Add 9460377092 in Your Class Whatsapp Group to Get all ...formation of cuprous oxide and iron oxide. The role of silica (as a flux) in this process is to remove the iron oxide as slag

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Praganya Prakashan

CBSE Class-12th

Chemistry Model Paper 2

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General Instructions1. All questions are compulsory.2. Questions number 1 to 5 are very short answer questions and carry 1 mark each.3. Questions number 6 to 12 are short answer questions and carry 2 marks each.4. Questions number 13 to 24 are also short answer questions and carry 3 marks each.5. Questions number 25 to 27 are long answer questions and carry 5 marks each.6. Use log tables, if necessary. Use of calculators is not allowed.Time : 3 Hrs Max. Marks : 70

SECTION A

QUE 1.1

Children should take more protein rich diets. Why?

Ans :

Growing children should take adequate amount of protein rich diets, as protein is essential for their growth and development.

(1)

QUE 1.2

Iodine solids are not an electrical conductor. Why?

Ans :

Iodine is non-polar solid molecule, in which iodine molecules are held together by London force or dispersion force. Therefore, has no free electrons. (1)

QUE 1.3

Name the alkene which will yield 1-chloro-1-methycyclohexane by its reaction with HCl. Write the reaction.

Ans :

Two alkenes are possible. These are:

(1)

QUE 1.4

What happens when white phosphorus is heated with conc. NaOH solution in an insert atmosphere of CO2?

Ans :

P H O NaOH PH NaH PO3 3COPhosphine

4 2 3 2 22+ + +

3

(1)

QUE 1.5

Write the name of the product formed when anilinium hydrogen sulphate is heated at 453-473 K.

Ans :

When anilinium hydrogen sulphate is heated at 453-473K then sulphanilic acid is formed.

(1)

SECTION B

QUE 1.6

A mole of complex compound, ( )Co NH Cl3 5 3 gives 3 moles of ions, when dissolved in water. One mole of the same complex reacts with two moles of AgNO3 solution to form two moles of AgCl( )S . What is the structure of complex?

Ans :

According to the given statement, ( )Co NH Cl3 5 3 gives 3 ions.So, [ ( ) ] [ ( ) ]Co NH Cl Cl Co NH Cl Cl2

( ) ions3 5 2 3 5

1

2

3$ ++ − (1)

QUE 1.7

Find out the molar conductivity of an aqueous solution of BaCl2 at infinite when ionic conductances of Ba2+ and Cl− ions are

. S cm mol127 30 2 1− and . S cm mol76 34 2 1− , respectively.

Ans :

By applying Kohlrausch’s law,

( )BaClm 2Λ 2Ba cr2λ λ= ++ −

Given, Ba2λ + . S cm mol127 30 2 1= −

Clλ − . S cm mol76 34 2 1= −

( )BaClm 2Λ . 2S cm mol127 30 2 1= +−

. S cm mol76 34 2 1# −

. .S cm mol S cm mol127 30 152 682 1 2 1= +− −

. S cm mol279 98 2 1= − (2)

QUE 1.8

Predict the product of the following reaction.

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1.

2.

Ans :

1.

(2)

2.

QUE 1.9

Shivam and sudesh are two workers go to their work with a mask before going inside the coal mines. Why?

Ans :

Gas mask is made up of charcoal. Workers go to their work with a mask and before going inside the coal mines they wear the mask. They use the mask for their safety. Adsorption is the process involved in working function of mask. (2)

QUE 1.10

What happen when−1. SO2 is passed through an aqueous solution of Fe(III) salts?2. Iron reacts with HCl?

Ans :

1. SO2 acts as a reducing agent and hence reduces an aqueous solution of Fe (III) to Fe (II).

Fe SO H O Fe SO H2 2 2 4Ferric ion Ferrous ion

32 2

242$+ + + ++ + − + (1)

2. When iron reacts with HCl then it forms FeCl2 and H2 Fe HCl FeCl H2 2 2$+ + Liberation of hydrogen prevents the formation of ferric

chloride. (1)

OR QUE

Write the balanced chemical equations for the following.1. Reaction of silver nitrate with hypophosphorous acid in

aqueous medium.2. Concentrated H SO2 4 is added to sucrose ( )C H O12 22 11 .

