a radioactive substance with half life period 500 year, the time for complete decay of 100milligram of it would be:(A) 1000 year (B) 100 × 500 year (C) 500 year (D) Infinite time
40. The decay constant of a radioactive element is defined as the reciprocal of the time interval afterwhich the number of atoms of the radioactive element falls to nearly:(A) 50% of its original number (B) 36.8% of its original number(C) 63.2% of its original number (D) 75% of its original number
41. The radioactive decay rate of a radioactive element is found to be 103 dps at a certain time. If thehalf life of element is 1sec, the decay rate after 1 sec. is..... and after 3 sec. is......(A) 500, 125 (B) 125, 500 (C) 103, 103 (D) 100, 10
42. The counting rate observed from a radioactive source at t = 0 seconds was 1600 counts/sec and att = 8 sec it was 100 counts/sec. The counting rate observed as count per sec at t = 6 sec will be:(A) 400 (B) 300 (C) 200 (D) 150
43. A freshly prepared radioactive source of half life 2 hr. emits radiations of intensity which is 64 timesthe permissible safe level. The minimum time after which it would be possible to work safely with thissource is:(A) 6 hr (B) 12 hr (C) 24 hr (D) 128 hr
44. One mole of A present in a closed vessel undergoes decay as zAm z–4B
m–8 + 2 2He4. Thevolume of He collected at NTP after 20 days is (t1/2 = 10 day):(A) 11.2 litre (B) 22.4 litre (C) 33.6 litre (D) 67.2 litre
45. The number of-particle emitted during the change aXcdY
b is:
(A)4
ba (B) d +
2
ba + c (C) d +
2
bc– a (D) d +
2
ba– c
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Problem 1 :Rate of a reaction A + B Products; is given below as a function of different initialconcentrations of A and B[A] mol/litre [B] mol litre–1 Initial rate mol litre–1 time–1
0.01 0.01 0.0050.02 0.01 0.0100.01 0.02 0.005Determine the order of the reaction with respect to A and with respect to B. What isthe half life of A in the reaction ?
Solution :Let order with respect to A is m and with respect to B is ‘n’.Rate = K[A]m [B]n
0.005 = K [0.01]m [0.01]n … (i)0.010 = K [0.02]m [0.01]n … (ii)0.005 = K[0.01]m [0.02]n … (iii)Dividing Eq. (i) by (ii), we get
m
02.0
01.0
010.0
005.0
m
2
1
2
1
m = 1
Thus, order with respect to A is one.Dividing Eq. (i) by (ii), we get
n
02.0
01.0
005.0
005.0
n
2
11
n = 0
Thus, order with respect to B is zero.Substituting the values of m and n in Eq. (i), we get
0.005 = K [0.01]1 [0.01]0 5.001.0
005.0
2/1t (Half life period)5.0
693.0
K
693.0 = 1.386 time
Problem 2 :The decomposition of Cl2O7 at 440K in the gas phase to Cl2 and O2 is a first orderreaction.(i) After 55 seconds at 400 K the pressure of Cl2O7 falls from 0.062 to 0.044 atm.,
calculate that rate constant.(ii) Calculate the pressure of Cl2O7 after 100 sec. of decomposition at this
temperature.Solution :
(i) In the present equation we may apply the formula
t
0
p
plog
t
303.2K
044.0
062.0log
55
303.2 13 sec102364.6
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(ii) Pressure after 100 sec may be calculated as
3
t
2.303 0.0626.2364 10 log
100 p
0332.0p t atm
Problem 3 :Rate law for the following reaction; is
Ester + H+ Acid + Alcohol; is Kdtdx [ester]1 [H+]0
What would be the effect on the rate if(i) Concentration of ester is doubled ?(ii) Concentration of H+ ion is doubled ?
Solution :The rate law expression in this question, suggests that concentration of acid is nothing to playwith velocity.(i) When concentration of ester is doubled; velocity of the reaction will become double.(ii) When concentration of H+ ion is doubled velocity will be unaffected.
Problem 4 :The reaction 2A + B + C D + 2E; is found to be first order in A; second order in B andzero order in C.(i) Give the rate law for the above reaction in the form of a differential equation.(ii) What is the effect on the rate of increasing the concentration of A, B and C two
times ?Solution :
(i) The rate law according to given information may be given as, 021 ]C[]B[]A[Kdt
dx
(ii) When concentration of A, B and C are doubled then rate will be
2 0dxK[2A][2B] [C]
dt 02 ]C[]B][A[K8
i.e., rate becomes 8 fold, the original rate.
