Chapter 13Chemical Kinetics

CHEMISTRY

• Kinetics is the study of how fast chemical reactions occur.• There are 4 important factors which affect rates of

reactions:– reactant concentration,– temperature,– action of catalysts, and– surface area.

• Goal: to understand chemical reactions at the molecular level.

Factors that Affect Reaction Rates

Change of Rate with Time• Consider:

C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)

Reaction Rates

• For the reaction A B there are two ways of measuring rate:– the speed at which the products appear (i.e. change in moles of B per unit time), or– the speed at which the reactants disappear (i.e. the change in moles of A per unit

time).

Reaction Rates

t

A of molesA respect to with rate Average

Reaction Rate and Stoichiometry• In general for:

aA + bB cC + dD

Reaction Rates

tdtctbta

D1C1B1A1Rate

• Rate is (-) if reagent is consumed.• Rate is (+) if compound is produced.

• Rate will ultimately be (+) because change in concentration will be negative. Two (-)’s become (+).

Take Note:

• Rate must ALWAYS be a positive value!

Take Note!

• Since the direction of equilibrium changes as more product is produced, rates have to be determined as soon as the experiment has begun.

2N2O5 4NO2 + O2

[N2O5] (mol/L) Time (sec)

0.100 0

0.0707 50

0.0500 100

0.0250 200

0.0125 300

0.00625 400

Sample Problem

• A. How is the rate at which ozone (O3) disappears related to the rate at which O2 appears in the reaction: 2 O3 (g) 3 O2 (g)?

• B. If the rate at which O2 appears, D[O2]/Dt, is 6.0 x 10-5 M/s at a particular instant, at what rate is O3 disappearing at this same time, -D[O3]/Dt?

Answers

• A. “Related to” means compare, so write the rate expression comparing the compounds.

• B. 4.0 x 10-5 M/s

Sample Problem

• The decomposition of N2O5 proceeds according to the following equation: 2 N2O5 (g) 4 NO2 (g) + O2 (g)

If the rate of the decomposition of N2O5 at a particular instant in a reaction vessel is 4.2 x 10-7 M/s, what is the rate of appearance of: (a) NO2, (b) O2 ?

Differential Rate Law

- is a rate law that expresses how rate is dependent on concentration

Example:Rate = k[A]n

Differential First Order Rate Law

• First Order Reaction–Rate dependent on concentration

– If concentration of starting reagent was doubled, rate of production of compounds would also double

Using Initial Rates to Determines Rate Laws

• A reaction is zero order in a reactant if the change in concentration of that reactant produces no effect.

• A reaction is first order if doubling the concentration causes the rate to double.

• A reacting is nth order if doubling the concentration causes an 2n increase in rate.

• Note that the rate constant does not depend on concentration.

Concentration and Rate

Differential Rate Law

• For single reactants: A C

Rate = k[A]n

• For 2 or more reactants: A + B C

Rate = k[A]n[B]m

Rate = k[A]n[B]m[C]p

Problem

• NH4+ + NO2

- N2 + 2H2O

• Give the general rate law equation for rxn.• Derive rate order.• Derive general rate order.• Solve for the rate constant k.

To Determine the Orders of the Reaction (n, m, p, etc….)

• 1. Write Rate law equation.• 2. Get ratio of 2 rate laws from successive

experiments. • Ratio = rate Expt.2 = k2[NH4

+]n[NO2-]m

rate Expt.1 k1[NH4+]n[NO2

-]m

• 3. Derive reaction order.• 4. Derive overall reaction order.

