Chemical Kinetic Note 03

8/3/2019 Chemical Kinetic Note 03

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S S C 2 4 5 3 –

C H E M I C A L K I N E T I C S A N DE L E C T R O C H E M I S T R Y

CHEMICAL KINETICS



The concentrations of N2O5 are 1.24e-2 and 0.93e-2 M at

600 and 1200 s after the reactants are mixed at the

appropriate temperature. Evaluate the reaction rates for2 N2O5 4 NO2 + O2



3

DIFFERENTIAL RATE LAWS

Dependence of reaction rate on the concentrations of reactants is called the rate law ,

which is unique for each reaction.

For a general reaction, a A + b B + c C products

the rate law has the general form

order wrt A, B, and C , determined experimentally reaction rate = k [ A ]X [ B ]Y [ C ]Z

the rate constant

For example, the rate law is

rate = k [Br- ] [BrO3- ] [H+ ] for

5 Br- + BrO3- + 6 H+ 3Br2 + 3 H2O

The reaction is 1st order wrt all three reactants, total order 3.

Use differentials to

express rates



4

VARIATION OF REACTION RATES AND

ORDER

First order, rate = k [A]

k = rate , 0th order

[A]

rate

2nd order, rate = k [A]2

The variation of reaction rates as functions of concentration for various order is

interesting.

Mathematical analysis is an important scientific tool, worth noticing.

[A] = ___?



5

DIFFERENTIAL RATE LAW

DETERMINATION

Estimate the orders and rate constant k from the results observed for

the reaction? What is the rate when [H2O2 ] = [I- ] = [H+ ] = 1.0 M?

H2O

2+ 3 I- + 2 H+ I

3- + 2 H

2O

Exprmt [H2O2 ] [I- ] [H+ ] Initial rate M s-1

1 0.010 0.010 0.0050 1.15e-6

2 0.020 0.010 0.0050 2.30e-6

3 0.010 0.020 0.0050 2.30e-6

4 0.010 0.010 0.0100 1.15e-6

Learn the strategy to determine the rate law from this example.

Figure out the answer without writing down anything.



6

DIFFERENTIAL RATE LAW DETERMINATIONEstimate the orders from the results observed for the reaction

H2O

2+ 3 I- + 2 H+ I

3- + 2 H

2O

Exprmt [H2O2 ] [I- ] [H+ ] Initial rate M s-1

1 0.010 0.010 0.0050 1.15e-6

2 0.020 0.010 0.0050 2.30e-6 1 for H2O2 3 0.010 0.020 0.0050 2.30e-6 1 for I-

4 0.010 0.010 0.0100 1.15e-6 0 for H+

1.15e-6 = k [H2O2 ]x [I- ]y [H+ ]z

1.15e-6 k (0.010)x(0.010)y (0.0050)z exprmt 1 1 1

----------- = ------------------------------------- ---- = ---

2.30e-6 k (0.020)x(0.010)y (0.0050)z exprmt 2 2 2

x = 1

( )x



DIFFERENTIAL RATE LAW DETERMINATION

Other orders are determined in a similar way as shown before.

Now, lets find k and the rate

Thurs, rate = 1.15e-6 = k (0.010)(0.010) from exprmt 1

k = 1.15e-6 M s-1 / (0.010)(0.010) M3 = 0.0115 M-1 s-1

And the rate law is therefore,

– d [H2O2 ] k

rate = ————— = 0.0115 [H2O2 ] [I- ] a differential rate law

d t total order 2

The rate when [H2O2 ] = [I- ] = [H+ ] = 1.0 M:

The rate is the same as the rate constant k, when concentrations of

reactants are all unity (exactly 1), doesn’t matter what the orders are.



15 Chemical Kinetics 8


The reaction rate – d [H2O2 ]/dt = 0.0115 [H2O2 ] [I – ],

for

H2O2 + 3 I- + 2 H+ I3- + 2 H2O

What is – d [I – ]/dt when [H2O2 ] = [I – ] = 0.5?

Note the reaction rate expression and the stoichiometry of equation.





The reaction rate – d [H2O2 ]/dt = 0.0115 [H2O2 ] [I – ], for

H2O2 + 3 I- + 2 H+ I3- + 2 H2O

What is – d [I – ]/dt when [H2O2 ] = [I – ] = 0.5?

