Question

The combustion reaction of ethane is as follows.

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

Using Hess's law and the reaction enthalpies given below, find the change in enthalpy for this reaction.

reaction (1):	C(s) + O2(g) → CO2(g)	ΔH = −393.5 kJ/mol
reaction (2):	H2(g) + 1/2 O2(g) → H2O(l)	ΔH = −285.8 kJ/mol
reaction (3):	2 C(s) + 3 H2(g) → C2H6(g)	ΔH = −84.0 kJ/mol

Answer 1

Write down the three given equations:

C (s) + O₂ (g) --------> CO₂ (g); ΔH = -393.5 kJ/mol …….(R1)

H₂ (g) + ½ O₂ (g) -------> H₂O (l); ΔH = -285.8 kJ/mol …..(R2)

2 C (s) + 3 H₂ (g) -------> C₂H₆ (g); ΔH = -84.0 kJ/mol ……(R3)

Reverse (R3); multiply (R1) by 2 and (R2) by 3. Add all these to obtain

C₂H₆ (g) + 2 C (s) + 2 O₂ (g) + 3 H₂ (g) + 3/2 O₂ (g) ---------> 2 C (s) + 3 H₂ (g) + 2 CO₂ (g) + 3 H2O (l)

Cancel and combine common terms to obtain

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

The above is our given equation. The enthalpy change for the reaction is given by

ΔH = -ΔH (R3) + 2ΔH (R1) + 3ΔH (R2) (note that we reversed R3; this reverses the sign of ΔH; also ΔH terms are additive and hence we take twice the ΔH for R2 and thrice the ΔH for R3). Therefore,

ΔH = -(-84.0 kJ/mol) + 2*(-393.5 kJ/mol) + 3*(--285.8 kJ/mol) = (84.0 – 787 – 857.4) kJ/mol = (84.0 – 1644.4) kJ/mol = -1560.4 kJ/mol.

Ans: The enthalpy change for the reaction is -1560.4 kJ/mol.