Given: Given equations are,
a. C2H6(g) + (\frac72)O2(g) → 2CO2(g) + 3H2O(l), ΔH° = –1560 kJ .......(1)
b. H2(g) + (\frac12) O2(g) → H2O(l),ΔH° = –285.8 kJ ........(2)
c. C(graphite) + O2(g) → CO2(g),ΔH° = –393.5 kJ. .........(3)
To find: Standard enthalpy of the given reaction
Calculation: Reversing equation (1),
2CO2(g) + 3H2O(l) → C2H6(g) + (\frac72)O2(g), ΔH° = –1560 kJ .......(4)
Multiplying equation (2) by 3 and (3) by 2, then adding to equation (4)
ΔH° = 1560 + (-857.4) + (-787.0) = -84.4 kJ