Chemical Kinetics

Chemical kinetics is concerned with the speed or rate of chemical reactions.

Factors affecting reaction rates

The physical state of the reactants:

For example, reactions between gases and solids are limited to their area of contact.

Also, the surface area of solids is extremely important; as surface area increases, rate increases.

The concentration of the reactants:

As concentration increases, rate increases. This is because the frequency with which reactant molecules collide increases.

The temperature at which the reaction occurs:

As temperature increases, rate increases. With an increase in temperature, the kinetic energy of reactant molecules increases and they move more rapidly, collide more frequently and with higher energy.

The presence of a catalyst:

Catalysts are chemical agents that affect the rate of a reaction without being used up.

They affect the kinds of collisions (i.e. the mechanism) that lead to the reaction.

Overall, reaction rates depend on the frequency of collisions between molecules.

The greater the frequency of collisions, the greater the rate of reaction. Also, for a collision to lead to a reaction, it must occur with sufficient energy.

Instantaneous rate (i.e. rate at time T1)

[Calculate the gradient of the tangent line]

T1

Speed – defined as the change that occurs in a given interval of time

The speed of a chemical reaction is called its reaction rate and is defined as the change in concentration of reactants or products per unit of time.

Units for rate – usually molarity per second [(mol dm-3 s-1) or (M s-1)]

Reaction Rate = change in conc. of reactant (or product) M s-1

time

Rate generally decreases as a reaction proceeds because the concentration of the reactants is decreasing (i.e. reactants are being used up).

Instantaneous rate – the rate at a particular moment in the reaction

This is the tangent to the curve of a concentration vs time plot:

The instantaneous rate at time t = 0 is called the initial rate of the reaction.

In general for the reaction A B:

Rate = -Δ[A] or rate = Δ[B]Δt Δt

[Note: The negative sign means that A is the reactant, which is being used up so its concentration decreases with time; B is the product whose concentration is increasing]

Expt. No. Initial [NH4+] (M)Initial [NO2-] (M)Observed Initial Rate (M s-1)

x 2

i.e. 21

The Rate Law:

For the general reaction: A + B C + D

Rate = k[A]m[B]n where k = the rate constant

m & n = the orders of the reaction

Overall order of the reaction = m + n (i.e. the sum of the individual orders)

e.g. (1) For the reaction: NH4+ + NO2

- N2 + 2H2O

Rate = k[NH4+]m[NO2-]n

Given the following experimental data:

Expt. No. Initial [NH4+] (M) Initial [NO2-] (M) Observed Initial Rate (M s-1)

1 0.01 0.20 5.4 x 10-7

2 0.02 0.20 10.8 x 10-7

3 0.20 0.02 10.8 x 10-7

4 0.20 0.04 21.6 x 10-7

Consider Expts. 1 & 2:

1 0.01x 2

0.20 5.4 x 10-7

2 0.02 0.20 10.8 x 10-7

3 0.20 0.02 10.8 x 10-7

4 0.20 0.04 21.6 x 10-7

When the conc. of NH4+ is multiplied by 2, the rate is also multiplied by 2 (or 21)

Therefore, the order of the reaction with respect to NH4+ is 1 (or 1st order)

3 0.20 0.02x 2

0.04

10.8 x 10-7

21.6 x 10-7

x 24 0.20 i.e. 21

Consider Expts. 3 & 4:

Expt. No. Initial [NH4+] (M) Initial [NO2-] (M) Observed Initial Rate (M s-1)

1 0.01 0.20 5.4 x 10-7

2 0.02 0.20 10.8 x 10-7

When the conc. of NO2- is multiplied by 2, the rate is also multiplied by 2 (or 21)

Therefore, the order of the reaction with respect to NO2- is 1 (or 1st order)

The rate law for the reaction then becomes:

Rate = k[NH4+]1[NO2

-]1 or simply:

Rate = k[NH4+][NO2-]

Overall order of the reaction = 1 + 1 = 2 (or 2nd order)

To determine the value of k:

Choose one experiment and substitute values into the rate law:

E.g. Using experiment 1:

5.4 x 10-7 M s-1 = k(0.01 M)(0.20 M)

k = 5.4 x 10-7 M s-1

(0.01 M)(0.20 M)

k = 2.7 x 10-4 M-1 s-1

Therefore: Rate = 2.7 x 10-4 M-1 s-1 [NH4+][NO2-]

Expt. No. Initial [A] (M) Initial [B] (M)