Ans :

1. When silver nitrate reacts with hypophosphorous acid in acqueous medium, metallic silver is formed.

AgNO H O H PO Ag4 2 4Metallic silver

3 2 3 2 $+ +

HNO H PO4 3 3 4+ + (1)2. Concentrated sulphuric acid is a strong dehydrating agent.

Sulphuric acid removes water from organic compounds. C H O C H O12 11.

Sucrose

Conc H SO12 22 11 2

2 4 + (1)

QUE 1.11

Why is freezing point depression of . M NaCl0 1 solution is nearly twice than that of . M C H O0 1 6 12 6 solution?

Ans :

Colligative properties are directly proportional to the number of particles present in solution.NaCl being a strong electrolyte completely dissociate into ions. One molecule of NaCl dissociate into two ions. But glucose being non-electrolyte does not dissociate and hence, present as a single molecule that’s why, freezing point depression of 0.1M NaCl is nearly twice to that of 0.1M glucose ( )C H O6 12 6 solution. (2)

QUE 1.12

Gelatin which is a peptide is added in ice-creams. What can be its role?

Ans :

Emulsifying agent is added to stabilise the emulsion. Emulsifying agent form a layer between suspended particles and the medium and hence stabilises the emulsion. Ice cream (emulsion) is stabilised by emulsifying agent like gelatin. (2)

SECTION C

QUE 1.13

Explain the role of1. NaCN in the extraction of gold from gold ore.2. SiO2 in the extraction of copper from copper matte.3. iodine in the refining of zirconium.Also, write the chemical equation for the involved reactions.

Ans :

1. The roasted ore of gold is treated with a solution of sodium cyanide (NaCN) in the presene of oxygen for few days.

The role of NaCN in this process is to dissolve the gold to form an aurocyanide complex. The metal is obtained by displacement from this complex.

[ ( ) ]Au NaCN H O O Na Au CN4 8 2 4Aurocyanide complex

2 2 2$+ + +

NaOH4+ [ ( ) ] [ ( ) ]Na Au CN Zn Na Zn CN Au2 22 2 2$ .+ + (1)2. Copper matte contains Cu S2 and FeS. In the blast furnace,

copper matte is added with powdered coke and silica. Therefore, the oxidation of ore takes place resulting in the




formation of cuprous oxide and iron oxide. The role of silica (as a flux) in this process is to remove the iron oxide as slag. Thus, FeO combines with silica (flux) to form iron silicate, FeSiO3 (slag).

FeO SiO FeSiOFlux Slag

2 3+T (1)

3. In van-Arkel method for refining of zirconium, iodine role is to remove oxygen and nitrogen present in the form of impurities. The impure metal is heated in an empty vessel with iodine. The metal iodide formed will volatilises to give zirconium tetraiodide.

Zr I ZrI2Impure

2 4$+

Zirconium tetraiodide is then decomposed on a tungsten filament, at a temperature of 1800 K, to obtain pure metal.

ZrI Zr I2Pure

4 2+T (1)

QUE 1.14

1. Calculate the boiling point of one molar aqueous solution (density . g mL1 04 1= − ) of potassium chloride ( . )K K kg mol0 52b 1= − .

2. When two different solutions of sucrose of same molality prepared in different solvents, will they have the smae depressions is freezing point?

Ans :

1. Calculation of molality of the solution, Calculation of elevation in boiling point and Calculation of boiling point of solution Boiling point of the solution

Step I Calculation of molality of the solution1 molar solution means 1 g mole of KCl is dissolved in 1 L of the solution.

Mass of 1 L solution V d#=

( ) ( . )mL g mL1000 1 04 1#= −

g1040=

Mass of KCl (1mol) .g g39 35 5= + . g74 5=

Mass of water .g g1040 74 5= −

. g965 5= . kg0 9655=

Molality of solution ( )Mass of solvent kg

Number of moles of solute=

( . )

( )kg

mol0 9655

1= . mol kg1 0357 1= −

Step II Calculation of elevation in boiling point ( )TbTThe chemical equation of dissociation of KCl in its aqueous solution is ( ) ( ) ( )KCl s K aq Cl aq

( )aq++ −

van’t Hoff factor, i Number of particles originally presentNumber of particles after dissociation=

12= 2=

Kb . K kg mol0 52 1= −

m . mol kg1 0357 1= −

i 2= ,

TbT i K mb# #=

( ) ( . )K kg mol2 0 52 1#= −

( . )mol kg1 0357 1# −

. K1 077= ( )11 2

Step III Calculation of boiling point of solutionBoiling point of the solution

Tb T Tob bT= + .K K373 1 077= +

. K374 077=2. According to the definition of depression in freezing point,

TfT k mf= where,Kf = freezing point depression constant. Value of

Kf depends upon nature of solvent. Hence, despite having same molality, two solutions of sucrose, prepared in different solvents, will have different depression in freezing point.

