Question

For the reaction H₂F₂(g) → H₂(g) + F2(g), ΔU = –59.6 kJ mol^–1 at 27°C. The enthalpy change for the above reaction is (–) _______ kJ mol–1 [nearest integer] Given: R = 8.314 J K–1 mol^–1.

Answer 1

Correct Answer: 57

To find the enthalpy change, ΔH, for the reaction H₂F₂(g) → H₂(g) + F₂(g), we use the relation ΔH = ΔU + Δ(nRT), where ΔU is the change in internal energy, Δn is the change in moles of gas, R is the gas constant, and T is the temperature in Kelvin.
1. Calculate T in Kelvin: T = 27°C + 273 = 300 K.
2. Determine Δn: For the reaction, 1 mole of H₂F₂ becomes 1 mole of H₂ + 1 mole of F₂. Thus, Δn = (1+1) - 1 = 1 mole.
3. Compute Δ(nRT): Δ(nRT) = 1 mole × 8.314 J K^–1 mol^–1 × 300 K = 2494.2 J = 2.4942 kJ (Note: 1 kJ = 1000 J).
4. Calculate ΔH: ΔH = ΔU + Δ(nRT) = (–59.6 kJ) + 2.4942 kJ = –57.1058 kJ.
Rounding to the nearest integer, ΔH ≈ –57 kJ mol^–1.
The result lies within the range 57,57.