7. Consider the reaction:

BrO_{3}^{-} + 5 B^{}r^{- } + 6 H^{+} Ž 3 Br_{2} + 3 H_{2}O

[1]a. How might you define the rate of the reaction?

rate = -d[BrO_{3}^{-}] / dt = -d[Br^{-}] / 5dt = -d[H^{+}] / 6dt = d[Br_{2}] / 3dt = d[H_{2}O] / 3dt

[1]b. How does the rate of disappearance of BrO_{3}^{-} anion compare with the rate of appearance of Br_{2} molecules?

Br_{2} molecules appear 3 times as fast as BrO_{3}^{-} ions disappear.

[6]c. The following initial rate data were obtained. Assuming that the rate equation has the format: Rate = k [BrO_{3}^{-}]^{x} [Br^{-}]^{y }[H^{+}]^{z},

Find k, x, y, and z.

Experiment # |
[BrO_{3}^{-}] initial
(mol/L) |
[Br^{-}] initial
(mol/L) |
[H^{+}] initial
(mol/L) |
Initial rate
(mol/L.s) |

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 [BrO_{3}^{-}] [Br^{-}] [H^{+}]^{2}

from which k = 12 L^{3} mol^{-3} s^{-1}.

So Rate = 12 L^{3} mol^{-3} s^{-1 }[BrO_{3}^{-}] [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.

[reactant] (M) | 1.00 | 0.75 | 0.60 | 0.50 | 0.37 | 0.25 |

time (s) | 0.0 | 83.0 | 147.4 | 200.0 | 286.9 | 400.0 |

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 (k_{2} / k_{1}) = E_{a} (T_{1}^{-1} - T_{2}^{-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 Ce^{4+} + Mn^{2+} _{}¾ Ce^{3+} + Mn^{3+} fast

step 2 Ce^{4+} + Mn^{3+} _{}Ž Ce^{3+} + Mn^{4+} slow

step 3 Mn^{4+} + Tl^{+} Ž Mn^{2+} + Tl^{3+} fast

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

2 Ce^{4+} + Tl^{+} Ž 2 Ce^{3+} + Tl^{3+}

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

Mn^{3+}, Mn^{4+}. 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, Mn^{2+}. 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 [Ce^{4+}] [Mn^{3+}]

but the [Mn^{3+}] is not directly measurable.

Use the approximation that the first, fast step is an equilibrium:

K = [Ce^{3+}] [Mn^{3+}] [Ce^{4+}]^{-1} [ Mn^{2+}]^{-1}, from which [Mn^{3+}] = K [Ce^{4+}] [Mn^{2+}] [Ce^{3+}]^{-1}

and rate = k [Ce^{4+}] K [Ce^{4+}] [Mn^{2+}] [Ce^{3+}]^{-1} = k' [Ce^{4+}]^{2} [Mn^{2+}] [Ce^{3+}]^{-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}.