( )11 2

QUE 1.15

A strip of nickel metal is dipped in a 1 molar solution of ( )Ni NO3 2 and a strip of silver metal is dipped in a 1 molar

solution of AgNO3. An electrochemical cell is created when the two solutions are joined by salt bridge and two strips are joined by wire to a voltmeter.Answer the following questions.1. Write the balanced equation for the overall cell reaction

and calculate the cell potential.2. Calculate the cell potential ( )Ecell at C25c for the cell if the

initial concentration of ( )Ni NO3 2 is 0.100 molar and the initial concentration of AgNO3 is 1.00 molar.

[ . VE 0 25/Ni Ni2 =−+ ; . VE 0 80/Ag Ag =+ ; ]log 10 11 =−−

Ans :

1. The half-cell reaction At anode

( )Ni s ( )Ni M e1 22$ ++ −

At cathode

( )Ag M e2 1 2++ − ( )Ag s2$ Overall reaction

Ni Ag2+ + Ni Ag22$ ++

Number of electrons (n) 2=

Ecell E Ecathode anode= −

. ( . )V V0 80 0 25= − −

Ecell . V1 05= ( )11 2 By using Nernst equation,

Ecell .

[ ][ ]

logAgNi

E n0 0591

cell 2

2

= − ++

Ecell .. ( )log1 05 2

0 0591 1= −

. .1 05 20 0591 0#= −

Ecell . V1 05=2. Given,

[ ]Ni2+ . ,M0 100= [ ]Ag+ . M1 00=

Ecell .

[ ][ ]

logn AgNi

E 0 0591cello 22

= − ++

. .( . )( . )

log1 05 20 0591

1 00 1

2= −

. . ( )1 05 0 0295 1= − −

. .1 05 0 0295= +




Ecell . V1 08= ( )11 2

QUE 1.16

Explain the following behaviours1. Noble gases have very low melting and boiling points.2. Neon is used in illuminating warning signals.3. ICl is more reactive than I2.

Ans :

1. Noble gas elements exist as monoatomic gases and the only attractive force in between them are the weak van der Waal’s forces. As a result, the melting and boiling points of these elements are very low. (1)

2. Neon lights (generally orange-red) are visible from long distance and can be seen even in foggy weather and mist. Hence, for illuminating warning signals, neon is generally used. (1)

3. ICl is polar ( )I Clδ δ+ − due to the electronegativity difference between the two participating halogen atoms. It is therefore, more reactive than I2 which is completely non-polar. (1)

QUE 1.17

1. Give the electronic configuration of [ ( ) ]Fe CN 6 4− on the basis of crystal field splitting theory.

2. Arrange the following complexes in the increasing order of conductivity of their solution.

[ ( ) ]Co NH3 3 , [ ( ) ]Co NH Cl Cl3 4 2 [ ( ) ]Co NH Cl3 6 3, [ ( ) ]Cr NH Cl Cl3 5 23. What type of isomerism shown by the following complexes? [ ( ) ]Co NH NO3 5 2 2+, [ ( ) ]Co NH SCN3 5 2+

Ans :

1. In [ ( ) ]Fe CN 6 4− complex, iron has +2 oxidation state. The cyanide ion ( )CN− being strong field ligand pair up the t g2 electrons before filling eg set.

Fe2+ ( )d t eg g6 26 0=

Hence, the configuration is t eg g26 0. (1)2. Ions or molecules present outside the coordination sphere are

ionisable. A complex which gives more ions on dissolution is more conducting.

