Q42 / 180 · COMEDK 2026 (9 May Shift 2) · Chemistry
The standard enthalpies of formation of CH$_4$(g), CO$_2$(g) and H$_2$O(l) are $-74.8$ kJ mol$^{-1}$, $-393.5$ kJ mol$^{-1}$ and $-285.8$ kJ mol$^{-1}$ respectively. Then the enthalpy change for the given reaction in kJ mol$^{-1}$ will be:
$2\text{CH}_4(g) + 4\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 4\text{H}_2\text{O}(l)$
$2\text{CH}_4(g) + 4\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 4\text{H}_2\text{O}(l)$
- A$-890.3$
- B$+1780.6$
- Correct answerC$-1780.6$
- D$+890.3$
Solution
The enthalpy of reaction is calculated using the standard enthalpies of formation of the products and reactants.
$\Delta_{r}H^{\circ} = \sum \Delta_{f}H^{\circ}(\text{products}) - \sum \Delta_{f}H^{\circ}(\text{reactants})$
For the given reaction:
$\Delta_{r}H^{\circ} = [2 \Delta_{f}H^{\circ}(\text{CO}_{2}) + 4 \Delta_{f}H^{\circ}(\text{H}_{2}\text{O})] - [2 \Delta_{f}H^{\circ}(\text{CH}_{4}) + 4 \Delta_{f}H^{\circ}(\text{O}_{2})]$
Substituting the given values:
$\Delta_{r}H^{\circ} = [2(-393.5) + 4(-285.8)] - [2(-74.8) + 4(0)]$
$\Delta_{r}H^{\circ} = [-787.0 - 1143.2] - [-149.6]$
$\Delta_{r}H^{\circ} = -1930.2 + 149.6 = -1780.6 \text{ kJ mol}^{-1}$
Answer: $-1780.6$
$\Delta_{r}H^{\circ} = \sum \Delta_{f}H^{\circ}(\text{products}) - \sum \Delta_{f}H^{\circ}(\text{reactants})$
For the given reaction:
$\Delta_{r}H^{\circ} = [2 \Delta_{f}H^{\circ}(\text{CO}_{2}) + 4 \Delta_{f}H^{\circ}(\text{H}_{2}\text{O})] - [2 \Delta_{f}H^{\circ}(\text{CH}_{4}) + 4 \Delta_{f}H^{\circ}(\text{O}_{2})]$
Substituting the given values:
$\Delta_{r}H^{\circ} = [2(-393.5) + 4(-285.8)] - [2(-74.8) + 4(0)]$
$\Delta_{r}H^{\circ} = [-787.0 - 1143.2] - [-149.6]$
$\Delta_{r}H^{\circ} = -1930.2 + 149.6 = -1780.6 \text{ kJ mol}^{-1}$
Answer: $-1780.6$
Hence, the correct answer is option: 'C'.