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chapter 13: chemical kinetics
chapter 13
chemical kinetics
Problem Categories
Biological: 13.66, 13.115, 13.122, 13.127, 13.133, 13.138.
Conceptual: 13.29, 13.30, 13.37, 13.47, 13.65, 13.67, 13.68, 13.75, 13.76, 13.77, 13.81, 13.85, 13.86, 13.87, 13.88, 13.91, 13.92, 13.104, 13.105, 13.107, 13.110, 13.114, 13.131.
Descriptive: 13.97, 13.98, 13.119, 13.123.
Environmental: 13.101, 13.102.
Industrial: 13.72, 13.90, 13.111, 13.119, 13.125, 13.142.
Difficulty Level
Easy: 13.13, 13.17, 13.18, 13.25, 13.27, 13.28, 13.41, 13.42, 13.43, 13.44, 13.45, 13.55, 13.65, 13.70, 13.71, 13.77, 13.78, 13.82, 13.85, 13.91, 13.96, 13.99, 13.100, 13.113, 13.117, 13.124.
Medium: 13.14, 13.15, 13.16, 13.19, 13.26, 13.29, 13.30, 13.37, 13.38, 13.40, 13.46, 13.47, 13.56, 13.58, 13.66, 13.67, 13.68, 13.69, 13.72, 13.73, 13.76, 13.79, 13.83, 13.86, 13.87, 13.88, 13.90, 13.92, 13.93, 13.94, 13.95, 13.102, 13.103, 13.104, 13.105, 13.106, 13.107, 13.108, 13.109, 13.110, 13.114, 13.116, 13.118, 13.123, 13.128, 13.130, 13.131, 13.133, 13.134, 13.135.
Difficult: 13.20, 13.39, 13.57, 13.74, 13.75, 13.80, 13.81, 13.84, 13.89, 13.97, 13.98, 13.101, 13.111, 13.112, 13.115, 13.119, 13.120, 13.121, 13.122, 13.125, 13.126, 13.127, 13.129, 13.132, 13.136.
13.5 In general for a reaction aA + bB ® cC + dD
(a)
(b)
Note that because the reaction is carried out in the aqueous phase, we do not monitor the concentration of water.
13.6 (a)
(b)
13.7
(a)
(b)
13.8 Strategy: The rate is defined as the change in concentration of a reactant or product with time. Each “change in concentration” term is divided by the corresponding stoichiometric coefficient. Terms involving reactants are preceded by a minus sign.
Solution:
(a) If hydrogen is reacting at the rate of -0.074 M/s, the rate at which ammonia is being formed is
or
(b) The rate at which nitrogen is reacting must be:
Will the rate at which ammonia forms always be twice the rate of reaction of nitrogen, or is this true only at the instant described in this problem?
13.13 rate = = (3.0 ´ 10-4 /M×s)(0.26 M)(0.080 M) = 6.2 ´ 10-6 M/s
13.14 Assume the rate law has the form:
rate = k[F2]x[ClO2]y
To determine the order of the reaction with respect to F2, find two experiments in which the [ClO2] is held constant. Compare the data from experiments 1 and 3. When the concentration of F2 is doubled, the reaction rate doubles. Thus, the reaction is first-order in F2.
To determine the order with respect to ClO2, compare experiments 1 and 2. When the ClO2 concentration is quadrupled, the reaction rate quadruples. Thus, the reaction is first-order in ClO2.
The rate law is:
rate = k[F2][ClO2]
The value of k can be found using the data from any of the experiments. If we take the numbers from the second experiment we have:
Verify that the same value of k can be obtained from the other sets of data.
Since we now know the rate law and the value of the rate constant, we can calculate the rate at any concentration of reactants.
rate = k[F2][ClO2] = (1.2 M-1s-1)(0.010 M)(0.020 M) = 2.4 ´ 10-4 M/s
13.15 By comparing the first and second sets of data, we see that changing [B] does not affect the rate of the reaction. Therefore, the reaction is zero order in B. By comparing the first and third sets of data, we see that doubling [A] doubles the rate of the reaction. This shows that the reaction is first-order in A.
rate = k[A]
From the first set of data:
3.20 ´ 10-1 M/s = k(1.50 M)
k = 0.213 s-1
What would be the value of k if you had used the second or third set of data? Should k be constant?