Solution : Please note the stoichiometry of equation and how the rate changes. – d [I – ]/dt = – 3 d [H2O2 ]/dt = 3* 0.0115 [H2O2 ] [I

– ]

= 0.0345 * 0.5 * 0.5

= 0.0086 M s-1

In order to get a unique rate constant k, we evaluate k for the reaction

a A + b B productthis way

rate = -1/a d [A]/dt = -1/b d [B]/dt = k [A]x [B]y

Note the reaction rate expression and the stoichiometry of equation.



10


From the following reaction rates observed in 4 experiments, derive

the rate law for the reaction

A + B + C products where reaction rates are measured as soon as the reactants are mixed.

Expt 1 2 3 4

[A]o 0.100 0.200 0.200 0.100

[B]o 0.100 0.100 0.300 0.100

[C]o 0.100 0.100 0.100 0.400 rate 0.100 0.800 7.200 0.400

This example illustrates the strategy to determine, and a reliable method

to solve rate-law experimentally.



11


From the following reaction rates , derive the rate law for the reaction

A + B + C

products where reaction rates are measured as soon as the reactants are mixed.

Expt 1 2 3 4 order

[A]o 0.100 0.200 0.200 0.100 3 from expt 1 & 2

[B]o 0.100 0.100 0.300 0.100 2 expt 1, 2 & 3 [C]o 0.100 0.100 0.100 0.400 1 expt 1 & 4

rate 0.100 0.800 7.200 0.400

0.800 k 0.2x 0.1 y 0.1z ----- = ----------------------

0.100 k 0.1x 0.1 y 0.1z Therefore 8 = 2x

log 8 = x log 2x = log 8 / log 2

= 3

Assume rate = k [A]x [B] y [C]z




INTEGRATED RATE LAWSCONCENTRATIONS AS FUNCTIONS OF TIME

One reactant A decomposes in 1st or 2nd order rate law.

Differential rate law Integrated rate law

– d[A] / dt = k [A] = [A]o – k t

d [A]

– —— = k [A] [A] = [A]o e – k t or ln [A] = ln [A]o – k t

d t

d [A] 1 1 [A] conc at t – —— = k [A]2 —— – —— = k t

d t [A] [A]o [A]o conc at t = 0

Describe, derive and apply the integrated rate laws

Learn the strategy to determine rate-law



13

CONCENTRATION AND TIME OF 1ST

ORDER REACTION

Describe the features of plot of [A] vs. t and ln[A] vs. t for 1st order reactions. Apply the technique to evaluate k or [A] at

various times.[A]

t

ln[A]

t

[A] = [A]o e – k t ln [A] = ln [A]o

– k t

t½



14

HALF LIFE & K OF FIRST ORDER

DECOMPOSITION

The time required for half of A to decompose is called half life t 1/2.

Since [A] = [A]o e – k t or ln [A]

= ln [A]o – k t

When t = t 1/2, [A] = ½ [A]o Thus ln ½ [A]o = ln [A]o – k t 1/2

– ln 2 = – k t 1/2

k t 1/2 = ln 2 = 0.693 relationship between k and t 1/2

Radioactive decay usually follow 1st order kinetics, and half life of an

isotope is used to indicate its stability.

Evaluate t½ from k or k from t½



15

1ST ORDER REACTION CALCULATION

N2O5 decomposes according to 1st order kinetics, and 10% of itdecomposed in 30 s. Estimate k, t½ and percent decomposed in 500 s.

Solution next

If the rate-law is known, what are the key parameters?




1ST ORDER REACTION CALCULATION

N2O5 decomposes according to 1st order kinetics, and 10% of it decomposedin 30 s. Estimate k, t ½ and percent decomposed in 500 s.

Solution: Assume [A]o = [N2O5]o = 1.0, then [A] = 0.9 at t = 30 s or

0.9 = 1.0 e – k t apply [A]o = [A] e – k t

ln 0.9 = ln 1.0 –

k 30 s

– 0.1054 = 0 – k * 30

k = 0.00351 s – 1

t ½ = 0.693 / k = 197 s apply k t ½ = ln 2

[A] = 1.0 e – 0.00351*500 = 0.173

Percent decomposed: 1.0 – 0.173 = 0.827 or 82.7 %

After 2 t ½ (2*197=394 s), [A] = (½)2 =¼, 75% decomposed.

After 3 t ½ (3*197=591 s), [A] = (½)3 =1/8, 87.5% decomposed.

Apply integrated rate law to solve problems



17

TYPICAL PROBLEM : 1ST ORDER

REACTION

The decomposition of A is first order, and [A] is monitored. The following

data are recorded:

t / min 0 2 4 8[A]/[M] 0.100 0.0905 0.0819 0.0670

Calculate k (What is the rate constant?

