Chapter 14 Assigned Homework Problems

Exercise 6, 12, 14, 16, 22, 24, 26, 28, 30, 32, 42, 44, 48, 56, 60, 62, 66

6. The average rate of reaction is defined as:

For the time interval between t = 0.0 min and t = 54.0 min:

For the time interval between t = 54.0 min and t = 107.0 min:

For the time interval between t = 107.0 min and t = 215.0 min:

For the time interval between t = 215.0 min and t = 430.0 min:

12. (a) Given the following reaction stoichiometry:

C₂H₄(g) + 3 O₂(g) ¾ ¾ ® 2 CO₂(g) + 2 H₂O(g)

The reaction rate is defined as the following in terms of each species in the reaction:

The problem gives the rate of reaction as:

Using the relationships above, the rates in change in concentrations of CO₂and H₂O are:

(b) Given the reaction stoichiometry:

N₂H₄(g) + H₂(g) ¾ ¾ ® 2 NH₃(g)

The reaction rate is defined as the following in terms of each species in the reaction:

The problem gives the rate of the reaction as 45 torr/hr. Solving for the rate of change of NH₃ pressure:

During the course of this reaction, 2 moles of gaseous reactants form 2 moles of gaseous products. Because there are equal numbers of moles of gas on both sides of the reaction, the total number of moles of gas will not change as the reaction proceeds. Therefore, the total pressure in the vessel will not change as the reaction proceeds.

14. (a) Since [A] does not appear in the rate law, doubling [A] will not change the rate. Therefore, the rate will not change. The only change that affects the value of the rate constant is temperature. Therefore, doubling [A] does not change the rate constant.

(b) The reaction is zero order with respect to [A], second order with respect to [B] and second order overall.

(c) To calculate the units of the rate constant, take the rate law, solve for k, and look at the units:

16. (a) A reaction that is first order in H₂ and second order in NO will have a rate law that looks like:

Rate = k[H₂][NO]²

(b) Placing value for the rate constant and concentrations into the rate law, the reaction rate is equal to:

(c) Because the reaction is second order in NO, when [NO] is doubled, the reaction rate will increase by a factor of 2² = 4. Therefore, the reaction rate is equal to 6.0 M s^-1 (= 1.5 M s^-1 × 4).

22. (a) Judging from the table headings, the rate law is of the form:

Rate = k[ClO₂]^m[OH¯]ⁿ

To find the value of the rate exponent m, find two experiments in which only [ClO₂] is changing. Looking at experiments 2 and 1 (in that order), [ClO₂] increases from 0.020 M to 0.060 M, which is a factor of 3 . The rate increases from 0.00276 M/s to 0.00248 M/s, which is a factor of 9 . [OH¯] is the same in both experiments 2 and 1. According to the rate law, 3^m = 9. Therefore m = 2.

To find the value of the rate exponent n, find two experiments in which only [OH¯] is changing. Looking at experiments 2 and 3, [OH¯] increases from 0.030 M to 0.090 M, which is a factor of 3 . The rate increases from 0.00276 M/s to 0.00828 M/s, which is a factor of 3. [ClO₂] is the same in both experiments. According to the rate law, 3ⁿ = 3. Therefore, n = 1.

Writing the rate law for the reaction:

Rate = k[ClO₂]²[OH¯]

(b) Solve the rate law for the rate constant:

Using the data from experiment 1:

(c) Placing the rate constant and the given concentrations into the rate law, the rate of reaction can be calculated:

24. (a) The table headings imply that the rate law is of the form: Rate = k[BF₃]^m[NH₃]ⁿ.

Looking at experiments 5 and 4 (in that order):

[BF₃] doubles from 0.175 M to 0.350 M

[NH₃] remains constant at 0.100 M

Initial rate doubles from 0.0596 M/s to 0.1193 M/s

Using the above rate law: 2^m = 2. Therefore, m = 1

Looking at experiments 2 and 1 (in that order):

[BF₃] remains constant at 0.250 M

[NH₃] doubles from 0.125 M to 0.250 M.

Initial rate doubles from 0.1065 M/s to 0.2130 M/s

Using the above rate law: 2ⁿ = 2. Therefore, n = 1

The rate law is: Rate = k[BF₃][NH₃]

(b) This rate law is overall second order (the sum of 1 and 1 is equal to 2).

(c) Placing the data from experiment 1 into the rate law and solving for the rate constant, k:

26. (a) The table headings imply that the rate law is of the form: Rate = k[S₂O₈^2-]^m[I¯]ⁿ

Looking at experiments 1 and 2:

[S₂O₈^2-] increases by a factor of 1.5

[I¯] remains constant at 0.036 M.

The Initial rate increases by a factor of 1.5

Using the above rate law: 1.5^m = 1.5. Therefore m = 1.

Looking at experiments 1 and 3:

[S₂O₈^2-] doubles. This will double the rate from experiment 1 to 5.2 × 10^-6 M/s

[I¯] increases by a factor of 1.5

The rate increases by a factor of 1.5 due to the increase in [I¯]

Using the above rate law: 1.5ⁿ = 1.5. Therefore n = 1.

