JEE Main 2026 Chemistry question
If 3.365 g of ethanol (l) is burnt completely in a bomb calorimeter at 298.15 K, the heat produced is 99.472 kJ. The |Δ°| of ethanol at 298.15 K is ×10² kJ mol⁻¹. (Nearest integer) Given: Standard enthalpy for combustion of graphite = −393.5 kJ mol⁻¹, standard enthalpy of formation of water(l) = −285.8 kJ mol⁻¹.
Answer: 3
Solution
Moles ethanol = 3.365/46 = 0.07315; Δ = −99.472/0.07315 ≈ −1360 kJ/mol. Δ=−1, so Δ = Δ + ΔnRT ≈ −1362.5
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