Chemical Kinetics

Kinetics

Kinetics in chemistry is concerned with how quickly a reaction proceeds

Factors that affect rate Physical state of the reactants Concentration of the reactants Temperature at which the reaction occurs The presence of a catalyst

Reaction Rates

Reaction rates depend on the frequency of collisions between molecules

Reaction rate = speed of a reaction (M/s)

A → B Δ[B]/Δt = -Δ[A]/Δt Average Rate

Change of Rates with Time

C4H

9Cl(aq) + H

2O(l) →

C4H

9OH(aq) + HCl(aq)

What is happening to the rate as the reaction proceeds?

Graphs of the data allow you to find the instantaneous rate

Reaction Rates and Stoichiometry

Stoichiometry will affect the rates of disappearance and formation

2HI(g) → H2(g) + I

2(g)

-1/2Δ[HI]/Δt = Δ[H2]/Δt = Δ[I

2]/Δt

For any general reaction aA + bB → cC + dD

Practice

How is the rate of disappearance of ozone related to the rate of appearance of oxygen in the following equation: 2O3(g) → 3O2(g)? If the rate of appearance of oxygen is 6.0x10- 5 M/s at a particular instant what is the value of the rate of disappearance of ozone at this same time?

Answer: 4.0x10- 5 M/s

Practice

The decompostion of N2O5 proceeds according to the following equation:

2N2O5(g) → 4NO2(g) + O2(g)

If the rate of decomposition of dinitrogen pentoxide at a particular instant in a reaction vessel is 4.2x10- 7 M/s, what is the rate of appearance of NO

2 and O

2?

Answer: 8.4x10-7 M/s, 2.1x10-7 M/s

Beer's Law

Spectroscopic methods are useful in seeing how concentration changes with time

2HI(g) → H2(g) + I

2(g)

A = abc A : absorbance a : molar absorptivity b : path length c : concentration

Concentration and Rate

To determine the effect of concentration of rate, you can vary the concentration of reactants and monitor the change in initial rate

NH4

+(aq) + NO2

-(aq) → N

2(g) + 2H

2O(l)

What happens to the initial rate when the concentrations are changed?

Rate Law

Rate law shows the rate of a reaction is related to the concentrations of the reactants Rate = k[NH

4+][NO

2-]

For a general reaction: aA + bB → cC + dD Rate = k[A]m[B]n

k = rate constant The magnitude of k is affected by changes in temp.

If we know the rate law we can calculate k From exp. 1: r = 5.4x10-7 M/s = k(0.0100M)(0.200M) k = 2.7x10-4 M-1·s-1

Reaction Orders

Rate = k[A]m[B]n

m and n are reaction orders Rate = k[NH

4+][NO

2-]

Each compound is 1s t order but the overall order is 2n

d (just add the exponents) Reaction orders must be determined

experimentally 2N

2O

5(g) → 4NO

2(g) + O

2(g) : Rate =

k[N2O

5]

2HI(g) → H2(g) + I

2(g) : Rate = k[H

2]

[I2]

Units of Rate Constants

Units of the rate constant depend on the overall order of the rate law

What are the overall reaction orders and units of the rate constant for the following reactions? 2N2O5(g) → 4NO2(g) + O2(g)

Rate = k[N2O

5]

2HI(g) → H2(g) + I2(g)

Rate = k[H2][I

2]

CHCl3(g) + Cl2(g) → CCl4(g) + Hcl(g)

Rate = k[CHCl3][Cl]1/2

Using Initial Rates

Observing the effect of changing the initial concentrations of the reactants on the initial rate allows us to determine reaction orders

Exponents will commonly be 0, 1, 2 What effect will a reactant with a reaction order of 0

have on the reaction? 1? 2? Remember only the rate depends on

concentration

Practice

The initial rate of a reaction A + B → C was measured for several different starting concentrations of A and B. Using this data determine the rate law for the reaction; the magnitude of the rate constant; and the rate when [A] = 0.050M and [B] = 0.100M