Problem 5 :At 27°C it was observed, during a reaction of hydrogenation that the pressure of H 2
gas decreases from 2 atm to 1.1 atm in 75 min. Calculate the rate of reaction (molarity/sec). Given (R = 0.082/litre atom K–1 mole–1)
Solution :
Rate =Decrease in pressure
Time duration012.0
75
1.12
atm/min
Rate in atm/sec 410260
012.0 atm/sec.
Answer is required in molarity per second PV = nRTT
V
n
RT
P
P / sec n/ sec
RT V
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/V
n
sec = molarity per second
sec/V
n
3000821.0
102 4
61012.8 . Rate in molarity/sec = (n/V)/sec
Problem 6 :In presence of an acid N-chloro acetanilide changes slowly into p-chloro acetanilide.Former substance liberated iodine from KI and not the later and hence progress ofreaction can be measured by titrating iodine liberated with Na 2S2O3 solution, theresults obtained were as follows :Time (hours) 0 1 2 4 6 8(a–x)i.e., hypo 45 32 22.5 11.3 3.7 2.9show that reaction is uniomlecular and find out the fraction of N-chloroacetanilidedecomposed after three hours.
Solution :The present reaction is :
NCl
CH3
O
catalystacid
NH CH3
O
Cl
Let us apply the kinetics of first order reaction.
After 1 hour
32
45log
1
303.2K = 0.34098 hour–1
After 2 hours
5.22
45log
2
303.2K = 0.34660 hour–1
After 4 hours
3.11
45log
4
303.2K = 0.3455 hour–1
After 6 hours
7.3
45log
6
303.2K = 0.4164 hour–1
After 8 hours
9.2
45log
8
303.2K = 0.3428 hour–1
Average value of constant = 0.3584 hours–1
Since, on applying first order kinetics, we get almost same values of rate constant after differenttime intervals, hence, the reaction is of first order.Let us see the fraction decomposed after 3 hours.
xa
alog
t
303.2K
0.3584 =xa
alog
3
303.2
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98.2xa
a
3412.0a
xa
3412.0a
x1
658.0a
x [fraction decomposed]
Problem 7 :For the decomposition of dimethyl ether, A in the Arrhenius equation K = Ae–E/RT hasa v a l u e o f 1 . 2 6 1013 and Ea value of 58.5 kcal. Calculate half life period for firstorder decomposition at 527°C.
Solution :Taking logarithm of Arrhenius equationK = Ae–E/RT we get
log K = log A –RT303.2
E… (i)
Given A = 1.26 1013
E = 58.5 kcalT = 527 + 273 = 800 KSubstituting these value in Eq. (i), we get
log K = log (1.26 1013) –800987.1303.2
105.58 2
= 13.1003 – 15.9799 = –2.8796
K = 1.3194 10–3 sec–1
32/1 10
3194.1
693.0t sec = 525 sec
Problem 8 :For the reaction ,2NO(g) + H2(g) N2O(g) + H2O(g) at 900 K, the following data are obtained :Initial pressure Initial pressure Rateof NO (atm) of H2(atm) (atm min–1)0.150 0.400 0.0200.075 0.400 0.0050.150 0.200 0.010Find the rate law and the value of rate constant.
Solution :Let order with respect to NO(g) is ‘m’ and order with respect to H
2(g) is ‘n’
Then, Rate =2
m nNO HK[P ] [P ] … (1)
0.020 = K [0.15]m [0.40]n … (2)0.005 = K [0.075]m [0.40]n … (3)0.010 = K [0.15]m [0.2]n … (4)Dividing Eq. (2) by (3), we get
m0.020 0.15
0.005 0.075
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m = 2
Dividing Eq. (2) by (4), we getn
2.0
4.0
010.0
020.0
n = 1
Rate =2
2 1NO H[P ] [P ]
Substituting the values of m, n in Eq. (2), we get0.020 = K (0.15)2 [0.40]K = 2.22 atm–2 min–1
Problem 9 :10 gram atoms of an -active radio isotope are disintegrating in a sealed container..In one hour the helium gas collected at STP is 11.2 cm3. Calculate the half-life of theradio-isotope.
