COMEDK 2025 Morning Solved Paper

Concept:In a bomb calorimeter, heat released at constant volume equals the change in internal energy ($\Delta U$).The enthalpy change ($\Delta H$) is then corrected using the change in moles of gas and the gas constant.Explanation:First, calculate the heat released by the $0.4\ \mathrm{g}$ sample of propane.Heat capacity of calorimeter: $C = 24\ \mathrm{kJ\,K^{-1}}$; temperature rise: $\Delta T = 0.4\ \mathrm{K}$.$q_v = C \cdot \Delta T = 24\ \mathrm{kJ/K} \times 0.4\ \mathrm{K} = 9.6\ \mathrm{kJ}$.Combustion is exothermic, so heat released is negative: $q_v = -9.6\ \mathrm{kJ}$ for the sample.Find moles of propane: molar mass $\mathrm{C_3H_8} = 44\ \mathrm{g/mol}$.$n = \frac{0.4\ \mathrm{g}}{44\ \mathrm{g/mol}} = 0.00909\ \mathrm{mol}$.Internal energy change per mole: $\Delta U = \frac{-9.6\ \mathrm{kJ}}{0.00909\ \mathrm{mol}} \approx -1056\ \mathrm{kJ/mol}$.Balanced combustion reaction: $\mathrm{C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)}$.Change in moles of gas: $\Delta n_g = 3 - (1+5) = -3$.$RT$ at $300\ \mathrm{K}$: $8.314\ \mathrm{J\,mol^{-1}K^{-1}} \times 300\ \mathrm{K} = 2494\ \mathrm{J/mol} = 2.494\ \mathrm{kJ/mol}$.Now, $\Delta H = \Delta U + \Delta n_g RT$$\Delta H = -1056\ \mathrm{kJ/mol} + (-3) \times 2.494\ \mathrm{kJ/mol}$$\Delta H = -1056 - 7.482 = -1063.482\ \mathrm{kJ/mol} \approx -1063.5\ \mathrm{kJ/mol}$.Thus, the enthalpy change for the reaction (per mole of propane) is $-1063.5\ \mathrm{kJ}$.Answer:$-1063.5\ \mathrm{kJ}$ (Option B)