Answer: r = k [A]2, 4.0x10- 3M- 1s- 1,

r = 1.0x10- 5M/s

[A] (M) [B] (M)1 0.100 0.100 4.00E-052 0.100 0.200 4.00E-053 0.200 0.100 1.60E-04

Experiment Number

Initial Rate (M/s)

Practice

The following data were measured for the reaction of nitric oxide with hydrogen:

2NO(g) + 2H2(g) → N

2(g) + 2H

2O(g)

Determine the rate law for this reaction, the value of the rate constant and the rate when [NO] = 0.050M and [H

2] = 0.150M

[NO] (M) [B] (M)1 0.100 0.100 1.23E-032 0.100 0.200 2.46E-033 0.200 0.100 4.92E-03

Experiment Number

Initial Rate (M/s)

Answer: r = k[NO]2[H2], k = 1.2 M- 2s- 1, r = 4.5x10-

4 M/s

Changing Concentration with Time

Rate laws tell us how the rate of a reaction changes at a given temperature as concentration changes.

Rate laws can be converted to tell us what the concentration of a substance is at any time during a reaction.

There are two special cases and you must be able to use them.

First Order Reactions

For a first order reaction that proceeds A → products the rate law is Rate = -Δ[A]/Δt = k[A]

After some math magic (involving integration) ln[A]

t – ln[A]

0 = -kt

With some rearrangement we get something similar to y = mx + b ln[A]

t = -kt + ln[A]

0

What would you graph to see if the reaction was first order?

Practice

The first order rate constant for the decomposition of a certain insecticide in water at 12°C is 1.45 yr -

1. A quantity of the insecticide is washed into a lake on June 1, leading to a concentration of 5.0x10- 7g/mL. What is the concentration of insecticide after 1 year? How long will it take for the concentration to drop to 3.0x10- 7g/mL?

Answer: 1.2x10- 7g/mL, 0.35 years

Practice

The decomposition of dimethyl ether, (CH3)

2O, at 510°C is

a first order process with a rate constant of 6.8x10 - 4 s- 1:

(CH3)

2O(g) → CH

4(g) + H

2(g) + CO(g)

If the initial pressure of dimethyl ether is 135 torr, what is the partial pressure after 1420s?

Answer: 51 torr

Second Order Reactions

For a reaction that proceeds A → products or A+B → products that are second order in just one reactant A: Rate = - Δ[A]/Δt = k[A]2

After some calculus magic this becomes: 1/[A]

t = kt + 1/[A]

0

If plotting 1/[A]t creates a straight line the reaction is second order

Practice

The following data were obtained for the gas phase decomposition of nitrogen dioxide at 300°C, NO

2(g) →

NO(g) + ½ O2(g):

Is the reaction first or second order in NO

2? What is k? If the initial

concentration of NO2 is 0.0500M, what

is the concentration after 0.500hr?

Answer: 2nd order r = k[NO2]2, k =

0.543 M- 1s- 1, 1.00x10- 3M

Time (s)0.0 0.01000

50.0 0.00787100.0 0.00649200.0 0.00481300.0 0.00380

[NO2] (M)

Half Life

Half life (t½) is the time it takes for the

concentration of a reactant to drop to one half of its initial value

Using algebra you can find the half life of a 1st order reaction t

½ = 0.693/k

Half life for a second order reaction depends on concetration t

½ = 1/k[A]

0

Temperature and Rate

Rate of most chemical reactions increase as temperature increases

Increasing temperature increases the rate constant and thus the rate

Glow stick fun What effect did the ice and hot water have on the

reaction?

Collision Model

Molecule must collide with enough energy to react What does increasing

temperature do? Collision Theory

Orientation Factor Activation Energy

Ea

Activation Energy

Higher activation energy the lower the rate

Only a fraction of molecules have energy to generate products upon collision f = e-Ea/RT

Arrhenius Equation

Rates depend on Fraction of molecules with an energy of Ea or greater Collisions per second Fraction of collision with proper orientation

Arrhenius used these ideas to relate k and Ea k = Ae- E a / R T

Taking the natural log of both sides ln k = (-Ea/R)T + ln A So we can graph ln k versus 1/T to find Ea

We can relate the rate constants of a reaction at different temperatures ln(k

1/k

2) = Ea/R (1/T

2 - 1/T

1)

Practice

The following table shows the rate constants for the rearrangement of methyl isonitrile at various temperatures. From these data, calculate the activation energy for the reaction. What is the value of the rate constant at 430.0K?