[ ( ) ] [ ( ) ] [ ( ) ]Co NH Cl Co NH Cl Cl Cr NH Cl Cl< <( ) ( ) ( )ion ions ions

3 3 31

3 4 22

3 5 23

[ ( ) ]Co NH Cl<( )ions

3 64

3

(1) As the number of ions increases conductivity also increases.3. Coordination compounds containing a ligand with more

than one non-equivalent binding position (ambident ligand) show linkage isomerism. The complex [ ( ) ]Co NH NO3 5 2 2+ contains NO2 which has two donor sites N and O which is shown by arrow ( )" as:

The complex [ ( ) ]Co NH SCN3 5 2+ contains SCN which has two different donor sites S and N which is shown by arrow ( )" as : S C N" !/−

Hence, both [ ( ) ]Co NH NO3 5 2 2+ and [ ( ) ]Co NH SCN3 5 2+ shows linkage isomerism. (1)

QUE 1.18

Ramu, a domestic helper of Mr. Pankaj, fainted while mopping the floor. Mr. Pankaj immediately took him to a nearby hospital where, he was diagnosed with pernicious anaemia. The doctor prescribed iron rich diets and multivitamins supplements to him. Mr. Pankaj helped him financially to get the medicines. After some time, Ramu was diagnosed to be normal.Answer the following questions.1. Name the vitamin whose deficiency causes pernicious

anaemia.2. Write down the components of iron rich diet.3. What values are shown by Mr. Pankaj?

Ans :

1. Vitamin B12 deficiency causes pernicious anaemia. (1)2. Meat, fish, egg and curd are the iron rich diets. (1)3. Mr. Pankaj is kind hearted man and has helping nature. (1)

QUE 1.19

1. Which of the following compounds will react faster in SN 1 reaction with the hydroxide ( )OH− ion?

CH CH Cl or C H CH Cl3 2 6 5 2− − − −2. Out of o and p-dibromobenzeme, which one has higher

melting point and why?3. Arrange the following compounds in the increasing order of

their densities.

Ans :

1. S 1N mechanism depends upon the stability of carbocation that is formed as intermediate in the mechanism.

C H CH Cl6 5 2− − will form C H CH6 5 2−+

carbocation as an intermediate.

This carbocation is resonance stabilised and will react faster in S 1N reaction.




While carbocation formed in CH CH Cl3 2 is CH CH3 25

which is not resonance stabilised. Hence, this carbocation is highly unstable and not give S 1N reaction with OH− ion but relatively at slower rate. (1)

2.

p-dibromobenzene has higher melting point than its o-isomer due to symmetry. Due to symmetry, p-isomer fits in the crystal lattice better than o-isomer. Hence, p-dibromobenzene has higher melting point. (1)

3. Density is directly related to molecular mass. Higher the molecular mass, higher will be the density of the compound. Among the four given compounds, the order of molecular mass is Benzene (a)< chlorobenzene (b)< dichlorobenzene (c) < bromochlorobenzene (d).

Therefore, increasing order of their densities are same as above. (1)

QUE 1.20

Chromium metal crystallises in a body centerd cubic lattice. The length of the unit cell edge is 287 pm. Evaluate the atomic radius of chromium.

Ans :

In body centred cubic lattice, r a43=

where, r and a are radius and edge length, respectively.

Given, Edge length, a pm287=

Hence, r pm43 287#=

. 41 73 287#= . pm124 1275=

(3)

QUE 1.21

1. Give one chemical test to distinguish between the following pairs of compounds.

(a) Ethyl amine and aniline(b) Aniline and N-methylaniline

2. How will you convert aniline to 2, 4, 6-tribromoflurorbenzene?

Ans :

1.(a) Ethyl amine and aniline Ethyl amine ( )CH CH NH3 2 2 is primary ( )1c aliphatic

amine and aniline is 1c aromatic amine. They can be distinguished by azo dye test.

CH CH OH N H O CH CH OH3 2 2 2 3 2- $+ + ( )N H O No orange dye2 2-+ +

(1)(b) Aniline and N-methylaniline Aniline is 1c aromatic amine, gives carbylamine test,

whereas N-methylaniline is 2c aromatic amine and does not give carbylamine test but gives Liebermann’s nitroso reaction. (1)

2. Anilie to 2, 4, 6-tribromofluorobenzene




(1)

QUE 1.22

1. Explain the following reactions.(a) Stephen reaction(b) Etard reacton

2. Arrange the following compounds in the increasing order of their acidic strength.

( ) , ( ) ,CH CHCOOH CH CH CH Br COOH3 2 3 2 ( )CH CH Br CH COOH3 2

Ans :

1. (a) Stephen reactionm : In this reaction nitriles are reduced to correspondin imine with stannous chloride in the presence of hydrochloric acid, which on hydrolysis gives corresponding aldehyde.

