Chem 150
Suggested Problems Week 10
12.16 Rate constants for the decomposition of gaseous dinitrogen pentoxide are 3.7 * 105 -1
s at 25oC and 1.7 * 10-3 s-1 at 55oC.
2 N2O5 (g)  4 NO2 (g) + O2 (g)
a. What is the activation energy for this reaction in kJ/mol?
b. What is the rate constant at 35oC?
12.84 Reaction of the anti-cancer drug cisplatin with water is described by the equation
Pt(NH3)2Cl2 (aq) + H2O(l)  Pt(NH3)2(H2O)Cl+(aq) + Cl-(aq)
The rate of this reaction increases by a factor of 15 on raising the temperature
from 25oC to 50oC. What is the value of the activation energy in kJ/mol?
12.90 What effect does a catalyst have on the rate, mechanism, and activation energy of
a chemical reaction?
12.115 I’m not going to solve this one because it solves a lot like the other ones above,
but I like the idea. The problem gives you two sets of experimental data: temperature in
Celsius, initial [reactant], and rate of decomposition. You’ll use the rate and initial
concentration data to solve for k at each of the two temperatures, which you can them use
in the logarithmic form of the Arrhenius equation.
13.46 If Kc = 7.5 * 10-9 at 1000 K for the reaction N2 (g) + O2 (g) (insert arrow both
ways) 2 NO (g), what is Kc at 1000 K for the reaction 2 NO (g) (insert arrow both ways)
N2(g) + O2 (g)?
13.54 At 500 K, Kc = 0.575 for the reaction PCl5 (g) (insert arrows) PCl3 (g) + Cl2 (g).
What is Kp at the same temperature?
13.61 Which of the following reactions goes almost all the way to completion, and
which proceeds “hardly at all” (I feel like this is Southern grammar)?
a. N2 (g) + O2 (g) (arrows) 2 NO (g)
b. 2 NO (g) + O2 (g) (arrows) 2 NO2 (g)
Kc = 2.7 * 10-18
Kc = 6.0 * 1013
12.16 Use the Arrhenius equation
ln (k2/k1) = (-Ea/R)(1/T2 – 1/T1)
a. where k2 = 1.7 * 10-3 s-1 T2 = 328 K
k1 = 3,7 * 10-5 s-1
T1 = 298 K
ln (1.7 * 10-3 s-1/3.7 * 10-5 s-1) = (-Ea/8.314 J/K*mol)(1/328 K – 1/298 K)
3.8275 = (-Ea/8.314 J/K*mol)(-3.069 * 10-4 K-1)
-Ea/8.314 J/K*mol = -12471.48 K
-Ea = -103688 J/mol = -104 kJ
Ea = 104 kJ/mol
b. This time, we know Ea and we need to sole for k2 if T2 = 35oC
ln k2 – ln k1 = (-Ea/R)(1/T2 – 1/T1)
And ln k2 = (-Ea/R)(1/T2 – 1/T1) + ln k1
If k1 = 3.7 * 10-5 s-1
T1 = 298 K
T2 = 308 K
ln k2 = (-104 kJ/mol/8.314 * 10-3 kJ/K*mol)(1/308 K – 1/298 K) + ln 3.7 *
10-5 s-1
= -8.84172
k2 = e-8.84172
= 1.4 * 10-4 s-1
12.84 If T1 =25oC = 298 K and T2 = 50oC = 323 K
then we can also say that k1 = 1 and k2 = 15 (since the value of k increases 15fold)
Use these numbers in the same equation used in the previous problem, solving for
the activation energy.
ln (k2/k1) = (-Ea/R)(1/T2 – 1/T1)
ln 15 = (-Ea/8.314 J/K*mol)(1/323 K – 1/298 K)
2.70805 = (-Ea/8.314 J/K*mol)(-2.5973 * 10-4 K-1)
-10426 K = (-Ea/8.314 J/K*mol)
Ea = 87 kJ/mol
-Ea = -86685 J/mol
12.90 A catalyst increases the rate of a chemical reaction by changing the mechanism in
some way – the mechanism is different from that of the original reaction. The
activation energy of a catalyzed reaction is lower than the uncatalyzed.
13.46 Kc for the reverse reaction = 1/Kc for the forward reaction
=1/7.5 * 10-9 = 1.3 * 108
13.54 All reactants and products are gaseous. Therefore, Δn = +1.
Kp = Kc(RT) = 0.575(0.08206)(500 K) = 23.6
13.61 a. This is a small Kc value. Meaning – at equilibrium, more reactant, less product.
The reaction has not proceeded very far at equilibrium.
b.This is a larger Kc. Meaning – at equilibrium, more product, less reactant. The
reaction has proceeded more toward product at equilibrium.
In both of the above cases, you can think of the equation for Kc, in which
products are the numerator and reactants are the denominator. Larger
number on top will yield a larger Kc, etc.