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Chemistry · Thermodynamics & Thermochemistry

JEE Main 2026 — 4 April, Evening Shift — Question 62

If 3.365 g of ethanol ( ll ) burnt completely in a bomb calorimeter at 298.15 K , the heat produced is 99.472 kJ . The ∣ΔHf∘∣\left|\Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ}\right| of ethanol at 298.15 K is ____\_\_\_\_ ×102 kJ mol−1\times 10^{2} \mathrm{~kJ} \mathrm{~mol}^{-1}. (Nearest integer)

Given : Standard enthalpy of combustion of graphite =−393.5 kJ mol−1=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}

Standard enthalpy of formation of water ( ll ) =−285.8 kJ mol−1=-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1} Molar mass in gmol−1\mathrm{gmol}^{-1} of C,H,O\mathrm{C}, \mathrm{H}, \mathrm{O} are 12,1 and 16 respectively.

Answer: 282

Numerical answer — enter this value.

Step-by-step solution

ΔUcombustion∘=−qmoles=−99.4723.365/46⇒1359.8 kJ/mol\Delta U_{\mathrm{combustion}}^\circ = \frac{-q}{\text{moles}} = \frac{-99.472}{3.365/46} \Rightarrow 1359.8\,\mathrm{kJ/mol}

C2H5OH(ℓ)+3O2(g)→2CO2(g)+3H2O(ℓ)ΔHC∘=ΔU∘+ΔngRT=−1359.8+(−1)(8.314)(298.15)1000=−1362.27 kJ/mol−1362.27=[2(−393.5)+3(−285.8)]−ΔHf∘[C2H5OH(ℓ)]−1362.27=−787−857.4−ΔHf∘[C2H5OH(ℓ)]ΔHf∘[C2H5OH(ℓ)]=−2.82×102 kJ/mol≃−3×102 kJ/mol\begin{aligned} \mathrm{C_2H_5OH}(\ell) + 3\mathrm{O_2}(g) &\rightarrow 2\mathrm{CO_2}(g) + 3\mathrm{H_2O}(\ell) \\ \Delta H_{\mathrm{C}}^\circ &= \Delta U^\circ + \Delta n_{\mathrm{g}}RT \\ &= -1359.8 + \frac{(-1)(8.314)(298.15)}{1000} \\ &= -1362.27\,\mathrm{kJ/mol} \\ -1362.27 &= [2(-393.5) + 3(-285.8)] - \Delta H_{\mathrm{f}}^\circ[\mathrm{C_2H_5OH}(\ell)] \\ -1362.27 &= -787 - 857.4 - \Delta H_{\mathrm{f}}^\circ[\mathrm{C_2H_5OH}(\ell)] \\ \Delta H_{\mathrm{f}}^\circ[\mathrm{C_2H_5OH}(\ell)] &= -2.82 \times 10^2\,\mathrm{kJ/mol} \\ &\simeq -3 \times 10^2\,\mathrm{kJ/mol} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
4 April, Evening Shift
Subject
Chemistry
Chapter
Thermodynamics & Thermochemistry
Topic
Thermochemistry and Enthalpy Changes

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