RCN SnCl HCl RCH NH RCHOminAlkylcyanide a e

H O2

3$+ + =+

(b) Etard reaction : In this reaction chromyl chloride oxidises methyl group of toluene to a chromium complex, which on hydrolysis gives corresponding benzaldehyde.

( )11 22. The increasing order of acidic strength is ( ) ( )CH CHCOOH CH CH Br CH COOH



1. Terylene is a polyester which is used as synthetic fibre. (1)2. It is a condensation polymer of ethylene glycol and

terephthalic acid. (1)

3. Condensation polymerisation takes place during the

formation of the polyster from terephthalic acid and ethylene glycol. (1)

QUE 1.24

1. Why would tranquiliser not be taken regularly for very long period.

2. Why medicines should not be taken without consulting doctors.

3. Why do we require artificial sweetening agents.

Ans :

1. It is because body becomes habitual of these drugs. (1)2. If doses taken higher than recommended may cause harmful

effect and act as poison. Thus, a doctor should always be consulted before taking medicine. (1)

3. (a) To control intake of calories. (b) As a substitute of cane sugar for diabetic patients. (1)

SECTION D

QUE 1.25

1. How many alcohols with molecular formula C H O4 10 are chiral in nature?

2. How are the following conversions carried out?(a) Phenol to acetophenone(b) Ethanol to methanol

3. Give the structure and the IUPAC name of the major product obtained in the following reaction

Ans :

1. Three isomers of butanol are possibel.(a) CH CH CH CH OH

tanBu ol3 2 2

12− −

− −

No chiral carbon in this compound as none of the four carbons is bonded to four different substituents.

(b)

This compound have one chiral carbon.(c)

Here again no carbon is chiral in nature. So, only one alcohol is chiral in nature. (2)

2. (a) Phenol to acetophenone

(b) Ethanol to methanol

(2)3.




(1)

OR QUE

1. Write the mechanism of the reaction of acid ( )H SO2 4 with ethene.

2. Identify A and B in the following reactions.

(a)

(b)

3. Give the Structure and IUPAC name of the major product obtained in the following reaction.

Ans :

1. The complete reaction is

CHEthene

2 = CH H O CH CH OHH

Ethanol2 2 3 2?+ − −

+

Mechanism Step I Protonation of alkene to form carbocation by

electrophilic attack of H O3 +. H O H H O2 3$+ + +

Step II Nucleophilic attack on carbocation.

Step III Deprotonation to form an alcohol.

(2)

2. (a)

(b)

(2)

3.

(1)

QUE 1.26

1. A violet compound of manganese W decomposes on heating to liberate oxygen and compounds X and Y of manganese are formed. Compound Y reacts with KOH in the presence of oxygen to give compound X .

2. Transition elements exhibit variable oxidation states. Explain, why?

Ans :

1. Since, compound Y on reacting with conc. H SO2 4 and




NaCl gives Cl2 gas, so it is manganese diodide ( )MnO2 (violet) is heated. (1)

Thus, W KMnO4=

X K MnO2 4=

Y MnO2= ,

Z MnCl2=

KMnO2( )W

4 K MnO MnO O( ) ( )X Y

2 4 2 2+ +T

MnO KOH O2 4( )Y

2 2+ + K MnO H O2 2( )X

2 4 2$ +

4MnO NaCl H SO4( )Y

2 2 4+ + MnCl NaHSO4( )Z

2 4$ +

H O Cl2 2 2+ + (2)2. Transition metals have incompletely filled d-orbitals. As a

result, they exhibit variable oxidation states. (2)

OR QUE

Give reasons for the following.1. Iron is a transition metal sodium is not.2. Cd2+ salts are white.3. Zn, Cd and Hg are quite soft and have low meting points.4. ( )La OH 3 is more basic than ( )Lu OH 3.5. Third inonisation enthalpy of manganese is exceptionall

high.