13.16 Strategy: We are given a set of concentrations and rate data and asked to determine the order of the reaction and the initial rate for specific concentrations of X and Y. To determine the order of the reaction, we need to find the rate law for the reaction. We assume that the rate law takes the form
rate = k[X]x[Y]y
How do we use the data to determine x and y? Once the orders of the reactants are known, we can calculate k for any set of rate and concentrations. Finally, the rate law enables us to calculate the rate at any concentrations of X and Y.
Solution:
(a) Experiments 2 and 5 show that when we double the concentration of X at constant concentration of Y, the rate quadruples. Taking the ratio of the rates from these two experiments
Therefore,
or, x = 2. That is, the reaction is second order in X. Experiments 2 and 4 indicate that doubling [Y] at constant [X] doubles the rate. Here we write the ratio as
Therefore,
or, y = 1. That is, the reaction is first order in Y. Hence, the rate law is given by:
rate = k[X]2[Y]
The order of the reaction is (2 + 1) = 3. The reaction is 3rd-order.
(b) The rate constant k can be calculated using the data from any one of the experiments. Rearranging the rate law and using the first set of data, we find:
Next, using the known rate constant and substituting the concentrations of X and Y into the rate law, we can calculate the initial rate of disappearance of X.
rate = (10.6 M-2s-1)(0.30 M)2(0.40 M) = 0.38 M/s
13.17 (a) second order, (b) zero order, (c) 1.5 order, (d) third order
13.18 (a) For a reaction first-order in A,
Rate = k[A]
1.6 ´ 10-2 M/s = k(0.35 M)
k = 0.046 s-1
(b) For a reaction second-order in A,
Rate = k[A]2
1.6 ´ 10-2 M/s = k(0.35 M)2
k = 0.13 /M×s
13.19 The graph below is a plot of ln P vs. time. Since the plot is linear, the reaction is 1st order.
Slope = -k
k = 1.19 ´ 10-4 s-1
13.20 Let P0 be the pressure of ClCO2CCl3 at t = 0, and let x be the decrease in pressure after time t. Note that from the coefficients in the balanced equation that the loss of 1 atmosphere of ClCO2CCl3 results in the formation of two atmospheres of COCl2. We write:
ClCO2CCl3 ® 2COCl2
Time [ClCO2CCl3] [COCl2]
t = 0 P0 0
t = t P0 - x 2x
Thus the change (increase) in pressure (DP) is 2x - x = x. We have:
t(s) P (mmHg) DP = x
0 15.76 0.00 15.76 2.757 0.0635
181 18.88 3.12 12.64 2.537 0.0791
513 22.79 7.03 8.73 2.167 0.115
1164 27.08 11.32 4.44 1.491 0.225
If the reaction is first order, then a plot of ln vs. t would be linear. If the reaction is second order, a plot of 1/ vs. t would be linear. The two plots are shown below.
From the graphs we see that the reaction must be first-order. For a first-order reaction, the slope is equal to -k. The equation of the line is given on the graph. The rate constant is: k = 1.08 ´ 10-3 s-1.
13.25 We know that half of the substance decomposes in a time equal to the half-life, t1/2. This leaves half of the compound. Half of what is left decomposes in a time equal to another half-life, so that only one quarter of the original compound remains. We see that 75% of the original compound has decomposed after two half-lives. Thus two half-lives equal one hour, or the half-life of the decay is 30 min.
100% starting compound 50% starting compound 25% starting compound
Using first order kinetics, we can solve for k using Equation (13.3) of the text, with [A]0 = 100 and [A] = 25,
Then, substituting k into Equation (13.6) of the text, you arrive at the same answer for t1/2.
13.26 (a)
Strategy: To calculate the rate constant, k, from the half-life of a first-order reaction, we use
Equation (13.6) of the text.
Solution: For a first-order reaction, we only need the half-life to calculate the rate constant. From
Equation (13.6)
(b)
Strategy: The relationship between the concentration of a reactant at different times in a first-order reaction is given by Equations (13.3) and (13.4) of the text. We are asked to determine the time required for 95% of the phosphine to decompose. If we initially have 100% of the compound and 95% has reacted, then what is left must be (100% - 95%), or 5%. Thus, the ratio of the percentages will be equal to the ratio of the actual concentrations; that is, [A]t/[A]0 = 5%/100%, or 0.05/1.00.