Solution :
No. of atoms of helium is 11.2 cc at NTP =22400
2.11 6.02 1023 = 3.01 1020 atoms
Since, helium atom corresponds to -particle. Thus, Rate of disintegration = 3.01 1020 perhour.We know, Rate = Rate constant concentration in atom3.01 1020 = K 10 6.02 1023
K = 0.05 10–3 hour–1
32/11005.0
693.0
K
693.0T
= 13860 hours
58.136524
13860T 2/1
years
Problem 10 :A carbon radio isotope ZXA (half life 10 days) decays to give Z–2Y
A–4. If 1.00 gm atom of
ZXA is kept in a sealed tube, how much helium will accumulate in 20 days ? Expressthe result in cm3 at STP.
Solution :Initial concentration (N
0) of radio-isotope is 1 gm atom. Concentration remained after 20 days
may be calculated as
Nt = N
0
n
2
1
where n = n0 of half lives = 20/10 = 2 =
4
1
2
11
2
Concentration decayed to -particles = 1 –4
3
4
1 gm atom. An -particle takes 2 electron
from air and from helium gas. Thus,
Helium formed4
3 gm atom 22400
4
3 cc 16800 cc
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Problem 11 :Radioactive decay is a first order process. Radioactive carbon wood sample decayswith a half life of 5770 years. What is the rate constant in (years)–1 for the decay?What fraction would remain after 11540 years?
Solution :
5770
693.0
T
693.0K
2/1
= 1.201 10–4 year–1
t
0
N
Nlog
t
303.2K
1.201 10–4
t
0
N
Nlog
11540
303.2
4.002 =t
0
N
N
0
t
N
N (Remaining fraction) =
002.4
1
Problem 11 :A sample of 53I
131, as iodide ion, was administered to a patient in a carrier consistingof 0.10 mg of stable iodide ion. After 4.00 days, 67.7% of the initial radioactivity wasdetected in the thyroid gland of the patient. What mass of the stable iodide ion hadmigrated to the thyroid gland? Given T1/2 I
131 = 8 days.Solution :
We know,
N
Nlog
t
303.2
T
693.0 010
2/1
N
Nlog
4
303.2
8
693.0 010
707.0N
N
0
70.7 % of initial activity is present. Given that 67.7% activity is migrated to thyroidgland.
Thus, weight of I– migrated to thyroid gland may be calculated as
7.70
7.67 100 = 95.75%
i.e, 0.1 100
75.95 = 0.09575 mg
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Problem 1 :If a reaction A + B C is exothermic to the extent of 30 kJ/mol and the forwardreaction has an activation energy 70 kJ/mol , the activation energy for the reverse reactionis(A) 30 kJ/mol (B) 40kJ/mol (C) 70 kJ/mol (D) 100 kJ/molSolution : (D)
EnergyEnergy 100kJ
30kJ
70kJ
Progress of reaction
By seeing the curve, activation energy for backward reaction = 100 kJ
Problem 2 :The rate constant , the activation energy and the Arrhenius parameter of a chemicalreaction at 25°C are 3.0 10–4 s–1, 104.4 kJ mol–1 and 6.0 1014 s–1 respectively thevalue of the rate constant as T is :(A) 2.0 1018 s–1 (B) 6.0 1014 s–1 (C) (D) 3.6 1030 s–1
Solution : (B)RT/EaAeK
When TK AA = 6 1014 s–1
Problem 3 :The inversion of cane sugar proceeds with halflife of 500 minute at pH 5 for anyconcentration of sugar. However if pH = 6, the halflife changes to 50 minute. Therate law expression for the sugar inversion can be written as(A) r = K[sugar]2[H]6 (B) r = K[sugar]1[H]0
(C) r = K[sugar]0[H+]6 (D) r = K[sugar]0[H+]1
Solution : (B)Since t
1/2 does not depends upon the sugar concentration means it is first order respect to sugar
concentration. t1/2 [sugar]1.