Calculate the half life (What is the half life?

Calculate [A] when t = 5 min.Calculate t when [A] = 0.0100. (Estimate the time required for 90% of A

to decompose.)

Work out all the answers



18

TYPICAL PROBLEM WRT 1ST ORDER

REACTION

The decomposition of A is first order, and [A] is monitored. The following

data are recorded:

t / min 0 2 4 8[A]/[M] 0.100 0.0905 0.0819 0.0670

Calculate k (What is the rate constant? k = 0.0499)

Calculate the half life (What is the half life? Half life = 13.89)

Calculate [A] when t = 5 min. (What is the concentration when t = 5 min?)Calculate t when [A] = 0.0100.

Work out all the answers



19

A 2ND ORDER EXAMPLE

Dimerization of butadiene is second order:

2 C4H6(g) C8H12(g).

The rate constant k at some temperature is 0.100 /min. The initial concentration of

butadiene [B] is 2.0 M.

Calculate the time required for [B] = 1.0 and 0.5 M

Calculate concentration of butadiene when t = 1, 5, 10, and 30.

Solution next

If the rate-law is known, what are the key parameters?

Apply the right model and work out all the parameters.



20

A 2ND ORDER

EXAMPLE

Dimerization of butadiene is second order:

2 C4H6(g) = C8H12(g).

The rate constant k at some temperature is

0.100 /min. The initial concentration of butadiene [B] is 2.0 M.

Calculate the time t required for [B] = 1.0 and

0.5 M

Calculate concentration of butadiene when t

= 1, 5, 10, and 30.

1 1

—— – —— = k t

[B] [B]o

1 1—— – ——

[B] [B]ot = ————————

k [B]o

[B] = ——————

[B]o k t + 1

t = 1 5 10 15 30 35

[B] = 1.671.0

0.670.50

0.290.25

Work out the formulas and then

evaluate values




HALF LIFE OF 2ND ORDER CHEMICAL

KINETICS

t = 1 5 10 15 30 35

[B] = 1.67 1.0 0.67 0.50 0.29 0.25

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

1/[B]

How does half life vary in 2nd order reactions?




PLOT OF [B] VS. T & 1/[B] VS. T FOR 2ND

ORDER REACTIONS

t = 1 5 10 15 30 35

[B] = 1.67 1.0 0.67 0.50 0.29 0.25[B]1 — [B]

t t

1 1

—— – —— = k t

[B] [B]o

[B]o [B] = ——————

[B]o k t + 1

What kind of plot is linear for 1st and 2nd reactions?



EXERCISES

The decomposition of N2O5 is an important process in troposphericchemistry. The half-life for the first order decomposition of thiscompound is 2.05 x 104 s.

How long will it take for a sample of N2O5 to decay to 60% of its initialvalue?



EXERCISES

Carbon-14 is a radioactive nucleus with a half-life of 5760 years. Livingmatter exchanges carbon with its surroundings (for example, throughCO2) so that a constant level of 14C is maintained, corresponding to15.3 decay events per minute. Once living matter has died, carbon

contained in the matter is not exchanged with the surroundings, andthe amount of 14C that remains in the dead material decreases withtime due to radioactive decay. Consider a piece of fossilized woodthat demonstrates 2.4 14C decay per minute.

How old is the wood?



Consider the experiment where the concentrationof products are given in the following table:

Time (minutes): 0 4 6 10 15 20(a-x) (M): 8.04 5.30 4.58 3.50 2.74 2.22

(i) Prove that the reaction is second order (ii) Determine the rate constant

EXERCISES

Assume (a-x) = [A]t



Consider the dimerization reaction: 2C4H6(g) C8H12(g)The concentration of C4H6 at time intervals, t was determined as follows:

[C4H6] (M) Time (s)0.01000 0

0.00625 10000.00476 18000.00370 28000.00313 36000.00270 44000.00241 5200

0.00208 6200(i) Using graphical method, determine the order of the reaction(ii) What is the value of the rate constant?(iii) What is the half-life of the reaction?

EXERCISES



METHOD OF INITIAL RATES

Establishing the Order of a reaction by the Method of Initial Rates. Use the data provided establish the order of the reaction with respectto HgCl2 and C2O2

2- and also the overall order of the reaction



Notice that concentration changes between reactions are by a factor of 2.Write and take ratios of rate laws taking this into account.

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Chemical Kinetic Note 03