What is the enthalpy of combustion of 0.2 moles of C2H6(g)?

Question

C(s) + O₂(g) → CO₂(g) 

 ΔH_f = −394 kJ

H₂(g) +1/2O₂(g) → H₂O(l) 

 ΔH_f = −242 kJ

2C(s) + 3H₂(g) → C₂H₆(g)

ΔH =−84 kJ

The combustion of C₂H₆ is shown by the following equation:

C₂H₆(g) +7/2 O₂(g) → 2CO₂(g) + 3H₂O(l)

What is the enthalpy of combustion of 0.2 moles of C₂H₆(g)?

Answer 1

Sum the ΔH_f of the products and subtract from that theΔ sum of the ΔH_f of the reactants.

∑products = (2 x -394) + (3 x -242) = -788 + -726 = -1514 kJ

∑reactants = -84 + 0 = -84 kJ

ΔH_rxn = -1514 + -84 = -1598 kJ for combustion of 1 mol C₂H₆

0.2 mol x -1598 kJ/mol = -320 kJ