Kinetics
2000D
By: Charles Liao
2000D
a. Referring to the data in the table below, calculate
the standard enthalpy change, for the reaction at
25˚C.
Standard
enthalpy of
formation, DH
at 25°C (kJ
mol-1)
O3(g)
NO(g)
NO2(g)
143 kJ
mol-1
90 kJ
mol-1
33 kJ
mol-1
O3(g) + NO(g)  O2(g) + NO2(g)
∆H =33 (kJ mol-1) -(143+90) (kJ mol-1) =-200 (kJ mol-1)
2000D
b.
Make a qualitative prediction about the
magnitude of the standard entropy change,
∆S˚, for the reaction at 25˚C. Justify your
answer.
A very small change is present because both
sides of the equation contain 2 moles of gas.
Because Entropy is the measure of disorder and
molecules in their gas state contain the least
amount of order, very little change is observed
from 2 moles to 2 moles.
2000D
c. On the basis of your answers to parts (a) and (b),
predict the sign of the standard free-energy
change, for the reaction at 25˚C. Explain your
reasoning.
Because the reaction’s ∆S˚ is very little and the
equation to determine free energy change is
∆G˚= ∆H˚-T ∆S˚, it can be assumed that with a
negative ∆H˚ and at 25˚C or 298˚K, that the
reaction is spontaneous. By having a spontaneous
reaction, ∆G is inherently Negative
2000D
d. Use the information in the table below to write the
rate-law expression for the reaction, and explain how
you obtained your answer.
k= [O3][NO]
Based on Experiments 1 and 2, by doubling the [NO]
concentration, the rate doubles. Therefore the
reaction is first order with respect for [NO]. The
doubling of the [O3] concentration also doubles the
rate therefore making the reaction a first order with
respect to [O3] as well.
Experiment
Number
Initial [O3]
(mol L-1)
Initial [NO]
(mol L-1)
Initial Rate of Formation of
[NO2]
(mol L-1 s-1)
1
0.001
0.001
x
2
0.001
0.002
2x
3
0.002
0.001
2x
4
0.002
0.002
4x
2000D
e. The following three-step mechanism is proposed
for the reaction. Identify the step that must be
the slowest in order for this mechanism to be
consistent with the rate-law expression derived in
part (d).
 Step I:
O3 + NO  O + NO3
 Step II:
O+ O3  2 O2
 Step III:
NO3 + NO  2 NO2
In Part D, it was found that the Rate-law expression
was k= [O3][NO]. Therefore the slow step of the
reaction must be Step 1. The Slow step is the
rate-determining step in equations.