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JEE Main2026Apr 4, Shift 2Chemistry
Q.

If 3.365 g of ethanol () burnt completely in a bomb calorimeter at 298.15K, the heat produced is 99.472 kJ. Thec of ethanol at 298.15 K is ____________ kJ mol. (Nearest integer)

Given:

Standard enthalpy of combustion of graphite kJ mol

Standard enthalpy of formation of water () kJ mol

Molar mass in g mol of C, H, O are 12, 1 and 16 respectively.

Solution

$\Delta U^\circ_{\text{combustion}} = \dfrac{-q}{\text{moles}} = \dfrac{-99.472}{3.365/46} \Rightarrow 1359.8\ \text{kJ/mol}$

$\mathrm{C_2H_5OH(\ell)} + 3\mathrm{O_2(g)} \rightarrow 2\mathrm{CO_2(g)} + 3\mathrm{H_2O(\ell)} \;;\; \Delta H_r^\circ = \Delta H_C^\circ$

$\Delta H_C^\circ = \Delta U^\circ + \Delta n_gRT$

$= -1359.8 + \dfrac{(-1)\times8.314\times298.15}{1000}$

$= -1362.27\ \text{kJ/mol}$

$-1362.27 = \left[2\times(-393.5) + 3\times(-285.8)\right] - \Delta H_f^\circ[\mathrm{C_2H_5OH(\ell)}]$

$-1362.27 = -787 - 857.4 - \Delta H_f^\circ[\mathrm{C_2H_5OH(\ell)}]$

$\Delta H_f^\circ[\mathrm{C_2H_5OH(\ell)}] = -2.82\times10^2\ \text{kJ/mol}$

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