7. Consider the reaction:
BrO3- + 5 Br- + 6 H+ Ž 3 Br2 + 3 H2O
a. How might you define the rate of the reaction?
rate = -d[BrO3-] / dt = -d[Br-] / 5dt = -d[H+] / 6dt = d[Br2] / 3dt = d[H2O] / 3dt
b. How does the rate of disappearance of BrO3- anion compare with the rate of appearance of Br2 molecules?
Br2 molecules appear 3 times as fast as BrO3- ions disappear.
c. The following initial rate data were obtained. Assuming that the rate equation has the format: Rate = k [BrO3-]x [Br-]y [H+]z,
Find k, x, y, and z.
|1||0.10||0.10||0.10||1.2 x 10-3|
|2||0.20||0.10||0.10||2.4 x 10-3|
|3||0.10||0.30||0.10||3.6 x 10-3|
|4||0.20||0.10||0.20||9.6 x 10-3|
From 1 and 2: x = 1 (double conc leads to double rate)
From 1 and 3: y = 1 (triple conc leads to triple rate)
From 2 and 4: z = 2 (double cons leads to quadruple rate)
So rate = k [BrO3-] [Br-] [H+]2
from which k = 12 L3 mol-3 s-1.
So Rate = 12 L3 mol-3 s-1 [BrO3-] [Br-] [H+]2.
8. Use the half life relation to estimate whether the following process is 0th , 1st or 2nd order: State why you choose the answer you do.
Time to first half life (to 0.5 M) = 200 seconds.
Time from 0.5 M to 0.25 M (second half life) = 200 seconds.
The half life is thus constant which is true if the process is 1st order.
9. The half life for cobalt-60 is 5.26 years. Calculate what fraction of a sample of cobalt-60 will remain 12 years after it was formed. Radioisotopes decompose following first order kinetics.
For first order reactions:
ln (a / (a - x)) = kt, and t½ = (ln 2) / k
For cobalt-60, k = (ln 2) / 5.26 yrs = 0.1318 yrs-1.
So in this case: ln (initial amount / final amount) = 0.1318 yrs-1 * 12 yrs = 1.581.
Leading to (initial amount / final amount) = 4.860.
Fraction required = (final amount / initial amount) = 0.206.
10. Calculate Ea for a reaction having k = 2.61 x 10-5 at 190 ºC
and k = 3.02 x 10-3 at 250 ºC.
ln k = ln A - Ea / RT
ln (k2 / k1) = Ea (T1-1 - T2-1) / R
In this case:
ln (2.61 x 10-5 / 3.02 x 10-3) = Ea (523-1 - 463-1) / R
From which Ea = 159 kJ.
11. A possible mechanism for a gas-phase reaction is:
step 1 Ce4+ + Mn2+ ¾ Ce3+ + Mn3+ fast
step 2 Ce4+ + Mn3+ Ž Ce3+ + Mn4+ slow
step 3 Mn4+ + Tl+ Ž Mn2+ + Tl3+ fast
a. Write the balanced equation for the reaction.
2 Ce4+ + Tl+ Ž 2 Ce3+ + Tl3+
b. Indicate any intermediates in the reaction:
Mn3+, Mn4+. These two ions are formed in a step and react in a later step.
c. Is a catalyst involved, if so, what is it?
Yes, Mn2+. This ion is used in a step and reformed in a later step.
d. Write the theoretical rate law (in terms of measurable concentrations) for the reaction.
From the rds: rate = k [Ce4+] [Mn3+]
but the [Mn3+] is not directly measurable.
Use the approximation that the first, fast step is an equilibrium:
K = [Ce3+] [Mn3+] [Ce4+]-1 [ Mn2+]-1, from which [Mn3+] = K [Ce4+] [Mn2+] [Ce3+]-1
and rate = k [Ce4+] K [Ce4+] [Mn2+] [Ce3+]-1 = k' [Ce4+]2 [Mn2+] [Ce3+]-1
Integrated rate equations: Corresponding half-life equations
0th Order x = kt t½ = 2k / a
1st Order ln(a / (a-x)) = kt t½ = (ln 2) / k
2nd Order x / a(a-x) = kt t½ = 1 / (ak)
ln k = ln A - Ea / RT
Gas constant, R = 8.314 J K-1 mol-1.