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Question:

The entropy (So) of the following substances are: CH4(g) = 186.2 J K⁻¹mol⁻¹, O2(g) = 205.0 J K⁻¹mol⁻¹, CO2(g) = 213.6 J K⁻¹mol⁻¹, H2O(l) = 69.9 J K⁻¹mol⁻¹. The entropy change (ΔSo) for the reaction CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) is:

-0.5 JK⁻¹mol⁻¹

-42.8 JK⁻¹mol⁻¹

-0.1 JK⁻¹mol⁻¹

-0.6 JK⁻¹mol⁻¹

Solution:

Given:
CH4(g) = 186.2 J K⁻¹mol⁻¹
O2(g) = 205.0 J K⁻¹mol⁻¹
CO2(g) = 213.6 J K⁻¹mol⁻¹
H2O(l) = 69.9 J K⁻¹mol⁻¹
The entropy change (ΔSo) for the reaction CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Entropy of a reaction = entropy of products - entropy of reactants
ΔSo = [So(CO2(g)) + 2So(H2O(l))] - [So(CH4(g)) + 2So(O2(g))]
ΔSo = [(213.6 J K⁻¹mol⁻¹) + 2(69.9 J K⁻¹mol⁻¹)] - [(186.2 J K⁻¹mol⁻¹) + 2(205.0 J K⁻¹mol⁻¹)]
ΔSo = (213.6 + 139.8) - (186.2 + 410)
ΔSo = 353.4 - 596.2
ΔSo = -242.8 J K⁻¹mol⁻¹
Therefore, the entropy change for the reaction is -242.8 J K⁻¹mol⁻¹.