Ans :

1. This is because iron has unpaired electrons in the d-subshell while sodium does not have. (1)

2. This is because Cd2+ ion has completely filled d-orbitals ( ) .d4 10 Hence, in the absence of d-d transition, they show only white colour. (1)

3. Zn, Cd, Hg metals have completely filled d-orbitals ( ) .d 10

It means that d-electrons are not readily available for metallic bond formation. Since, the metallic bonds are weak. Therefore, these metals are quite soft and also have low melting points. (1)

4. ( )Lu OH 3 has greater covalent character than ( )La OH 3 on account of lanthoid contraction. Thus, the liberation of hydroxide ions from ( )Lu OH 3 is difficult and it is less basic than ( )La OH 3. (1)

5. Mn2+ has the configuration, [ ]Ar d3 5 with highly symmetrical configuration. Hence, the removal of third electron is very difficult. Therefore, third ionisation enthalpy of the metal Mn is exceptionally high. (1)

QUE 1.27

1. Consider a first order gas phase decomposition reaction given below:

( ) ( ) ( )X g Y g Z g$ + The initial pressure of the system before decomposition of

X was pi. After lapse of time t , total pressure of the system increased by a units and became px, What will be the rate constant ( )k for the reaction?

2. In a reaction X and Y , the initial rate of reaction ( )r0 was measured for different initial concentration of X and Y as given below:

/mol LX 1− 0.20 0.20 0.40

/mol LY 1− 0.30 0.10 0.05

/mol L sr0 1 1− − .5 07 10 5# − .5 07 10 5# − .1 43 10 4# −

What is the order of the reaction with respect to X and ?Y

Ans :

1. ( )gX ( )gY$ + ( )gZ Initially Pi 0 0 At time t , aPi − a a

Pt a a aPi= − + + aPi= − For first order reaction, a P Pt i= −

k . logt aPP2 303

i

i= − .

( )logt P P P

P2 303i t i

i= − −

. logt P PP2 303

2 i ti= − ( )1

12

2. Rate law states that,

Rate [ ] [ ]k X Ya b=

( )Rate 1 [ . ] [ . ]k 0 20 0 30a b= .5 07 10 5#= − ...(i)

( )Rate 2 [ . ] [ . ]k 0 20 0 10a b= .5 07 10 5#= − ...(ii)

( )Rate 3 [ . ] [ . ]k 0 40 0 05a b= .1 43 10 4#= − ...(iii) Dividing Eq. (i) by Eq. (ii)

( )( )RateRate

2

1 [ . ] [ . ][ . ] [ . ]

kk

0 20 0 100 20 0 30

a b

a b

= ..

5 07 105 07 10

5

5

##= −

− [ ]1 3 b= [ ]3 0=

Hence,b 0= (1) Now, dividing Eq. (iii) by Eq. (ii)

( )( )RateRate

2

3 [ . ] [ . ][ . ] [ . ]

kk

0 20 0 100 40 0 05

a

a

0

0

= ..

5 07 101 43 10

5

4

##= −

−

2a= .2 8205= Taking log on both sides,

.log 2 8205 loga 2= & a ..

0 30100 4503= .1 5=

Thus, order of the reaction with respect to X and Y is 1.5 and 0, respectively. ( )21 2

OR QUE

1. According to Arrhenius equation, rate constant k is equal

to Ae /E RTa− . Plot the graph of In k versus T1 .

2. For a first order reaction, show that time required for %99 completion is twice the time required for the completion of

%90 of reaction.

Ans :

1. According to Arrhenius equation, k Ae /E RTa= −

Taking log on both sides, In k ( )In Ae RTEa

=−

In k In A RTEa= −

& In k RE

T In A1a #=− +

Y mx c= + This equation can be related to equation of straight line as

shown below:




From the graph, it is very clear that slope of the plot REa= −

and intercept In A= . ( )21 2(ii) Case I

If a 100= ,( )a x− ( )100 99= − 1= For 99% completion of the reaction,

t %99 . logk

2 3031

100= . logk2 303 102= . k

2 303 2#=

t %99 .k

4 606= ...(i)

Case II

If a 100=

( )a x− ( )100 90= − 10= For 90% completion of the reaction

t %90 . logk

2 30310100=

. logk2 303 10= .k

2 303= ...(ii)

On dividing Eq. (i) by Eq. (ii), we get

tt

%

%

90

99 . .kk4 606

2 303#= 2=

It means that time required for 99% completion of the reaction is twice the time required to complete 90% of the reaction.

( )21 2


Documents

Add 9460377092 in Your Class Whatsapp Group to Get all ...formation of cuprous oxide and iron oxide. The role of silica (as a flux) in this process is to remove the iron oxide as slag