The rate law is: Rate = k[S₂O₈^2-][I¯]

(b) Calculate the value of the rate constant by solving the rate law for k using the data from each experiment. From experiment 1:

From experiment 2:

From experiment 3:

From experiment 4:

The average rate constant is equal to:

(c) Using the stoichiometry of the reaction, the rate of disappearance of I¯ is three times the rate of disappearance of S₂O₈^2-.

(d) The rate law measures the rate of disappearance of S₂O₈^2-. Calculating the rate:

The rate of disappearance of I¯ is three times this number, or 7.2 × 10^-6 M/s.

28. (a) A graph of 1/[A] versus t (time) will yield a straight line for a second order reaction.

(b) For a first order reaction, the half-life, t_1/2, is dependent on the value of the rate constant, k, only. For a second order reaction, the half-life is dependent on both the value of the rate constant, k, and the initial concentration, [A]₀.

30. (a) For a first order reaction, the half-life is related to the rate constant by the following expression:

(b) Using the relationship between half-life and the rate constant and solving for the rate constant:

32. The relationship between the concentration of N₂O₅ at any time, t, [N₂O₅]_t, and the initial concentration of N₂O₅, [N₂O₅]₀, is:

Calculating the initial concentration of N₂O₅:

(a) Given that k = 6.82 × 10^-3 s^-1, the concentration of N₂O₅ at t = 2.5 min = 150 s can be calculated using the equation above:

To calculate the number of moles of N₂O₅ remaining, multiply this concentration by the volume:

Moles of N₂O₅ = (0.0045 M)(2.0 L) = 0.0090 mol

After 2.5 min, 0.0090 mol of N₂O₅ remain.

(b) Calculate the concentration when there are 0.010 mol of N₂O₅ in 2.0 L:

Using the equation for a first order reaction and solving for t:

Converting 130 seconds to minutes:

(c) The half-life is equal to:

42. (a) The fraction of molecules that has an energy greater than E_a is given by equation 14.18:

Given that E_a = 160.0 kJ/mol = 1.600 × 10⁵ J/mol and T = 500 K:

(b) At a temperature of 510 K, the fraction is equal to:

The ratio of the fraction at 510 K to the fraction at 500 K is:

44. (a) The energy profile for the reaction will look like the following:

(b) The activation energy for the reverse reaction is the energy change from the products to the top of the hill which is equal to 73 kJ (= 66 + 7)

48. The relationship between activation energy and the rate constant is given by the Arrhenius equation:

Definine k_A as the rate constant for reaction A and k_B as the rate constant for reaction B, at 30°C (or 303 K) , the following is true given the activation energies of each reaction (converted to units of J/mol):

Setting k_A = k_B:

The expressions for k_A and k_B at 60°C (or 333 K):

Dividing k_A by k_B then substituting 3280A_B for A_A:

56. (a) Since there are two reactants, this is a bimolecular step. The rate is equal to: Rate = k[NO]²

(b) There is only one reactant, this is a unimolecular step. The rate is equal to:

Rate =k

(c) There is only one reactant, this is a unimolecular step. The rate is equal to: Rate = k[SO₃]

60. (a) Adding up the elementary steps:

NO(g) + NO(g) ¾ ¾ ® N₂O₂(g)

N₂O₂(g) + H₂(g) ¾ ¾ ® N₂O(g) + H₂O(g)

N₂O₂(g) appears on both sides, so it cancels. The sum of these two steps is:

2 NO(g) + H₂(g) ¾ ¾ ® N₂O(g) + H₂O(g)

(b) The rate law for the first step is: Rate = k[NO]². The rate law for the second step is:

Rate = k[N₂O₂][H₂].

(c) N₂O₂(g) is an intermediate in this reaction.

(d) If the observed rate law is rate = k[NO]²[H₂], then the second step must be the slow rate-determining step, and the first step is fast and reversible. If the first step is fast and reversible, we can assume it is in equilibrium and:

k₁[NO]² = k_-1[N₂O₂]

where k₁ and k_-1 are the rate constants for the forward and reverse of step 1. Solving for [N₂O₂]:

The rate law for the second step can now be written as:

62. (a) The overall reaction is the sum of the first and second steps plus two times the third step.

(b) The observed rate law is: rate = k[HBr][O₂]. This is also the rate for the first step. Therefore, the first step must be rate determining.

(c) The intermediates are HOOBr and HOBr.

(d) Intermediates are typically unstable molecules that exist in very low concentrations. Not being able to detect an intermediate does not necessarily disprove a mechanism.

66. (a) The overall reaction is the sum of twice the first step plus the second step. 2 must multiply the first step since only one NO₂ is produced in the first step but two NO₂'s are reacted in the second step. The overall reaction is:

2 N₂O(g) ¾ ¾ ® 2 N₂(g) + O₂(g)

(b) A catalyst appears as a reactant in a early step and then is regenerated as a product in a following step. An intermediate appears as a product in an early step and is consumed as a reactant in a following step.

(c) Not being able to detect the intermediate NO₂ in measurable quantities does not rule out the proposed mechanism. NO₂ may be consumed almost immediately upon being produced. In other words, the second step of the mechanism may be very fast which would result in NO₂ being consumed at the instant it is being made.