1 0.1 0.1

Observed Initial Rate (M s-1)4.0 x 10-5

x 2x 42 0.1 0.2

0.1

4.0 x 10-5

16.0 x 10-5

i.e. 22

3 0.2

Expt. No. Initial [A] (M) Initial [B] (M) Observed Initial Rate (M s-1)

1 0.1 x 2 x 1

i.e. 202 0.1

3 0.2 0.1

0.1

0.2

4.0 x 10-5

4.0 x 10-5

16.0 x 10-5

e.g. (2) For the reaction: A + B C

Expt. No. Initial [A] (M) Initial [B] (M) Observed Initial Rate (M s-1)

1 0.1 0.1 4.0 x 10-5

2 0.1 0.2 4.0 x 10-5

3 0.2 0.1 16.0 x 10-5

Rate = k[A]m[B]n

Consider Expts. 1 & 3:

When the conc. of A is multiplied by 2, the rate is multiplied by 4 (or 22)

Therefore, the order of the reaction with respect to A is 2 (or 2nd order)

Consider Expts. 1 & 2:

When the conc. of B is multiplied by 2, the rate is multiplied by 1 (or 20)

Therefore, the order of the reaction with respect to B is 0 (or 0 order)

The rate law for the reaction then becomes:

Rate = k[A]2[B]0 or simply:

Rate = k[A]2

Overall order of the reaction = 2 + 0 = 2 (or 2nd order)

To determine the value of k:

Using experiment 1:

4.0 x 10-5 M s-1 = k(0.1 M)2

k = 4.0 x 10-5 M s-1

(0.1 M)2

k = 4.0 x 10-3 M-1 s-1

Therefore: Rate = 4.0 x 10-3 M-1 s-1 [A]2

Summary:

To find the order of a reaction with respect to a particular reactant ALL other reactant concentrations must be the same. Therefore, compare the change in concentration of one reactant at a time (while keeping the others constant) with the change in the rate of the reaction.

If the reactant concentration changes but the rate does not change, then the order is 0.

If the reactant concentration changes and the rate changes by the same factor, then the order is 1.

e.g. if rate also doubles when concentration is doubled, the order = 1.

If the reactant concentration changes but the rate changes by the square of the factor, then the order is 2.

e.g. if rate quadruples (i.e. changes by a factor of 4) when the concentration is doubled, order = 2

Finding rate by graphical methods

Rate laws can also be converted into equations that show the relationship between the concentration of reactants or products with time. Graphical plots can then be made:

For a 0 order reaction: Rate vs. concentration gives the following plot:

This means that as concentration changes, rate does not change.

For a 1st order reaction:

Concentration vs. time i.e. [A] vs. t gives a curve.

But: ln[A]t = -kt + ln[A]0

So plot ln of concentration vs. time i.e. ln[A]t vs. t instead to get a straight line:

The gradient of the line gives –k, therefore we can find k

The y-intercept gives ln[A]0, therefore we can find [A]0 (i.e. the initial concentration of A)

For a 2nd order reaction: 1 = kt + 1 [A]t [A]0

Plot 1/ concentration vs. time to get a straight line Plot 1 vs. t:

[A]t

The gradient of the line gives k

The y-intercept gives 1/ ln [A]0, therefore we can find [A]0

Therefore, to distinguish between 1st and 2nd order:

If ln [A] vs. t is linear, the reaction is 1st order with respect to A.

If 1/ [A] vs. t is linear, reaction is 2nd order with respect to A.

Temperature and rate

Rate increases as temperature increases due to an increase in the rate constant (k).

Recall: As temperature increases, kinetic energy increases and molecules move faster leading to more frequent, higher-energy collisions.

In addition, reactant molecules must be oriented in a certain way during collisions for a reaction to occur. The orientation of molecules determines whether new bonds will be formed.

General Rule: If temperature increases by 10oC, rate approximately doubles.

Activation Energy (Ea)

To react, colliding molecules must have a total energy equal to or more than some minimum value. The minimum energy required to initiate a chemical reaction is called the activation energy, Ea.

The rate of a reaction depends on the magnitude of Ea.

Generally, the lower the Ea, the faster the rate of reaction.

The Boltzmann Distribution curve is used to show the impact of temperature on rate.

At higher temperatures, the curve shifts to the right so the fraction of molecules with kinetic energy greater than the activation energy increases and reaction rate increases.

T1 = lower temp.

T2 = higher temp.

Fraction ofmolecules

Energy

Catalysis

Catalyst – a substance that changes the rate of a chemical reaction without undergoing a permanent chemical change in the process

Catalysts are very common e.g. metals like nickel (Ni) and platinum (Pt)

Enzymes are also catalysts. They are examples of biological catalysts in living systems.

