Unit 9 Kinetics
Practice Problems (II)
Key
1. SO2Cl2(g)  SO2(g) + Cl2(g)
The reaction is first order and the k = 2.2 x 10-5 s-1.
a. Calculate the time to react 20.% of the SO2Cl2?
ln([A]o/[A]t) = kt
ln(1.0/0.80) = (2.2 x 10-5 s-1)(t)  t = 1.0 x 104 s
b. The initial concentration of SO2Cl2 = 0.16 M, how much is left after six hours?
ln([A]o/[A]t) = kt
ln(0.16/x) = (2.2 x 10-5s-1)(6 x 3600 s)
[SO2Cl2] = 0.10 M
c. What is the half-life of this reaction?
t½ = ln2/k = ln2/(2.2 x 10-5 s-1) = 3.2 x 104 s
2. C6H12O3(g)  3 C2H4O(g)
The reaction is first order and k = 3.0 x 10-4 s-1. If we start with 0.25 M C6H12O3, what will its concentration be after 10. min?
ln([A]o/[A]t) = kt
ln(0.25/[A]t) = (3.0 x 10-4 s-1)(10 x 60 s)  [A]t = 0.21 M
3. The decomposition of HI, 2 HI(g)  H2(g) + I2(g), is a second order reaction with a rate constant of 2.8 x 10 -4 s-1.
a. If [HI]o = 1.0 M, how long will it take for the concentration to decrease by 25 %?
1/[A] – 1/[A]o = kt
1/(0.75) – 1/(1.0) = (2.8 x 10-4 s-1)t  t = 1.2 x 103 s
b. If [HI]o = 1.0 M, what [HI] after 2.0 hours?
1/[A] – 1/[A]o = kt
1/[HI] – 1/(1.0) = (2.8 x 10-4 s-1)(2.0 x 3600)s
[HI] = 0.33 M
4. Explain why a reaction occurs at a faster rate at higher temperature.
A higher fraction of the reactants have sufficient KE to exceed the Ea threshold at the higher temperature  more
successful collisions.
5. Consider the following mechanism:
step 1:
NO(g) + Br2(g)  NOBr2(g)
slow
step 2: NOBr2(g) + NO(g)  2 NOBr(g)
fast
a. Write the overall equation for the reaction.
2 NO(g) + Br2(g)  2 NOBr(g)
b. Write the rate law for this reaction.
rate = k[NO][Br2]
6. Consider the following mechanism.
step 1: CH4(g) + Cl(g)  CH3(g) + HCl(g)
step 2: CH3(g) + Cl2(g)  CH3Cl(g) + Cl(g)
a. Write the overall equation of the reaction.
CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g)
b. Identify the following.
Cl—early reactant and later product
catalyst
intermediate
CH3—early product and later reactant
slow step
Step 1—contains catalyst (Cl is catalyst, used in first step regenerated in last step)
rate = k[CH4][Cl]
(Cl2 isn’t in the rate law because it occurs after the rate determining
step)
rate law
7. Consider the following mechanism.
step 1: NO2(g) + SO2(g)  NO(g) + SO3(g)
step 2: NO(g) + ½ O2(g)  NO2(g)
a. Write the overall equation of the reaction.
SO2(g) + ½ O2(g)  SO3(g)
b. Identify the following.
catalyst
NO2
intermediate
NO
slow step
Step 1 (NO2 is the catalyst, it’s used in first step and regenerated in last step)
rate law
rate = k[NO2][SO2]
c. How do you know that this a homogeneous catalyst?
The catalyst is in the same phase as the reactant.
8.
a.
Hydrogen peroxide decomposes to water and oxygen according to the following reaction.
2 H2O2(aq)  2 H2O(l) + O2(g)
0.100 mol/L of H2O2 is consumed in 72.0 min.
(1) What is the average rate of H2O2 consumption?
(-0.100 M)/72.0 min = -1.39 x 10-3 M/min
(2) What is the average rate of O2 formation?
(1.39 x 10-3 M/min)/2 = 6.94 x 10-4 M/min
b. Initial rate determination at 40oC for the decomposition given the following data:
[H2O2] (mol/L)
0.100
0.200
0.300
Rate (mol/L•min)
1.93 x 10-4
3.86 x 10-4
5.79 x 10-4
(1) What is the order of the reaction?
(rate2/rate1) = ([H2O2]2/[H2O2]1)n
(3.86 x 10-4)/(1.93 x 10-4) = (0.200/0.100)n
2 = 2n, n = 1
(2) Write the rate law for the reaction.
rate = k[H2O2]
(3) Calculate the rate constant k.
rate = k[H2O2]
1.93 x 10-4 M/min = k(0.100 M)  k = 1.93 x 10-3 min-1
c. If a 30.0 % solution is kept at 40oC, how long will it take for it to become 10.0 % H2O2?
ln([A]o/[A]t) = kt
ln(30.0/10.0) = (1.93 x 10-3)t  t = 569 min
d. The rate constants for the uncatalyzed and catalyzed reactions at 25 oC are 5.21 x 10-4 min-1 and 2.95 x 108 min-1 respectively.
Calculate the half-life for the
(1) uncatalyzed reaction?
t½ = ln2/k = ln2/5.21 x 10-4 min-1 = 1330 min
(2) catalyzed reaction?
t½ = ln2/k = ln2/2.95 x 108 min-1 = 2.35 x 10-9 min
e.
Hydrogen peroxide in basic solution oxidizes iodide ions to iodine. The proposed mechanism is
H2O2(aq) + I-(aq)  HOI(aq) + OH-(aq)
slow
HOI(aq) + I-(aq)  I2(aq) + OH-(aq)
fast
(1) Write the overall redox reaction.
H2O2(aq) + 2 I-(aq)  I2(aq) + 2 OH-(aq)
(2) Write a rate law consistent with this proposed mechanism.
rate = k[H2O2][I-]
9.
(a) e
(b) c
(c) a
(d) b
(e) f
(f) endothermic
(g) exothermic (h) b, c, d
10. (a) initial rate will increase because concentration of [H] increases, more collisions.
(b) initial rate will decrease because concentration of both reactants are decreasing.
(c) rate increases because an new pathway is provided with a lower Ea.
(d) larger proportion of molecules will have sufficient KE to reach Ea threshold.
(i) lowers Ea
(j) no
11. (a) 2=2n n=1, therefore the rate is first order with respect to H2. 4=2n n=2, therefore the rate is second order with respect to
NO
(b) rate= k[H2][NO]2
−4
(c) 𝑘 =
1.8𝑥10
2
[.0010][.0060]
= 5000
𝐿2
2
𝑚𝑜𝑙 ∙𝑚𝑖𝑛
(d) ln[A]=-(5000)(180min) + ln[.25] = essentially there is zero H2 left after the elapsed time, because the rate constant k is
very large.
.693
(e) 𝑡1/2 =
= 1.39𝑥10−4 𝑚𝑖𝑛
𝑘
12. (i) the rate determining step is the step that occurs immediately after the fast step in this case the rxn at equilibrium, therefore
reaction (II) is the rate determining step. rate = k[N2O2][H2].
(ii) 2NO + 2H2 N2 + 2H2O