COMEDK 2025 Evening Solved Paper

Concept:Hess's law: The enthalpy change of a reaction is the same regardless of the pathway, so the enthalpy of formation can be calculated from combustion enthalpies.Explanation:To find $\Delta H_f^\circ$ for $\mathrm{C_2H_6}(g)$, we consider the formation reaction: $2\mathrm{C}(\text{graphite}) + 3\mathrm{H_2}(g) \rightarrow \mathrm{C_2H_6}(g)$.Given data at $25^\circ\mathrm{C}$ and 1 atm:$\mathrm{H_2}(g) + \frac12\mathrm{O_2}(g) \rightarrow \mathrm{H_2O}(l)$, $\Delta H = -286.0\ \mathrm{kJ/mol}$.$\mathrm{C}(\text{graphite}) + \mathrm{O_2}(g) \rightarrow \mathrm{CO_2}(g)$, $\Delta H = -394.0\ \mathrm{kJ/mol}$.$\mathrm{C_2H_6}(g) + \frac72\mathrm{O_2}(g) \rightarrow 2\mathrm{CO_2}(g) + 3\mathrm{H_2O}(l)$, $\Delta H = -1560.0\ \mathrm{kJ/mol}$.We can combine these using Hess's law to get the desired enthalpy.Pathway 1: Form ethane from elements then combust it: $\Delta H = x + (-1560.0)$, where $x = \Delta H_f^\circ(\mathrm{C_2H_6})$.Pathway 2: Combust the elements directly: For carbon: $2 \times (-394.0) = -788.0\ \mathrm{kJ}$. For hydrogen: $3 \times (-286.0) = -858.0\ \mathrm{kJ}$. Total = $-788.0 + (-858.0) = -1646.0\ \mathrm{kJ}$.Both pathways lead to the same products ($2\mathrm{CO_2} + 3\mathrm{H_2O}$), so their enthalpies are equal:$x - 1560.0 = -1646.0$.Solving: $x = -1646.0 + 1560.0 = -86.0\ \mathrm{kJ/mol}$.Answer:The enthalpy of formation of ethane is $-86.0\ \mathrm{kJ/mol}$, which corresponds to option B.