Chemical KineticsChapter 14

Reminders

• Assignment 2 up on ACME, due Jan. 29 (in class)

• Assignment 3 up now and will be due Mon., Feb. 05

• Assignment 4 (Ch. 15) will not be due before Midterm 1, but Ch. 15 will be on the midterm

Answers: (a) 8.4 10 –7 M/s, (b) 2.1 10 –7 M/s

PRACTICE EXERCISEThe decomposition of N2O5 proceeds according to the following equation:

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

aA + bB cC + dD

Rate = −1a

[A]t

= −1b

[B]t

=1c

[C]t

1d

[D]t

=

SAMPLE EXERCISE Determining a Rate Law from Initial Rate Data

Using these data, determine (a) the rate law for the reaction, (b) the magnitude of the rate constant, (c) the rate of the reaction when [A] = 0.050 M and [B] = 0.100 M.

Solve: (a) As we move from experiment 1 to experiment 2, [A] is held constant and [B] is doubled. Thus, this pair of experiments shows how [B] affects the rate, allowing us to deduce the order of the rate law with respect to B. Because the rate remains the same when [B] is doubled, the concentration of B has no effect on the reaction rate. The rate law is therefore zero order in B (that is, n = 0).

In experiments 1 and 3, [B] is held constant so these data show how [A] affects rate. Holding [B] constant while doubling [A] increases the rate fourfold. This result indicates that rate is proportional to [A]2 (that is, the reaction is second order in A). Hence, the rate law is

SAMPLE EXERCISE Determining a Rate Law from Initial Rate Data

Using these data, determine (a) the rate law for the reaction, (b) the magnitude of the rate constant, (c) the rate of the reaction when [A] = 0.050 M and [B] = 0.100 M.

(b) Using the rate law and the data from experiment 1, we have

(c) Using the rate law from part (a) and the rate constant from part (b), we have

SAMPLE EXERCISE Using the Integrated First-Order Rate Law

The decomposition of a certain insecticide in water follows first-order kinetics with a rate constant of 1.45 yr–1 at 12°C. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 10–7 g/cm3. Assume that the average temperature of the lake is 12°C. (a) What is the concentration of the insecticide on June 1 of the following year? (b) How long will it take for the concentration of the insecticide to drop to 3.0 10–7 g/cm3?

Solve: (a) Using the appropriate integrated rate law, we have

We use the ln function on a calculator to evaluate the second term on the right, giving

Note that the concentration units for [A]t and [A]0 must be the same.

To obtain [insecticide]t = 1 yr, we use the inverse natural logarithm, or ex, function on the calculator:

SAMPLE EXERCISE Using the Integrated First-Order Rate Law

The decomposition of a certain insecticide in water follows first-order kinetics with a rate constant of 1.45 yr–1 at 12°C. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 10–7 g/cm3. Assume that the average temperature of the lake is 12°C. (a) What is the concentration of the insecticide on June 1 of the following year? (b) How long will it take for the concentration of the insecticide to drop to 3.0 10–7 g/cm3?

(b) Using the integrated rate law equation, with [insecticide]t = 3.0 10–7 g/cm3, gives

Solving for t gives

Temperature and Rate

• Generally, as temperature increases, so does the reaction rate.

• This is because k is temperature dependent.

Arrhenius Equation

Svante Arrhenius developed a mathematical relationship between k and Ea:

k = A e−Ea/RT

where A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction.

Arrhenius Equation

Taking the natural logarithm of both sides, the equation becomes

ln k = -Ea ( ) + ln A1RT

y = mx + b

Therefore, if k is determined experimentally at several temperatures, Ea can be calculated from the slope of a plot of ln k vs. 1/T.

Arrhenius Equation

12

1 11

2 TTR

E

k

k aln

• Ea can be evaluated in a nongraphical way if the rate constant is known at two different temperatures

• If we know the activation energy and rate constant at one temperature, then we can find the rate constant at a different temperature

SAMPLE EXERCISE 14.11 Determining the Energy of Activation

The following table shows the rate constants for the rearrangement of methyl isonitrile at various temperatures:

(a) From these data, calculate the activation energy for the reaction. (b) What is the value of the rate constant at 430.0 K?

Solve: (a) We must first convert the temperatures from degrees Celsius to kelvins. We then take the inverse of each temperature, 1/T, and the natural log of each rate constant, ln k. This gives us the table shown at the right:

SAMPLE EXERCISE

A graph of ln k versus 1/T results in a straight line.

The linear relationship is predicted by the Arrhenius equation giving a slope equal to – Ea/R.

The slope of the line is obtained by choosing two well-separated points, as shown, and using the coordinates of each:

Because logarithms have no units, the numerator in this equation is dimensionless. The denominator has the units of 1/T, namely, K–1. Thus, the overall units for the slope are K. The slope equals –Ea/R. We use the value for the molar gas constant R in units of J/mol-K.

(b) To determine the rate constant, k1, at T = 430.0 K, we can use the equation below with Ea = 160 kJ/ mol, and one of the rate constants and temperatures from the given data, such as k2 = 2.52 10–5s–1 and T2 = 462.9

12

1 11

2 TTR

E

k

k aln

SAMPLE EXERCISE

Thus,

Note that the units of k1 are the same as those of k2.

14.5

Reaction Mechanisms

The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions.

The sequence of elementary steps that leads to product formation is the reaction mechanism.

2NO (g) + O2 (g) 2NO2 (g)

N2O2 is detected during the reaction!

Elementary step: NO + NO N2O2

Elementary step: N2O2 + O2 2NO2

Overall reaction: 2NO + O2 2NO2

+

14.5

Elementary step: NO + NO N2O2

Elementary step: N2O2 + O2 2NO2

Overall reaction: 2NO + O2 2NO2

+

Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation.

An intermediate is always formed in an early elementary step and consumed in a later elementary step.

The molecularity of a reaction is the number of molecules reacting in an elementary step.

• Unimolecular reaction – elementary step with 1 molecule

• Bimolecular reaction – elementary step with 2 molecules

• Termolecular reaction – elementary step with 3 molecules

Unimolecular reaction A products rate = k [A]

Bimolecular reaction A + B products rate = k [A][B]

Bimolecular reaction A + A products rate = k [A]2

Rate Laws and Elementary Steps

14.5

Writing plausible reaction mechanisms:

• The sum of the elementary steps must give the overall balanced equation for the reaction.

• The rate-determining step should predict the same rate law that is determined experimentally.

The rate-determining step is the slowest step in the sequence of steps leading to product formation.

The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The reaction is believed to occur via two steps:

Step 1: NO2 + NO2 NO + NO3

Step 2: NO3 + CO NO2 + CO2

What is the equation for the overall reaction?

NO2+ CO NO + CO2

What is the intermediate?

NO3

What can you say about the relative rates of steps 1 and 2?

rate = k[NO2]2 is the rate law for step 1 so step 1 must be slower than step 2

14.5

Methyl bromide in the atomosphere

• Methyl bromide can react with ozone in the stratosphere• However, it takes some time for molecules from the earth’s

surface to diffuse through the troposphere to the stratosphere, where the ozone layer is located

• Decomposition in the troposphere competes with diffusion into the stratosphere: CH3Br + H2O -> CH3OH + HBr