JEE Main 2026 Chemistry question

ChemistryThermodynamicsJEE Main 2026

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

See the full step-by-step solution

Create a free account to read the complete working, and practise questions like this in a timed test.

Create a free account

Question ID VK-015405