Δᵥₐₚ H⊖ for water is +40.79 kJ mol⁻¹ at 1 bar and 100° C. Change in internal energy for this vapourisation under same condition is ________ kJ mol⁻¹. (Integer answer) (Given R=8.3 JK⁻¹ mol⁻¹)

Solution

Chemistry ma’am · 4 steps

Only the gas counts in Δ n_g: one mole of steam appears, so subtract RT from Δ H.

1. Link ΔH and ΔU. For a change at constant pressure, Δ H = Δ U + Δ n_g RT. Here Δ n_g is moles of gaseous products minus moles of gaseous reactants.
2. Find Δn_g. The change is H₂O(l) → H₂O(g). The liquid has no gas moles, so Δ n_g = 1 - 0 = 1.

   Δ U = Δ H - Δ n_g RT = 40.79 - (1 × 8.3 × 373)/1000

3. Put in numbers. With T = 373 K, RT = 8.3 × 373 = 3095.9 J mol⁻¹ = 3.096 kJ mol⁻¹. Then Δ U = 40.79 - 3.096 = 37.69 kJ mol⁻¹.
4. So. Δ U ≈ 37.69 kJ mol⁻¹, which rounds to 38.

The common slip: 40. Putting T = 100 instead of 373 K makes RT only 0.83 kJ, so Δ U comes out as 39.96, which rounds to 40.

The idea to keep. To go between Δ H and Δ U, count only gas moles for Δ n_g, use T in kelvin, and convert RT from J to kJ before you subtract.

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