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04/10/2021 1:33 pm
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Standard enthalpy of combustion of CH4 is -890 kJ mol-1 and standard enthalpy of vaporisation of water is 40.5 kJ mol-1. The enthalpy change of the reaction
CH4(g) + 2O2(g) → CO2(g) + H2O(g)
(a) -809.5 kJ mol-1
(b) -890 kJ mol-1
(c) 809 kJ mol-1
(d) -971 kJ mol-1
1 Answer
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04/10/2021 1:37 pm
Correct answer: (a) -809.5 kJ mol-1
Explanation:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
ΔH = -890 kJ .....(i)
2H2O(l) → 2H2O(g);
ΔH = 2 x 40.5 = 81 kJ .....(ii)
From (i) + (ii), we get
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
ΔH = -890 + 81 = -809 kJ