8-3-2011

It is clear from the first two experiments that when the concentration of O3 was doubled, the rate was doubled as well. Therefore, the reaction is first order with respect to O3

From experiments 2 and 3 keeping the concentration of O3 constant at 2.0x10-5 M, decreasing the concentration of NO2 by one-half results in a decrease of the initial rate by the same value. The reaction is therefore first order with respect to NO2

The rate law can be written as:

Rate = k [NO2][O3]

The rate constant can be calculated by substitution for the concentrations of reactants and corresponding rate value, taking the first experiment, for instance, will give:

0.022 mol L-1s-1 = k * 5.0x10-5 (mol L-1)* 1.0x10-5 (mol L-1)

k = 4.4x107 L mol-1 s-1

Example

In the reaction:

2 NOCl 2 NO + Cl2

The following data was collected at 27 oC:

What is the rate law of the reaction?

What is the value of the rate constant?

Determination of Rate Law

Experiment[F2][ClO2]Rate (M/s)

10.040.031.0x10-2

20.040.041.3x10-2

30.020.046.7x10-3

40.040.062.0x10-2

F2 (g) + 2ClO2 (g) 2FClO2 (g)

Determination of Rate Law

Experiments 1 & 4

As ]F2[ doubles, so does the rate

Experiments 2 & 3

As ]ClO2[ doubles, so does the rate

2:2 ratio…..1:1 ratio

x = 1 and y = 1

Rate = k ]F2[ ]ClO2[

Experiment]S2O82-[]I-[

Initial Rate (M/s)

10.080.0342.2 x 10-4

20.080.0171.1 x 10-4

30.160.0172.2 x 10-4

Determine the rate law and calculate the rate constant for the following reaction from the following data:S2O8

2- (aq) + 3I- (aq) 2SO42- (aq) + I3

- (aq)

rate = k ]S2O82-[x]I-[y

Double ]I-[, rate doubles (experiment 1 & 2)

y = 1

Double ]S2O82-[, rate doubles (experiment 2 & 3)

x = 1

k = rate[S2O8

2-][I-]= 2.2 x 10-4 M/s

(0.08 M)(0.034 M)= 0.08/M•s

rate = k [S2O82-][I-]

Relation between Reactant Concentration and Time

First Order Reaction- a reaction whose rate depends on the reactant concentration raised to the first power.

Reaction Type: A→B

Rate of: -Δ ]A[/Δt or k]A[

Combining and simplifying these equations brings us to the following rate equation:

ln[A]t = -kt + ln[A0]

Relation between Reactant Concentration and Time

Reaction Time

The reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88M to 0.14M?

ln]A[ = ln]A[0 - kt

kt = ln]A[0 – ln]A[

t = ln]A[0 – ln]A[

k= 66 s

]A[0 = 0.88 M]A[ = 0.14 M

ln]A[0

]A[k

=

ln0.88 M0.14 M

2.8 x 10-2 s-1=

Example

At 400 oC, the first order conversion of cyclopropane into propylene has a rate constant of 1.16x10-6 s-1. What will the concentration of cyclopropane be after 24 hours of reaction if the initial concentration was 0.0100 M.

Cyclopropane propylene

ln [cyclopropane]0/[cyclopropane]t = kt

ln 0.0100/[cyclopropane]t = 1.16x10-6 s-1 * (24 hr *3600 s/hr)

ln (0.0100 M)/[cyclopropane]t = 0.100

0.0100/[cyclopropane]t = e0.100

[cyclopropane]t = 9.01x10-3 M

The isomerization reaction CH3NC CH3CN

obeys the first order rate law rate = k ]CH3NC[

Measurements at 500 K reveal that in 520 seconds the concentration of CH3NC has decreased to 71% of its original value. Calculate the rate constant k of the reaction at 500 K.

ln[At] = -kt + ln[A0]

It is important to remember that for first order reactions, one does not need to know the absolute values of [At] and [A0] to determine the rate constant (if time is known) or the time (if the rate constant is given).

