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 |
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.
The combustion reaction of ethane is as follows. C2H6(g) + 7/2 O2(g) → 2 CO2(g) +...
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