Chemical Kinetics

Chemical KineticsChapter 14

Kinetics

• Studies the rate at which a chemical process occurs.

• Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs).

• Four main factors affect reaction rates:– Physical states of reactants– Concentrations of reactants– Temperature– Catalysts

Factors That Affect Reaction Rates

• Physical State of the Reactants– In order to react, molecules must come in

contact with each other.– The more homogeneous the mixture of

reactants, the faster the molecules can react.

Factors That Affect Reaction Rates

• Concentration of Reactants– As the concentration of reactants increases,

so does the likelihood that reactant molecules will collide.

Factors That Affect Reaction Rates

• Temperature– At higher temperatures, reactant

molecules have more kinetic energy, move faster, and collide more often and with greater energy.

Factors That Affect Reaction Rates

• Presence of a Catalyst– Catalysts speed up reactions by

changing the mechanism of the reaction.

– Catalysts are not consumed during the course of the reaction.

Chemical Kinetics

Thermodynamics – does a reaction take place?

Kinetics – how fast does a reaction proceed?

Reaction rate is the change in the concentration of a reactant or a product with time (M/s).

A B

rate = -[A]t

rate = [B]t

[A] = change in concentration of A over time period t

[B] = change in concentration of B over time period t

Because [A] decreases with time, [A] is negative.

14.1

A B

14.1

rate = -[A]t

rate = [B]t

Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)

time

393 nmlight

Detector

[Br2] Absorption39

3 nm

Br2 (aq)

14.1

Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)

average rate = -[Br2]t

= -[Br2]final – [Br2]initial

tfinal - tinitial

slope oftangent

slope oftangent slope of

tangent

instantaneous rate = rate for specific instance in time 14.1

rate [Br2]

rate = k [Br2]

k = rate[Br2]

14.1

= rate constant

= 3.50 x 10-3 s-1

2H2O2 (aq) 2H2O (l) + O2 (g)

PV = nRT

P = RT = [O2]RTnV

[O2] = PRT1

rate = [O2]t RT

1 Pt=

measure P over time

14.1

2H2O2 (aq) 2H2O (l) + O2 (g)

14.1

Reaction Rates and Stoichiometry

14.1

2A B

Two moles of A disappear for each mole of B that is formed.

rate = [B]t

rate = -[A]t

12

aA + bB cC + dD

rate = -[A]t

1a

= -[B]t

1b

=[C]t

1c

=[D]t

1d

Write the rate expression for the following reaction:

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)

rate = -[CH4]

t= -

[O2]t

12

=[H2O]

t12

=[CO2]

t

14.1

The Rate Law

14.2

The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers.

aA + bB cC + dD

Rate = k [A]x[B]y

reaction is xth order in A

reaction is yth order in B

reaction is (x +y)th order overall

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

rate = k [F2]x[ClO2]y

Double [F2] with [ClO2] constant

Rate doubles

x = 1

Quadruple [ClO2] with [F2] constant

Rate quadruples

y = 1

rate = k [F2][ClO2]

14.2

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

rate = k [F2][ClO2]

Rate Laws

• Rate laws are always determined experimentally.

• Reaction order is always defined in terms of reactant (not product) concentrations.

• The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation.

1

14.2

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)

Experiment [S2O82-] [I-] Initial Rate

(M/s)

1 0.08 0.034 2.2 x 10-4

2 0.08 0.017 1.1 x 10-4

3 0.16 0.017 2.2 x 10-4

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

[S2O82-][I-]

=2.2 x 10-4 M/s

(0.08 M)(0.034 M)= 0.08/M•s

14.2

rate = k [S2O82-][I-]

First-Order Reactions

14.3

A product rate = -[A]t

rate = k [A]

k = rate[A]

= 1/s or s-1M/sM=

[A]t = k [A]-

[A] is the concentration of A at any time t[A]0 is the concentration of A at time t=0

[A] = [A]0exp(-kt) ln[A] = ln[A]0 - kt

Decomposition of N2O5

14.3

(400 s, -0.34)

(2430 s, -1.50)

ot k)(t)( [A]ln[A]ln y m x b= +

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 0.88 M to 0.14 M ?

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

=ln

0.88 M0.14 M

2.8 x 10-2 s-1=

14.3