2 NO(g) + O2(g) 2 NO2(g)Determine the rate expression and the value of the rate

constant from the data below. [NO] (mol L-1) [O2](mol L-1) initial rate (mol L-1 s-1)

(1) 1.0 x 10-4 1.0 x 10-4 2.8 x 10-6

(2) 1.0 x 10-4 3.0 x 10-4 8.4 x 10-6

(3) 2.0 x 10-4 3.0 x 10-4 3.4 x 10-5

Rate = k [O2]m [NO]n

To determine the rate law from the data, first determine the dependence of the rate on each reactant separately.

rate2/rate1 = k [O2]2m [NO]2

n / k [O2]1m [NO]1

n

8.4 x 10-6 / 2.8 x 10-6 = (3.0 x 10-4)m/ (1.0 x 10-4)m

3= 3m => m = 1; 1st order in O2

Determining Rate Laws

rate3/rate2 = k [O2]3m [NO]3

n / k [O2]2m [NO]2

n

3.4 x 10-5 / 8.4 x 10-6 = (2.0 x 10-4)n/ (1.0 x 10-4)n

4= 2n => n = 2; 2nd order in NO

Rate = k [O2][NO]2

Order of reaction = 3

2.8 x 10-6 mol L-1s-1 = k [1.0 x 10-4 mol L-1] [1.0 x 10-4 mol L-1]2

k = 2.8 x 106 L2 mol-2s-1

Concentration and Time - Integrated Rate Laws

Integrated rate laws: variation of concentration of reactants or products at any time

Derived from the experimental rate laws

Zero order reactions

A P- d[A]/ dt = k d[A] = - k dt

[A]t = [A]o - k t integrated rate law[A]o : concentration of A at t = 0

Slope = -k d[A]=−kdt

0

t∫

[A]o

[A]t∫

First Order Reactions

Rate = - d[A]/ dt = k [A] first order reactionUnits of k for a 1st order reaction is time-1

d[A]/ dt = - k [A] d[A]/[A] = - k dt

Solution is:

ln [A]t

[A]o= - kt

[A]o where is the initial concentration of A at time t = 0

[A]t = [A]o e-kt

d[A]A[A]o

[A]t∫ =−kdt

0

t∫

ln[A]t = ln [A]o - k t [A]t = [A]o e-kt

N2O5(g) N2O4(g) + 1/2 O2(g)

Half life of a 1st order reaction

Half life : time it takes for the concentration of the reactant A to fall to half its initial value

t1/2 when [A]t = [A]o/2

ln[A]t = ln [A]o - ktln [A]o/2 = ln [A]o - k t1/2 ln(1/2) = - k t1/2 ln(2) = k t1/2 t1/2 = ln(2) / k

t1/2 =0.693

k

Mercury (II) is eliminated from the body by a first-order process that has a half life of 6 days. If a person accidentally ingests mercury(II) by eating contaminated grain, what percentage of mercury (II) would remain in the body after 30 days if therapeutic measures were not taken?

k = ln 2 / t1/2 = (ln 2) / (6 days)

Fraction remaining after 30 days

[A]t / [A]0 = e-kt = 0.03

Answer: 3% remaining in the body after 30 days.

Radioactive decay is a first order process

Nt = No e-kt

where Nt is the number of radioactive nuclei at time t

No is the initial number of radioactive nuclei

k is called the decay constant

t1/2 = ln 2/ k

Carbon-14 dating uses the decay of 14C

14C is produced in the atmosphere at an almost constant rate

As a result proportion of 14C to 12C is ~ constant14C enters living systems as 14CO2; all living systems have a fixed ratio of 14C to 12C; about 1 14C to 1012 12C

When the organism dies, there is no longer exchange of C with surroundings, but 14C already in the organism continues to decay with a constant half life, so ratio of 14C to 12C decreases.

714N + 01n → 6

14C +1 1p

614C → 7 14N +−1

0e +ν

Second order reactions

Rate = k[A]2

1

[A]t= 1

[A]o+kt

- d[A]/ dt = k [A]2

2nd order reaction for which the rate depends on one reactant

rate = k [A] [B] or rate = k [A]2

d[A]/[A]2 = - k dt

d[A][A]2[A]o

[A]t∫ = kdt

0

t∫

[A]t = [A]o

1+[A]okt

[A]t = [A]o

1+[A]okt

The half-life of a 2nd order reaction when

[A] = [A]o/2 t1/2=

1k[A]o

1

[A]t= 1

[A]o+ktln[A]t = ln [A]o - kt

1st order2nd order

slope = k

2C2F4 C4F8

ln[C2F4] vs time - nonlinear

rate = k [C2F4]2

1

[A]t= 1

[A]o+kt

Reactions often proceed in a series of steps - elementary reactionsFor example:

O2 + light O2*

O2* O. + O.

2(O. + O2 + M O3 + M)

Overall reaction: 3O2 + light 2O3

Reaction Mechanisms

O. is an intermediate species; involved in a step in the mechanism, but does not appear in the overall reaction

Determines rate laws; use experimental rate law to determine mechanism

The rate of an elementary reaction is directly proportional to the product of the concentrations of the reactants, each raised to a power equal to its coefficient in the balanced equation for that step

The number of reacting species in an elementary reaction - molecularity

Overall reaction

6 Fe2+(aq) + Cr2O72-(aq) + 14H+(aq)

6 Fe3+(aq) + 2 Cr3+(aq) + 7 H2O(l)

Unimolecular reaction; molecularity = 1

O2* O + ORate = k[O2*]

Bimolecular reaction; molecularity = 2

NO(g) + O3(g) NO2(g) + O2(g)Rate = k [NO] [O3]

Termolecular reaction; molecularity = 3

O + O2 + M O3 + M rate = k [O] [O2] [M]

Termolecular reactions are low probability reactions; require three species to come together simultaneously

Types of elementary reactions

From rate law to reaction mechanism

Products of a reaction can never be produced faster than the rate of the slowest elementary reaction - rate determining step

Experimental data for the reaction between NO2 and F2 indicate a second-order rate

Overall reaction: 2 NO2(g) + F2(g) 2FNO2(g)

Rate = k [NO2] [F2]

How can a mechanism be deduced from the rate law?

Rate law indicates that the reaction cannot take place in one step