Solution: The time required for 95% of the phosphine to decompose can be found using Equation (13.3) of the text.
13.27 (a) Since the reaction is known to be second-order, the relationship between reactant concentration and time is given by Equation (13.7) of the text. The problem supplies the rate constant and the initial (time = 0)
concentration of NOBr. The concentration after 22s can be found easily.
[NOBr] = 0.034 M
If the reaction were first order with the same k and initial concentration, could you calculate the concentration after 22 s? If the reaction were first order and you were given the t1/2, could you calculate the concentration after 22 s?
(b) The half-life for a second-order reaction is dependent on the initial concentration. The half-lives can be calculated using Equation (13.8) of the text.
For an initial concentration of 0.054 M, you should find . Note that the half-life of a second-order reaction is inversely proportional to the initial reactant concentration.
13.28
t = 3.6 s
13.29 (a) Notice that there are 16 A molecules at t = 0 s and that there are 8 A molecules at t = 10 s. The time of 10 seconds represents the first half-life of this reaction. We can calculate the rate constant, k, from the half-life of this first-order reaction.
(b) For a first-order reaction, the half-life is independent of reactant concentration. Therefore, t = 20 s represents the second half-life and t = 30 s represents the third half-life. At the first half-life (t = 10 s), there are 8 A molecules and 8 B molecules. At t = 20 s, the concentration of A will decrease to half of its concentration at t = 10 s. There will be 4 A molecules at t = 20 s. Because the mole ratio between A and B is 1:1, four more B molecules will be produced and there will be 12 B molecules present at t = 20 s.
At t = 30 s, the concentration of A will decrease to half of its concentration at t = 20 s. There will be 2 A molecules at t = 30 s. Because the mole ratio between A and B is 1:1, two more B molecules will be produced and there will be 14 B molecules present at t = 30 s.
13.30 (a) For a reaction that follows first-order kinetics, the rate will be directly proportional to the reactant concentration. In this case,
Rate = k[X]
Because the containers are equal volume, we can use the number of molecules to represent the concentration. Therefore, the relative rates of reaction for the three containers are:
(i) Rate = 8k
(ii) Rate = 6k
(iii) Rate = 12k
We can divide each rate by 2k to show that,
Ratio of rates = 4 : 3 : 6
(b) Doubling the volume of each container will have no effect on the relative rates of reaction compared to part (a). Doubling the volume would halve each of the concentrations, but the ratio of the concentrations for containers (i) – (iii) would still be 4 : 3 : 6. Therefore, the relative rates between the three containers would remain the same. The actual (absolute) rate would decrease by 50%.
(c) The reaction follows first-order kinetics. For a first-order reaction, the half-life is independent of the initial concentration of the reactant. Therefore, the half-lives for containers (i), (ii), and (iii), will be the same.
13.37 (1) The slope of a plot of ln k vs. 1/T equals –Ea/R. The plot with the steeper slope will have the greater activation energy. The reaction with the greater activation energy is the one represented by the red line in the graph.
(2) The slope of a plot of ln[A]t vs. t equals –k. The plot with the steeper slope has the greater rate constant (brown line) and therefore corresponds to the reaction that is run at the higher temperature. Reaction rate increases with increasing temperature.
13.38 Strategy: A modified form of the Arrhenius equation relates two rate constants at two different temperatures [see Equation (13.14) of the text]. Make sure the units of R and Ea are consistent. Since the rate of the reaction at 250°C is 1.50 ´ 103 times faster than the rate at 150°C, the ratio of the rate constants, k, is also 1.50 ´ 103 : 1, because rate and rate constant are directly proportional.
Solution: The data are: T1 = 250°C = 523 K, T2 = 150°C = 423 K, and k1/k2 = 1.50 ´ 103. Substituting into Equation (13.14) of the text,
Ea = 1.35 ´ 105 J/mol = 135 kJ/mol
13.39 First, we use the Arrhenius equation to calculate the frequency factor (A) for each reaction.