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ChemistryInteger+4 / -1Question 71 of 75

JEE Main 2026 · 4 Apr · Shift 2, Chemistry question 71

If 3.365 g of ethanol (ll) is burnt completely in a bomb calorimeter at 298.15 K, the heat produced is 99.472 kJ. The ∣ΔHf∘∣|\Delta H_f^\circ| of ethanol at 298.15 K is ________ ×102\times10^2 kJ mol−1^{-1}. (Nearest integer)
Given: Standard enthalpy for combustion of graphite = −393.5 kJ mol−1^{-1}
Standard enthalpy of formation of water (l) = −285.8 kJ mol−1^{-1}
Molar mass in g mol−1^{-1} of C, H, O are 12, 1 and 16 respectively

Answer and worked solution

Official answer: 3

Enthalpy of combustion. 3.3653.365 g of ethanol is 0.073150.07315 mol, so ΔcH=−99.4720.07315=−1359.8\Delta_cH=-\frac{99.472}{0.07315}=-1359.8 kJ mol⁻¹. (Bomb-calorimeter heat is taken as ΔH, as the question intends.)

Hess's law.
ΔfH=2(−393.5)+3(−285.8)+1359.8=−284.6 kJ mol−1.\Delta_fH=2(-393.5)+3(-285.8)+1359.8=-284.6\text{ kJ mol}^{-1}.

Answer. ∣ΔfH∣=2.85×102|\Delta_fH|=2.85\times10^2, nearest integer 3.

Answer: 3

Topics:Hess's law and enthalpies of reactionCalorimetry
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