Answer the following question.
Calculate ΔU at 298 K for the reaction,
C2H4(g) + HCl(g) → C2H5Cl(g), ΔH = - 72.3 kJ
How much PV work is done?
Enthalpy change = ΔH = –72.3 kJ
Temperature = T = 298 K
PV work done and internal energy change (ΔU)
1. W = - ΔngRT
2. ΔH = ΔU + ΔngRT
Δng = (moles of product gases) - (moles of reactant gases)
Δng = 1 – 2 = –1 mol
Using formula (i)
W = - ΔngRT
= - (- 1 mol) × 8.314 J K-1 mol-1 × 298 K
= 2477.57 J = 2.48 kJ
Now, using formula (ii) and rearranging,
ΔU = ΔH - Δ ngRT = ΔH + W = –72.3 kJ + 2.48 kJ = –69.8 kJ
∴ The PV work done is 2.48 kJ.
∴ The internal energy change (ΔU) is –69.8 kJ.