Saved Bookmarks
| 1. |
Calculate the enthalpy of hydrogenation of ethylene, given that the enthalpy of combustion of enthylene, hydrogen and ethane are – 1410.0, – 286.2 and – 1560.6 kJ mol–1 respectively at 298 K. |
|
Answer» We are given (i) C2H4 (g) + 3O2 (g) → 2CO2 (g) + 2H2O (l) ΔH = – 1410 kJ mol–1 (ii) H2 (g)+ 1/2 O2(g)→ H2O (l), ΔH = – 286.2 kJ mol–1 (iii) C2H6 (g) + 31/2 O2(g)→ 2CO2 (g)+ 3H2O (l), ΔH = – 1560.6 kJ mol–1 We aim at: C2H4 + H2 (g) → C2H6 (g), ΔH = ? Equation (i) + Equation (ii) – Equation (iii) gives C2H4 (g) + H2 (g) → C2H6 (g), ΔH = – 1410.0 + (–286.2) – (1560.6) = – 135.6 kJ mol–1 |
|