For the reaction
H₂ F₂( g) → H₂( g)+F₂( g)
Δ U=-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.

Solution

Chemistry ma’am · 4 steps

Count gas moles on each side first. That count alone decides how far Δ H moves away from Δ U.

1. Link ΔH and ΔU. For a reaction involving gases, the difference between Δ H and Δ U is the expansion work done by the change in the number of gas moles, Δ n_g.

   Δ H = Δ U + Δ n_g RT

2. Find Δn. Products have 1 + 1 = 2 mol of gas (H₂ and F₂). The reactant side has 1 mol (H₂F₂). So Δ n_g = 2 - 1 = +1.
3. Put in numbers. Use T = 27 + 273 = 300 K. Then Δ n_g RT = 1 × 8.314 × 300 = 2494 J mol⁻¹ = 2.494 kJ mol⁻¹.
4. So. Δ H = -59.6 + 2.494 = -57.1 kJ mol⁻¹, so the blank is 57.

The common slip: 62. Taking Δ n_g as reactants minus products (-1) gives -59.6 - 2.494 = -62.1, so 62, not 57.

The idea to keep. Δ H - Δ U = Δ n_g RT, and Δ n_g counts gas moles only: products minus reactants. Put T in kelvin and convert RT to kJ before adding.

Answer from the NTA final answer key. Solution by PrepWiser.