One can work with relative values since the above equation can be recast in the following way: ln([At]/[A0]) = -kt

Since the relationship between [At] and [A0] is known then the problem can be solved.

The problem states 71% of A is left after 520 seconds:

ln([At]/[A0]) = -kt

ln(0.71[A0]/[A0) = -k (520 seconds)

k = 0.00066 s-1

Reaction Half-life

As a reaction proceeds, the concentrations of the reactants decreases.

Another way to measure ]reactant[ over time is to use the half-life.

Half-life, t1/2 – the time required for the concentration of a reactant to decrease to half of its initial concentration.

Reaction Half-life

Considering the condition where one-half of A is consumed ([A]t = ½ [A]0), the time can be denoted as t½;

ln [A]0 / ½ [A]0 = kt½

ln 2 = kt½

t½ = (ln 2)/k

t½ = 0.693/k

A product

First-order reaction

# of half-lives [A] = [A]0/n

1

2

3

4

2

4

8

16

13.3

Reaction Half-life

The decomposition of RX has a rate law: rate = k[RX]. If, at 550 oC, k = 0.032 s-1, find:

1. The half life of the reaction

2. The initial concentration of RX provided that the concentration of RX after one min of reaction was 0.010 M.

a. Calculation of the half life of the reaction

t½ = (ln 2)/k

t½ = 0.693/0.032 s-1

t½ = 22 s

b. ln [A]0/[A]t = kt

ln [RX]0/(0.010 M) = 0.032 s-1 * (1 min * 60 s/min)

ln [RX]0/(0.010 M) = 1.92

[RX]0 = 0.068 M

Reaction Half-lifeWhat is the half-life of N2O5 if it

decomposes with a rate constant of 5.7 x 10-4 s-1?

t½ln2k=

0.6935.7 x 10-4 s-1=

= 1200 s

= 20 minutes

t½

t½

t½

Second-Order Reactions

• Second-order reaction- a reaction whose rate depends on the concentration of one reactant raised to the second power OR on the concentrations of two different reactants, each raised to the first power.

• Simple Type: A→B

rate = k]A[2 • Complex Type: A + B→C

rate = k]A[]B[

Second-order Reactions

• For A→B, the following expression is used:• rate = - [A]/t = k[A]2 • k = - [A]/([A]2 * t) M/M2s = 1/s = M-1s-1

• Integration gives:

• we can obtain t1/2, by setting [A]t = [A]o/2, therefore, • 1/([A]o/2) = 1/[A]o + kt1/2

• t1/2 = 1/([A]o* k).

1]A[t

= 1]A[0

+ kt

Half-life of a Second-order Reaction

Equation for half-life

What is the difference between this equation and the equation for half-life of first-order reactions?

1t½ =k[A]0

At 25oC with CCl4 as solvent, the reaction:

I + I I2 is second order with respect to the concentration of

iodine atoms. The rate constant has been measured as 8.2 x 109 L mol-1s-1. Suppose the initial concentration of I is 1.00 x 10-4 mol L-1. Calculate [I] after 2.0 x 10-6 s.

In this particular problem, k is given as well as t and [A0] while [At] is unknown. This is usually the case, the integrated form of the second order rate law has 4 variables so three should be given in order to solve for one .

The solution to this problem is straightforward, it merely involves substituting the known values for t, k and [A0] and then solving for [At].

(1/[At]) = kt + (1/[A0])(1/[At]) = (8.2 x 109 L mol-1 s-1) (2.0 x 10-6 s) +

(1/1.00 x 10-4 mol L-1)

(1/[At]) = (2.6 x 104 L mol-1)

[At] = 3.8 x 10-5 mol L-1

Zero-order Reactions

• Very rare reactions (dissociation of ammonia)

• Usually occur on metallic surfaces

• Half-life Equation:

• Reaction rate is described by:Rate = k

t½ = [A]02k