1. A reaction by which ozone is destroyed in the stratosphere is
O3(g) + O(g) ® 2 O2(g)
The activation energy for ozone destruction is 17.1 kJ per mole of O3 consumed. Use standard enthalpies of formation
from Table B of Appendix II to calculate the enthalpy change for this reaction. Is the reaction endothermic or exothermic?
Then construct an energy diagram for this reaction. Draw vertical arrows to indicate the relative sizes of DHE, Ea(forward),
and Ea(reverse) for this reaction.
DHE = 2[DHfE(O2(g))] – { [DHfE(O(g))] + [DHfE(O3(g))] }
DHE = 0 – [(1)(249.2 kJ/mol) + (1)(142.7 kJ/mol)]
DHE = – 391.9 kJ/mol; since DHE < 0 this reaction is exothermic
2. Use the graph given below to determine the activation energy (Ea) for the gas-phase decomposition of hydrogen iodide.
2 HI (g) ® H2 (g) + I2 (g)
The slope of the line which results from a plot of ln k vs. 1/T equals –Ea/R.
Since slope = – Ea/R we have Ea = – slope x R
Ea = – (- 2.24 x 104 K) x 8.314 x 10-3 kJ/molCK = 186 kJ/mol
3. Rate constants for the reaction
2 N2O5(g) ® 4 NO2(g) + O2(g)
are given below.
k (at 25EC) = 3.38 x 10 – 5 min-1
k (at 45EC) = 4.30 x 10 – 4 min-1
Calculate the activation energy (Ea) and Arrhenius frequency factor (A) for this reaction.
Activation Energy
ln(k2/k1) = {(Ea/R)[(T2 – T1)/(T2T1)]}
k1 = 3.38 x 10 – 5 min – 1 T1 = 25EC + 273.15 = 298.15 K
k2 = 4.30 x 10 – 4 min – 1 T2 = 45EC + 273.15 = 318.15 K
R = 8.314 x 10 – 3 kJ/mol×K
Substitute these quantities into the following equation in order to calculate the activation energy.
Ea = [ln(k2/k1) x R x T2 x T1]/( T2 – T1)
Ea = 1.00 x102 kJ/mol
Arrhenius frequency factor (A):
Rearrange the usual form of the Arrhenius equation to solve for A
A = k/(e – Ea/RT)
Use the following quantities to determine A.
k = 3.38 x 10 – 5 min – 1
Ea = 1.00 x 102 kJ/mol
R = 8.314 x 10 – 3 kJ/mol×K
T = 298.15 K
A = 1.14 x 1013 min – 1
4. Determine the molecularity of each of the following elementary reactions.
.
(a) CH4(g) + 2 O2(g) ® CO2(g) + 2 H2O(g)
3 reactant particles Þ molecularity = 3 Þ termolecular
(b) O3(g) + O(g) ® 2 O2(g)
2 reactant particles Þ molecularity = 2 Þ bimolecular
(c) N2(g) + 3 H2(g) ® 2 NH3(g)
4 reactant particles Þ molecularity = 4
(d) O3(g) ® O2(g) + O(g)
1 reactant particle Þ molecularity = 1 Þ unimolecular
5. Consider the following energy diagram.
(a) How many elementary reactions are in the reaction mechanism?
There are three elementary reactions in the reaction mechanism: A ® B; B ® C; and C ® D.
(b) How many intermediates are formed in the reaction?
There are two intermediates (B and C) formed in the reaction.
(c) Which step (or elementary reaction) is rate-limiting?
The second step (B ® C) is rate-limiting because it has the largest activation energy value.
(d) Is the overall reaction endothermic or exothermic?
The overall reaction is endothermic because the energy of the products (D) is higher than the energy of the reactants (A).
6. At temperatures below 500 K, the reaction between carbon monoxide and nitrogen dioxide
CO(g) + NO2(g) ® CO2(g) + NO(g)
exhibits the experimental rate law
rate = k[NO2]2.
Which of the following mechanisms is consistent with this rate law? Also, identify any reactive intermediates present.
Remember that the slowest elementary reaction (or step) of a mechanism determines the rate of reaction. All of these mechanisms give the overall chemical equation.
(a) NO2 + CO ® CO2 + NO
predicted rate law: rate = k[NO2][CO] Þ not consistent
(b) NO2 + NO2 ® NO3 + NO (slow)
NO3 + CO ® NO2 + CO2 (fast)
predicted rate law: rate = k[NO2]2 Þ consistent
NO3 is a reactive intermediate.
(c) NO2 ® NO + O (slow)
CO + O ® CO2 (fast)
predicted rate law: rate = k[NO2] Þ not consistent
O is a reactive intermediate.
7. The decomposition of ozone in the stratosphere can be catalyzed by Cl atoms and is postulated to occur by the following
mechanism.
O3 (g) + Cl (g) ® O2 (g) + ClO (g) Ea = 2.1 kJ/mol
ClO (g) + O (g) ® Cl (g) + O2 (g) Ea = 0.4 kJ/mol
(a) What is the overall chemical reaction?
Add the two elementary reactions together to obtain the overall reaction.
O3 (g) + O (g) ® 2 O2 (g)
(b) What is the rate law predicted by this mechanism?
The predicted rate law is given by the rate law for the slow step in the mechanism. Remember, the larger the activation energy (Ea) the smaller the rate constant and the slower the rate of reaction. Since the first step of this mechanism has the larger Ea value it is the slow step. Therefore, the rate law predicted by this mechanism is
rate = k1[O3][ Cl ]
(c) Which species is a catalyst?
Cl is a catalyst since it is consumed in the first elementary reaction and then regenerated in the second elementary reaction.
(d) Which species is a reactive intermediate?
CIO is a reactive intermediate since it is produced in the first elementary reaction and then consumed in the second elementary reaction.
(e) Calculate the fraction of ozone molecules in the stratosphere at 220 K that have an energy greater than or equal to the activation energy for the uncatalyzed pathway. (Refer to problem 1 for the value of Ea for this process).
The fraction of particles at a given temperature that have energy equal to or greater than a specified energy is given by the following expression.
f = e – E/RT
f = e – Ea/RT
– Ea/RT = – 17.1 kJ/mol/[(8.314 x 10-3 kJ/molCK)(220 K)] = – 9.35
f = e – 9.35 = 8.7 x 10-5
(f) Calculate the fraction of ozone molecules in the stratosphere at 220 K that have an energy greater than or equal to the activation energy for the rate-determining step of the catalyzed pathway.
f = e – Ea/RT
– Ea/RT = – 2.1 kJ/mol/[(8.314 x 10-3 kJ/molCK)(220 K)] = – 1.15
f = e – 1.15 = .32
(g) How many times faster is the catalyzed pathway than the uncatalyzed pathway at 220 K? (Assume that the frequency factors for the catalyzed and uncatalyzed reactions are identical.)
This is given by the ratio fcatalyzed/funcatalyzed
.32/(8.7 x 10-5) » 3700 times faster