There are two major types of catalysts:

Homogenous catalysts – where the catalyst is in the same physical state as the reactants

E.g. when a liquid catalyst is used for a reaction between 2 liquids

Heterogeneous catalysts – where the catalyst is in a different state from the reactants

E.g. when a solid catalyst is used for a reaction between 2 gases

Catalysts work by lowering the activation energy (Ea) for a chemical reaction. They do this by providing an alternative mechanistic pathway for the reaction to occur:

Questions

1) (a) The bromination of acetone is catalyzed by acid:

H+ catalystCH3COCH3 (aq) + Br2 (aq) + H2O (l) CH3COCH2Br (aq) + Br- (aq) + H3O+ (aq)

The rate of disappearance of bromine was measured for several different initial concentrations of acetone, bromine and water, as represented in the following table:

Run Initial Concentration (mol L-1) Initial rate of change of [Br2] (mol L-1 s-1) (x 10-5)

[CH3COCH3] [Br2] [H2O]

1 0.30 0.05 0.05 5.7

2 0.30 0.10 0.05 5.7

3 0.30 0.05 0.10 12.0

4 0.40 0.05 0.20 31.0

5 0.40 0.05 0.05 7.6

(i) Deduce the order with respect to each reactant and hence the overall order, showing all of your working. (4 marks)

(ii) Write the rate law for the reaction. (2 marks)

(iii) What is the numerical value of the rate constant? (2 marks)

(iv) If [H2O] is maintained at 0.050 M, whereas both [CH3COCH3] and [Br2] are 0.10 M, what is the rate of the reaction? (3 marks)

Solution:(i) Comparing experiments 1 & 5:

As the concentration of CH3COCH3 is multiplied by 1.33, the rate is multiplied by 1.33 or 1.331

Therefore, the order w.r.t. CH3COCH3 is 1 (or 1st order)

Comparing experiments 1 & 2:

As the concentration of Br2 is multiplied by 2, the rate is multiplied by 1 or 20

Therefore, the order w.r.t. Br2 is 0 (or 0 order)

Comparing experiments 1 & 3:

As the concentration of H2O is multiplied by 2, the rate is multiplied by 2 or 21

Therefore, the order w.r.t. H2O is 1 (or 1st order)

Overall order = 1 + 0 + 1 = 2 (or 2nd order)

(ii) Rate = k[CH3COCH3][H2O]

(iii) Using values for experiment

1: 5.7 x 10-5 = k (0.30)(0.05)

k = 0.0038

(iv) Rate = 0.0038[CH3COCH3][H2O]

Rate = 0.0038(0.10)(0.050) = 1.9 x 10-5 mol L-1 s-1

2) (a) The iodide ion reacts with the hypochlorite ion (the active ingredient in chlorine bleaches) in the following way:

OCl- + I- OI- + Cl-

This rapid reaction gives the following rate data:

Run [OCl-] (M) [I-] (M) Rate (M s-1)

1 1.5 x 10-3 1.5 x 10-3 1.36 x 10-4

2 3.0 x 10-3 1.5 x 10-3 2.72 x 10-4

3 1.5 x 10-3 3.0 x 10-3 2.72 x 10-4

(i) Deduce the order with respect to each reactant and also the overall order.

(ii) Give the rate law for the reaction.

(iii) What is the numerical value of the rate constant?

(iv) Calculate the rate when [OCl-] = 2 x 10-3 M and [I-] = 5 x 10-4 M.(4 x 2 marks)

Solution:

(a)(i) Comparing experiments 1 & 2:

As the concentration is multiplied by 2, the rate is multiplied by 2

or 21 Therefore, the order w.r.t. OCl- is 1 (or 1st order)

Comparing experiments 1 & 3:

As the concentration is multiplied by 2, the rate is multiplied by 2

or 21 Therefore, the order w.r.t. I- is 1 (or 1st order)

Overall order = 1 + 1 = 2 (or 2nd order)

(ii) Rate = k[OCl-][I-]

(iii) Using values for experiment 1:

1.36 x 10-4 M s-1 = k (1.5 x 10-3 M)(1.5 x 10-3

M) k = 60.44 M-1 s-1

(iv) Rate = 60.44 M-1 s-1 [OCl-][I-]

Rate = 60.44 M-1 s-1 (2 x 10-3 M)(5 x 10-4 M) = 6.044 x 10-5 M s-1