Q30 / 180 · COMEDK 2025 (10 May Shift 3) · Chemistry
The enthalpies of combustion of $\mathrm{H}_2, \mathrm{C}$ (graphite) and $\mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})$ are $-286.0,-394.0$ and $-1560.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $25^{\circ} \mathrm{C}$ and 1 atm pressure. The enthalpy of formation of ethane is:
- A$-97.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- B$-78.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- Correct answerC$-86.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- D$-92.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Solution
The combustion reactions for the given substances are:
1) $\mathrm{H}_2(g) + \dfrac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2\mathrm{O}(l)$, $\Delta H_1 = -286.0 \text{ kJ mol}^{-1}$
2) $\mathrm{C}(s) + \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g)$, $\Delta H_2 = -394.0 \text{ kJ mol}^{-1}$
3) $\mathrm{C}_2\mathrm{H}_6(g) + \dfrac{7}{2} \mathrm{O}_2(g) \rightarrow 2\mathrm{CO}_2(g) + 3\mathrm{H}_2\mathrm{O}(l)$, $\Delta H_3 = -1560.0 \text{ kJ mol}^{-1}$
The formation reaction for ethane is:
2$\mathrm{C}(s) + 3\mathrm{H}_2(g) \rightarrow \mathrm{C}_2\mathrm{H}_6(g)$, $\Delta H_f = ?$
Using Hess's Law, we can express the formation reaction as:
$\Delta H_f = 2 \times (\Delta H_2) + 3 \times (\Delta H_1) - (\Delta H_3)$
Substituting the given values:
$\Delta H_f = 2(-394.0) + 3(-286.0) - (-1560.0)$
$\Delta H_f = -788.0 - 858.0 + 1560.0$
$\Delta H_f = -1646.0 + 1560.0$
$\Delta H_f = -86.0 \text{ kJ mol}^{-1}$
Answer: $-86.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
1) $\mathrm{H}_2(g) + \dfrac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2\mathrm{O}(l)$, $\Delta H_1 = -286.0 \text{ kJ mol}^{-1}$
2) $\mathrm{C}(s) + \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g)$, $\Delta H_2 = -394.0 \text{ kJ mol}^{-1}$
3) $\mathrm{C}_2\mathrm{H}_6(g) + \dfrac{7}{2} \mathrm{O}_2(g) \rightarrow 2\mathrm{CO}_2(g) + 3\mathrm{H}_2\mathrm{O}(l)$, $\Delta H_3 = -1560.0 \text{ kJ mol}^{-1}$
The formation reaction for ethane is:
2$\mathrm{C}(s) + 3\mathrm{H}_2(g) \rightarrow \mathrm{C}_2\mathrm{H}_6(g)$, $\Delta H_f = ?$
Using Hess's Law, we can express the formation reaction as:
$\Delta H_f = 2 \times (\Delta H_2) + 3 \times (\Delta H_1) - (\Delta H_3)$
Substituting the given values:
$\Delta H_f = 2(-394.0) + 3(-286.0) - (-1560.0)$
$\Delta H_f = -788.0 - 858.0 + 1560.0$
$\Delta H_f = -1646.0 + 1560.0$
$\Delta H_f = -86.0 \text{ kJ mol}^{-1}$
Answer: $-86.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Hence, the correct answer is option: 'C'.