t1/2
× an-1 = k
1 n1/ 2 1 1
1 n1/ 2 22
t [H ]
t [H ]
1 n5
6
500 10
50 10
10 = (10)1-n
Hence n = 0
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Problem 4 :Two substances A and B are present such that [A0] = 4[B0] and halflife of A is 5minute and that of B is 15 minute. If they start decaying at the same time followingfirst order kinetics how much time later will the concentration of both of them wouldbe same.(A) 15 minute (B) 10 minute (C) 5 minute (D) 12 minute
Solution: (A)
Amount of A left in n1 halves =
1n
0
1[A ]
2
Amount of B left in n2 halves =
2n
0
1[B ]
2
At the end, according to the question
1 2
00
n n
[B ][A ]
2 2
1 20 0n n
4 1, [A ] 4[B ]
2 2
1
1 2
2
nn n 2
1 2n2 4 2 (2) n n 22
2 1n (n 2) …(1)
Also 1 1/ 2( )At n t= ´ 2 1/ 2( ) Bt n t
(Let concentration of both become equal after time t)
1 1/ 2( ) 1 1
2 1/ 2( ) 2 2
51 1 315
A
B
n t n nn t n n´ ´\ = Þ = Þ =´ ´ …(2)
For equation (1) and (2)n
1 = 3, n
2= 1
t = 3 5 = 15 minute
Problem 5 :Fill in the blank
235 1 92 192 0 36 0U + n ? + Kr + 3 n
(A) 14156 Ba (B) 139
56 Ba (C) 13954 Ba (D) 141
54 B
Solution: (A)92 + 0 = Z + 36 + 0 Z = 56235 + 1 A + 92 + 3
A = 144 Missing nuclide is 14156 Ba
Problem 6 :The rate of a reaction increases 4-fold when concentration of reactant is increased 16times. If the rate of reaction is 4 10–6 mole L–1 S–1 mole L–1 when concentration of thereactant is 4 10–4, the rate constant of the reaction will be(A) 2 10–4 mole1/2 L–1/2 S–1 (B) 1 10–2 S–1
(C) 2 10–4 mole–1/2, L1/2 S–1 (D) 25 mole–1 L min–1
Solution : (A)
Rate ionconcentrat , Rate = ionconcentratk
2
6
2/14
6
2/1 102
104
104
104
concen
Ratek
= 2 10–4 mole1/2 L-1/2 S–1
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Problem 7 :A catalyst lowers the activation energy of a reaction from 20 kJ mole –1 to10 kJ mole–1. The temperature at which the uncatalysed reaction will have the samerate as that of the catalysed at 27C is(A) –123C (B) 327C (C) 327C (D) + 23C
Solution : (B)
22
a
1 T
20
300
10
T
E
T
aE
T
2 = 600 K = 327 C
Problem 8 :The reaction , A(g) + 2B(g) C(g) + D(g) is an elementary process . In an experiment,the initial partial pressure of A and B are PA = 0.60 and PB = 0.80 atm. When PC = 0.2atm the rate of reaction relative to the initial rate is(A) 1/48 (B) 1/24 (C) 9/16 (D)1/6
Solution : (D)
]B][A[KR1 2 2]80.0][6.0[KAfter reactionA + 2B C + D0.6 – 0.2 0.8 – 0.4 0.2 0.20.4 0.4 0.2 0.2
6
1
)8.0)(6.0(
)4.0)(4.0(K
R
R2
2
1
2
Problem 9 :Thermal decomposition of a compound is of first order . If 50 % of a sample of thecompound is decomposed in 120 minutes , show how long will it take for 90 % of thecompound to decompose ?(A) 399 min (B) 410 min (C) 250 min (D) 120 min
Solution : (A)
120
6932.0K …(1)
a10.0
alog
t
303.2K = 10log
t
303.2…(2)
Equating (1) and (2)t
303.2
120
6932.0
t = 399 minutes
Problem 10 :If concentration are measured in mole/litre and time in minutes, the unit for the rateconstant of a 3rd order reaction are(A) mol lit–1min–1 (B) lit2 mol–2 min–1 (C) lit mol–1min–1 (D) min–1
