Chemical Energetics: Question 7

Worked solution

Part (a): Why $\Delta H_f^{\ominus}$ of ethyne cannot be measured directly

Carbon and hydrogen do not combine directly to give a pure, measurable yield of ethyne under standard conditions. The reaction between them does not go to completion, and other products (such as other hydrocarbons) may also form. Because a single clean reaction of the elements to ethyne alone cannot be carried out, $\Delta H_f^{\ominus}$ must instead be calculated indirectly, using a Hess’s-law cycle built from enthalpies that can be measured directly, such as enthalpies of combustion.

Part (b): Energy cycle and calculation of $\Delta H_f^{\ominus}$

The cycle links the direct formation route with an indirect route via the shared combustion products, $\text{CO}_2\text{(g)}$ and $\text{H}_2\text{O(l)}$.

Direct route (the formation reaction itself): $2\text{C(graphite)} + \text{H}_2\text{(g)} \xrightarrow{\ \Delta H_f^{\ominus}\ } \text{C}_2\text{H}_2\text{(g)}$

Indirect route, both branches finishing at the same combustion products:

1. The elements are burned separately, releasing $2\Delta H_c^{\ominus}[\text{C}] + \Delta H_c^{\ominus}[\text{H}_2]$.
2. The ethyne formed by the direct route is then burned, releasing $\Delta H_c^{\ominus}[\text{C}_2\text{H}_2]$.

By Hess’s law, since both routes start at the same elements and finish at the same combustion products, the two routes give the same total enthalpy change: $\Delta H_f^{\ominus} + \Delta H_c^{\ominus}[\text{C}_2\text{H}_2] = 2\Delta H_c^{\ominus}[\text{C}] + \Delta H_c^{\ominus}[\text{H}_2]$

Rearranging for the unknown, $\Delta H_f^{\ominus}$: $\Delta H_f^{\ominus} = 2\Delta H_c^{\ominus}[\text{C}] + \Delta H_c^{\ominus}[\text{H}_2] - \Delta H_c^{\ominus}[\text{C}_2\text{H}_2]$

Substituting the values from the table: $\Delta H_f^{\ominus} = \left[2\times(-394) + (-286)\right] - (-1300)$

First total the combustion of the elements: $2\times(-394) = -788$ $-788 + (-286) = -1074$

Then subtract $\Delta H_c^{\ominus}[\text{C}_2\text{H}_2]$: $\Delta H_f^{\ominus} = -1074 - (-1300) = -1074 + 1300 = 226\ \text{kJ mol}^{-1}$

Check (independent recomputation): $2\times(-394)=-788$; $-788+(-286)=-1074$; $-1074-(-1300)=-1074+1300$. Since $1300>1074$, the result is positive: $1300-1074=226$, giving $+226\ \text{kJ mol}^{-1}$, consistent.

Part (c): Exothermic or endothermic?

The formation of ethyne from its elements is endothermic, because $\Delta H_f^{\ominus} = +226\ \text{kJ mol}^{-1}$ is positive, meaning energy is absorbed from the surroundings as ethyne forms.

Final answers

• (a) Carbon and hydrogen do not react cleanly/completely to give ethyne alone, so $\Delta H_f^{\ominus}$ must be found indirectly via Hess’s law.
• (b) $\Delta H_f^{\ominus}[\text{C}_2\text{H}_2] = \boxed{+226\ \text{kJ mol}^{-1}}$
• (c) Endothermic ($\Delta H_f^{\ominus} > 0$).