Experimental Data

Expt. [NH4]initial [NO2-]initial Initial Rate

1 0.100 M 0.0050 M 1.35 x 10-7

2 0.100 M 0.010 M 2.70 x 10-7

3 0.200 M 0.010 M 5.40 x 10-7

Experiment Number

[A] (mol·L-1) [B] (mol·L-1) Initial Rate (mol·L-1·s-1)

1 0.100 0.100 4.0 x 10-5

2 0.100 0.200 4.0 x 10-5

3 0.200 0.100 1.6 x 10-4

A + B C

(a) Determine the differential rate law

(b) Calculate the rate constant

(c) Calculate the rate when [A]=0.050 mol·L-1 and [B]=0.100 mol·L-1

Use the data in table 12.5 to determine

1) The orders for all three reactants 2) The overall reaction order3) The value of the rate constant

Experiment Number

[NO] (mol·L-1) [H2] (mol·L-1) Initial Rate (mol·L-1·s-1)

1 0.10 0.10 1.23 x 10-3

2 0.10 0.20 2.46 x 10-3

3 0.20 0.10 4.92 x 10-3

(a) Determine the differential rate law

(b) Calculate the rate constant

(c) Calculate the rate when [NO]=0.050 mol·L-1 and [H2]=0.150 mol·L-1

2NO(g) + 2H2(g) N2(g) + 2H2O(g)

Consider the general reaction aA + bB cC and thefollowing average rate data over some time period Δt:

11 sLmol 0.0080Δt

Δ[A] 11 sLmol 0.0120Δt

Δ[B]

11 sLmol 0.0160Δt

Δ[C]

Determine a set of possible coefficients to balance thisgeneral reaction.

Sample Problem:.

Problem

• Reaction: A + B C obeys the rate law: Rate = k[A]2[B].

• A. If [A] is doubled (keeping B constant), how will rate change?

• B. Will rate constant k change? Explain.• C. What are the reaction orders for A & B?• D. What are the units of the rate constant?

You now know that….• The rate expression correlates consumption of

reactant to production of product. For a reaction: 3A 2B

- 1D[A] = 1D[B] 3 Dt 2 Dt

• The differential rate law allows you to correlate rate with concentration based on the format:

Rate = k [A]n

You also know that…

• 1. Rate of consumption of reactant decreases over time because the concentration of reactant decreases. Lower concentration equates to lower rate.

• 2. If a graph of concentration vs. time were constructed, the graph is not a straight line

0 0.002 0.004 0.006 0.008 0.01 0.012 0.0140

200

400

600

800

1000

1200

1400

Expt. 25

Series1

Concentration (M)

Tim

e in

seco

nds

0 200 400 600 800 1000 1200 14000

0.002

0.004

0.006

0.008

0.01

0.012

0.014

Expt. 25

Series1

Time in seconds

Conc

entr

ation

in M

olar

ity

How can we make the line straight?

What is the relationship between concentration and time?

By graphing concentration vs. 1/time?

0 0.002 0.004 0.006 0.008 0.01 0.012 0.0140

0.001

0.002

0.003

0.004

0.005

0.006

Concentration vs. Reciprocal of Time

Series1

Concentration (M)

Recip

roca

l of T

ime

(1/s

ec)

The Integrated Rate Law makes this possible!

0 200 400 600 800 1000 1200 1400 1600

-6.5

-6

-5.5

-5

-4.5

-4

-3.5

-3

First Order Reaction

Series1

Time in seconds

ln[c

once

ntra

tion]

Integrated Rate Law

• Expresses the dependence of concentration on time

Integrated Rate Laws

• Zero Order: [A]t = -kt + [A]o

• First Order: ln[A]t = -kt + ln[A]o

• Second Order: 1 = kt + 1 [A]t [A]o

where [A]o is the initial concentration and [A]t is the final concentration.

Integrated First-Order Rate Law

• ln[A]t = -kt + ln[A]0

• Eqn. shows [concn] as a function of time• Gives straight-line plot since equation is of

the form y = mx + b

Zero Order Reactions

• A plot of [A]t versus t is a straight line with slope -k and intercept [A]0.

The Change of Concentration with Time

First Order Reactions

• A plot of ln[A]t versus t is a straight line with slope -k and intercept ln[A]0.

The Change of Concentration with Time

Second Order Reactions

• A plot of 1/[A]t versus t is a straight line with slope k and intercept 1/[A]0.

The Change of Concentration with Time