Chemistry 2710 Spring 2000 Practice Problems: Answers and hints

1.

(a)

$4.2\times 10^{-20}\,\mathrm{J}$

(b)

$3.9\times 10^{-5}$

(c)

99.996%

(d)

The CO bond is very strong (Draw a Lewis structure to see this.) so vibrations won't contribute much to the energy. The molecule is linear so the rotational contribution is $2\times\frac{1}{2}k_BT = k_BT$. The translational contribution is always $\frac{3}{2}k_BT$ so the average total energy is approximately $\frac{5}{2}k_BT = (3.453\times 10^{-23}\,\mathrm{J/K})T$, which is in excellent agreement with the equipartition value.

2.

(a)

(b)

We first need to calculate $\Delta\bar{E} = -156.1\,\mathrm{kJ/mol}$. Then $\bar{E}_a^- = 271\,\mathrm{kJ/mol}$.

(c)

For the forward reaction, $\Delta\bar{n}^\ddagger = -1$ so that $\Delta\bar{H}^{\ddagger +} = 110\,\mathrm{kJ/mol}$. For the reverse reaction, $\Delta\bar{n}^\ddagger = 0$ so that $\Delta\bar{H}^{\ddagger -} = 269\,\mathrm{kJ/mol}$.

3.

The reverse reaction is a radical recombination recombination so it is all downhill. Accordingly, for the dissociation reaction, $\Delta\bar{H}^\ddagger = \Delta\bar{H} = 2\Delta\bar{H}^\circ_{f\mathrm{(CH_3)... ...H}^\circ_{f\mathrm{(C_2H_6)}} = 2(145.69) - (-83.85) = 375.23\,\mathrm{kJ/mol}$. The transition state is not fully dissociated so $\Delta\bar{n}^\ddagger\approx 0$. Therefore $\bar{E_a} = \Delta\bar{H}^\ddagger + RT = 377.71\,\mathrm{kJ/mol}$.

4.

(a)

$\bar{E}_a = 220\,\mathrm{kJ/mol}$, $k_\infty = 1.9\times 10^{16}\,\mathrm{s}^{-1}$.

(b)

As an initial assumption, I'm going to guess that the transition state is not fully dissociated, i.e. that it is still like an azomethane molecule, only with weakened bonds. Then $\Delta\bar{n}^\ddagger = 0$ and $\Delta\bar{S}^\ddagger = 53\,\mathrm{J\,K^{-1}mol^{-1}}$. Since the entropy change is positive and relatively large, there is an increase in disorder on going to the transition state which is probably due to a considerable weakening of the C-N bonds. The transition state is therefore likely more like two or three molecules than like one. We can revise our estimate of $\Delta\bar{S}^\ddagger$ on this basis. If we take the extreme case ( $\Delta\bar{n}^\ddagger = 3-1=2$), we get $\Delta\bar{S}^\ddagger = 70\,\mathrm{J\,K^{-1}mol^{-1}}$. Reality lies somewhere between these two extremes so we can report $\Delta\bar{S}^\ddagger = 62\pm 9\,\mathrm{J\,K^{-1}mol^{-1}}$. Whatever value we compute for $\Delta\bar{S}^\ddagger$, we conclude that the transition state is much more loosely bound than the reactant, which makes good chemical sense since this is a dissociation reaction.

5.

$k_\infty = 1.7\times 10^{13}\,\mathrm{s}^{-1}$, assuming that room temperature is about $20^\circ\mathrm{C}$. The condition $\Delta\bar{S}^\ddagger=0$ separates reactions in which the transition state is more orderly than the reactants (negative $\Delta\bar{S}^\ddagger$, $k_\infty < 1.7\times 10^{13}\,\mathrm{s}^{-1}$, corresponding to bond formation) from reactions in which the transition state is less orderly than the reactants (positive $\Delta\bar{S}^\ddagger$, $k_\infty > 1.7\times 10^{13}\,\mathrm{s}^{-1}$, corresponding to bond breaking on the way to the transition state).