Solution : (B)K = [conc.]1–n min–1
For 3rd order reaction = [mole/litre]1–3 min–1 = lit2.mole–2min–1
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Problem 11 :What is the activation energy for the decomposition of N2O5 as
N2O5 2NO2 +12
O2
If the values of the rate constants are 3.45 10–5 and 6.9 10–3 at 27°C and 67°Crespectively .(A) 102 102 kJ (B) 488.5 kJ (C) 112 kJ (D) 112.5 kJ
Solution : (D)
21
12
1
2
TT
TT
R303.2
Ea
K
Klog
400300
40
31.8303.2
Ea
1045.3
109.6log
5
3
Ea = 112.5 kJ
Problem 12 :Half life period for a first order reaction is 20 minutes. How much time is required tochange the concentration of the reactants from 0.08 M to 0.01M(A) 20 minutes (B) 60 minutes (C) 40 minutes (D) 50 minutes
Solution : (B)
xa
alog
t
303.2K
0.6932 2.303 0.08log
20 t 0.01 = 8log
t
303.2 =
t
2log3303.2
t
3
20
1
t = 60 minutes
Problem 13 :The kinetic datas for the reaction: 2A + B2 2AB are as given below:[A] [B2] Ratemol L–1 mol L–1 mol L–1 min–1
0.5 1.0 2.5 10–3
1.0 1.0 5.0 10–3
0.5 2.0 1 10–2
Hence the order of reaction w.r.t. A and B2 are, respectively,(A) 1 and 2 (B) 2 and 1 (C) 1 and 1 (D) 2 and 2
Solution : (A)2.5 10–3 = K[0.5] [1.0] …(1)5 10–3 = K[1.0] [1.0] …(2)1 10–2 = K[0.5] [2.0] …(3)
Dividing equation (1) and (2)
2
1
2
1
hence = 1Dividing equation (1) and (3)
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0.2
0.1
101
105.22
3
2
1
4
1
= 2
Problem 14 :For the first order reaction A(g) 2B(g) + C(g), the initial pressure is PA = 90 mm Hg,the pressure after 10 minutes is found to be 180 mm Hg. The rate constant of the reactionis :(A) 1.15 10–3 sec–1 (B) 2.3 10–3 sec–1
(C) 3.45 10–3 sec–1 (D) 6 10–3 sec–1
Solution : (A)A 2B + CP 0 0P – x 2x x
At equilibrium180 = P – x + 2x + x180 = 90 + 2x2x = 90, x = 45
xP
Plog
t
303.2K
=
4590
90log
10
303.2
=
10
6932.02log
10
303.2
= 0.6932 =60
06932.0 = 1.1555 10–3 sec–1
Daily Practice Problems 1
1. (D) 2. (C) 3. (B) 4. (B) 5. (A)
6.dt
]NO[ = 3.5 × 10–2 mol dm–3 s–1
dt
]OH[ 2 = 5.3 × 10–2 mol dm–3 s–1
7. 6.0 × 10–4 mol L–1 min–1 8. (a) 1.6 × 10–4 M/s ; (b) 3.2 × 10–4 M/s
9. (C) 10. (D) 11. (B) 12. (B) 13. (D) 14. (D)
15. (a) Rate = k[CH3Br] [OH–] (b) rate will decrease by factor of 5
(c) rate will increase by factor of 4
16. (a) k[Br–] [BrO3
–] [H+]2 (b) 4 (c) rate will increase by the factor of 9(d) rate will decrease by factor of 1/4
17. (i) 7.5 × 10–7 M min–1 (ii) rate increased by 64 times
18. order w.r.t. [NO] = 2 ; order w.r.t. [Cl2] = 1 19. 8 × 10–9 M–2 s–1
20. (i) 3/2 (ii) atm–1/2 time–1 (iii) increased by 64 times
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Daily Practice Problems 2
1. (B) 2. (C) 3. (B) 4. (C) 5. (B) 6. (A) 7 .(A) 8. (B) 9. (B) 10. (B) 11. (B)
12. (a) 2 N2O (g) 2 N
2(g) + O
2(g) (b) oxygen atom
(c) (i) unimolecular (ii) bimolecular (d) not defined
13. 3/2
15. (a) Rate = k[NH4
+] [NO2
–] (b) 3.0 × 10–4 M/s (c) 6.1 × 10–6 M/s
16. (i) r = k[A]–1 [B]1/2 (ii) wrt : A = –1, B = 1/2, C = 0 ,overall order = –1/2
(iii) k =3
