Reaction Rate and Rate Law

Chemistry 521 Exam I, Spring Quarter
7:30 - 9:00 PM, 25 April 2000

NAME
1. [10 points] What are the concentrations of HSO− , SO2− , and H+ in a 0.20 M KHSO4 solution? (Hint:
4
4
H2 SO4 is a strong acid; Ka = 1.3 × 10−2 for HSO− .)
4

2. [15 points] Calculate the pH of 1.00 L of the buffer 1.00 M CH3 COONa/1.00 M CH3 COOH (pKa = 4.74) before and after the addition of (a) 0.080 moles NaOH and (b) 0.12 moles HCl. (Assume there is no change in volume).

3. [10 points] The following reaction is found to be first order in A:
A −→ B + C
If half of the starting quantity of A is used up after 56 seconds, calculate the fraction that will be used up after 6.0 minutes.

4. [15 points] The rate law for the decomposition of ozone to molecular oxygen
3 O3 (g) −→ 3 O2 (g) is rate = k

[O3 ]2
.
[O2 ]

The mechanism for this process is k1 O3

O + O2 k−1 k

2
O + O3 −→ 2 O2

Derive a rate law from these elementary steps. Clearly state the assumptions you use in the derivation.
Explain why the rate decreases with increasing O2 concentration.

5. [15 points] Consider the following parallel reactions k 1
A −→ B

k

2
A −→ C

The activation energies are 45.3 kJ/mole for k1 and 69.8 kJ/mole for k2 . If the rate constants are equal at 320 K, at what temperature will k1 /k2 = 2.00?

6. [15 points] The pre-exponential factor and activation energy for the unimolecular reaction
CH3 NC(g) −→ CH3 CN(g) are 4.0 × 1013 s−1 and 272 kJ/mole, respectively. Calculate the values of ∆S ◦‡ , ∆H ◦‡ , and ∆G◦‡ at 300
K.

7. [20 points] An enzyme-catalyzed reaction (KM = 2.7 × 10−3 M) is inhibited by a competitive inhibitor
I (KI = 3.1 × 10−5 M). Suppose that the substrate concentration is 3.6 × 10−4 M. How much of an inhibitor is needed for 65% inhibition? How much does the substrate concentration have to be increased to reduce the inhibition to 25%?

Constants
R = 8.314 J-K−1 -mol−1
N0 = 6.022 × 1023 mol−1

kB = 1.381 × 10−23