ThermochemistryMedium

Question

Calculate the standard free energy of the reaction at 27°C for the combustion of methane using the given data

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l).

Species CH4(g) O2(g) C O2(g) H2O(l)

ΔfHo/(kJ mol–1) −74.5 0 −393.5 −286.0

S°/ (JK–1 mol–1) 186 205 216 70

Options

A.−891.0 kJ/mol
B.−240 kJ/mol
C.−819 kJ/mol
D.−963 kJ/mol

Solution

$\Delta H = \sum_{}^{}{(B.E.)_{\text{Reactants}} - \sum_{}^{}{(B.E.)_{\text{Products}} = (435 + 240) - (2 \times 430) = - 185\text{ KJ}}}$

$\Delta S = \sum_{}^{}{S_{\text{Products}} - \sum_{}^{}{S_{\text{Reactants}} = (2 \times 186) - (130 + 222) = 20\text{ J/K}}} $$$\text{Now, }\Delta G = \Delta H - T.\Delta S = ( - 185) - 300 \times \frac{20}{1000} = - 191\text{ kJ}$$

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