7. Consider the reaction:

BrO₃^- + 5 Br^- + 6 H⁺ Ž 3 Br₂ + 3 H₂O

[1]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

[1]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.

[6]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.

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.

[2]8. Use the half life relation to estimate whether the following process is 0^th , 1^st or 2^nd 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 1^st order.

[4]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 t_½ = (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.

[4]10. Calculate E_a 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 - E_a / RT

from which:

ln (k₂ / k₁) = E_a (T₁^-1 - T₂^-1) / R

In this case:

ln (2.61 x 10^-5 / 3.02 x 10^-3) = E_a (523^-1 - 463^-1) / R

From which E_a = 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

[2]a. Write the balanced equation for the reaction.

2 Ce⁴⁺ + Tl⁺ Ž 2 Ce³⁺ + Tl³⁺

[2]b. Indicate any intermediates in the reaction:

Mn³⁺, Mn⁴⁺. These two ions are formed in a step and react in a later step.

[2]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.

[4]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

Data:

Integrated rate equations: Corresponding half-life equations

0^th Order x = kt t_½ = 2k / a

1^st Order ln(a / (a-x)) = kt t_½ = (ln 2) / k

2^nd Order x / a(a-x) = kt t_½ = 1 / (ak)

Arrhenius equation

ln k = ln A - E_a / RT

Gas constant, R = 8.314 J K^-1 mol^-1.