Answer: Ea = 160 kJ/mol, k = 1.0x10- 6s - 1

189.7 0.0000252198.9 0.0000525230.3 0.0006300251.2 0.0031500

Temperature (°C) k (s-1)

Reaction Mechanisms

Reaction mechanisms show how a reaction occurs

Elementary Steps NO(g) + O

3(g) → NO

2(g) + O

2(g)

Molecularity Unimolecular Bimolecular Termolecular

Multistep Mechanisms

Many reactions do not happen in just one step NO

2(g) + CO(g) → NO(g) + CO

2(g)

NO2(g) + NO

2(g) → NO

3(g) + NO(g)

NO3(g) + CO(g) → NO

2(g) + CO

2(g)

Elementary steps must add up to give the overall reaction Intermediate

Practice

It has been proposed that the conversion of ozone into O2 proceeds via two elementary steps:

O3(g) → O

2(g) + O(g)

O3(g) → O(g) + 2O

2(g)

Describe the molecularity if each step in this mechanism. Write the equation for the overall reaction. Identify any intermediates.

For the reaction Mo(CO)6 + P(CH

3)

3 → Mo(CO)

5P(CH

3)

3 + CO

The proposed mechanism is

Mo(CO)6 → Mo(CO)

5 + CO

Mo(CO)5 + P(CH

3)

3 → Mo(CO)

5P(CH

3)

3

Is the proposed mechanism consistent with the equation for the overall reaction? Identify any intermediates.

Rate Laws for Elementary Steps

Rate laws cannot normally be predicted from the coefficients of balanced equations. Why?

For elementary steps the equation tells you the rate law. Rate law is determined

by its molecularity

Practice

If the following reaction occurs in a single elementary step, predict the rate law:

H2(g) + Br

2(g) → 2HBr(g)

Consider the following reaction: 2NO(g) + Br2(g)

→ 2NOBr(g). Write the rate law for the reaction assuming it involves a single elementary step. Is a single elementary step likely for this reactipon?

Rate Laws for Multistep Mechanisms

In multistep systems the rate laws is set by the slowest step Rate determining step

Mechanisms with slow first step

Step 1: NO2(g) + NO

2(g) → NO

3(g) + NO(g) (slow)

Step 2: NO3(g) + CO(g) → NO

2(g) + CO

2(g) (fast)

Overall: NO2(g) + CO(g) → NO(g) + CO

2(g)

Rate = k1[NO

2]2

Mechanisms with Initial Fast Step

You need to derive the rate law for a mechanism in which there is an intermediate.

2NO(g) + Br2(g) → 2NOBr(g) Rate = k[NO]2[Br

2]

Possible Mechanism NO(g) + Br

2(g) ↔ NOBr

2(g) (fast)

NOBr2(g) + NO(g) → 2NOBr(g) (slow)

Algebra fun When a fast step precedes a slow one we can solve

for the concentration of an intermediate by assuming that an equilibrium is established in the fast step.

Practice

Show that the following mechanism for the reaction producing NOBr also produces a rate law consistent with the experimentally observed one:

Step 1: NO(g) + NO(g) ↔ N2O

2(g) (fast)

Step 2: N2O

2(g) + Br

2(g) → 2NOBr(g) (slow)

The first step of a mechanism involving the reaction of bromine is: Br2(g) ↔ 2Br(g) (fast). What is the expression relating the concentration of Br(g) to that of Br2(g)

Catalysis

What does a catalyst do?

Types of catalysts Homogeneous – same

phase as the reactants Heterogeneous –

different phase from the reactants Adsorption happens

first

Enzymes