63 × 10–9 (iv) rate of reaction increases by 2 2 times
17. (i) k[A2]3 [B
2] (ii) order wrt A
2 = 3 ; O. wrt B
2 = 1 , overall order = 4
(iii) k = 7.5 × 103 (mol/lit)–3/sec (iv) rate of reaction decrease to161
18. (a) 2 and 1 (b) r = K[Cl2]1[NO]
2(c) 8 litre2 mol–2sec–1 (d) 0.256 mol litre–1sec–1
19. 211 torr, 13.57 min 20. Zero order
Daily Practice Problems 3
2. (i) a = R0K, b =
2
KR 20 , c =
6
KR 30
2. (i) t1/2
= 66.66 min, k = 0.01039 min–1 (ii) k = 0.06386 min–1 ; t = 36.06 min(iii) 0.03465 min–1 (iv) (i) 0.740 s (ii) 2.005 s
4. 0.09 5. (a) 3 (b) No change
6. K= 2.5 × 10–4 sec–1, 7. (a) 46.2 minute (b) 107.3 minute
8. (a) 2.232 ×10–2 min–1 (b) 62.12 minute 9. (i) 12.5 %, (ii) 6.25 %, (iii) 1.56%
10. 15.86 % 11. Time at this point is t1/2
because [A] = [B]: Alsodt
]A[d +
dt
]B[d = 0
12. 5.25 %, 128.33 hr 14. 200 minute 15. 3.59 × 102 , 1.1 × 103 sec
16. 8.6× 10–4 sec; 13.43 minute 17. 0.023 mol/L 18. K =t
303.2 log
10 PP3
P2
0
0
19. 26 sec 20. 85.64 21. 1.58 ×10–2, 233 atm
22. 130.41 23. 'N)CH( 223
P = 127.55 mm , 'HC 62
P = 72.45 mm
24. (i) 6.93 × 10–2 M/min.(ii) 3.465 ×10–2 M/min (iii) 1.7325 × 10–2 M/min.(iv) 8.67 × 10–3
25. (i) 250 seconds (ii) 0.01386 M sec–1
26. (i) 2079 sec (ii) 2772 sec, (iii) 1386 sec. (iv) 3465 sec.
Daily Practice Problems 41. k = 2.0 × 10–2 min–1 3. I, 3.2 × 10–2 min–1 5. 108.23 ml
8. k = 3.27 × 10–3 min–1 9. k = 3.12 × 10–3 min–1 10. r = k [sugar]1 [H+]0
11. 6.2 × 10–4 sec–1
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CHEMICAL KINETICS
Daily Practice Problems 51. 0.537 2. K
b = 1.7 × 10–5 sec–1 K
c = 1.35 × 10–4 (order = 1 for each path)
3. methyl cyclopentene =23%, cyclohexane = 77% 4. 1990 sec.
5. 7.5937 6. 1.88 7. EA = 209.6 kJ 8. Large value of E
9. k = 3.19 × 10–2 L mol–1s–1
10. k = 5.10 × 10–5 s–1 ; E = 92.011 kJ/mol ; A = 6.73 × 1011 s–1
11. k = 10–2 s–1 ; E = 87.297 kJ/mol ; A = 2.2 × 1014 s–1
12. k = 5 × 10–4 s–1 ; E = 94.78 kJ/mol ; A = 1015 s–1
13. t = 19 sec, Ea = 67.6 kJ mol–1 14. E
af = 50.80 kJ/mol ; E
ab = 27.61 kJ/mol
15. 63.18 kJ/mol 16. 6.4 × 109 17. 3000 seconds 18. 0.002
19. 24.7 kJ/mol 20. T = 310K 21. Ans. 311.2 22. 4.768
23. 12 kcal 25. k = A, but it is not reasonable since Ea can not be zero
26. 2033.8 27. 86.1ºC 28. 124.46 mol–1 29. 2.1645 × 104
30. 349.04 K
Daily Practice Problems 61. 144 2. 214, 83 3. (i) AD (ii) BCD4. (a) i-ii-, (b) i-, ii- 5. 56. B 7. C 8. D 9. A,D10. C 11. B 12. 3.7 ×1010 13. Bi14. A 15. C, D 16. A, D 17. A, B, C18. 4n+2, 4n, 4n+3, 4n+1, 4n+1
19. (i) C126 (ii) H
11 (iii) (iv) Kr
9236 (v) n
10
(vi) As7633 (vii) 16 n
10 (viii) He
42
20. (i)5,5 (ii) 3,6 (iii) 11,14 (iv),2 (v) 2(vi) 3, (vii) 4,2 (viii) 7,4 (ix) 7,14 (x) 5,4
Daily Practice Problems 71. C 2. C 3. B 4. D 5. D6. A 7. A 8. B 9. C 10. D11. C 12. D 13. C 14. A 15. D16. D 17. B 18. C 19. A 20. B21. D 22. A 23. D 24. D 25. A26. D 27. B 28. A 29. B 30. C31. B 32. A 33. B 34. D 35. B36. D 37. D 38. A 39. D 40. B41. A 42. C 43. B 44. C 45. C
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