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Example · Example 6

Q.The standard enthalpy of combustion of methane, CH4(g)\text{CH}_4(g), is −890.3 kJ mol−1-890.3\ \text{kJ mol}^{-1}. Given the standard enthalpies of formation of CO2(g)\text{CO}_2(g) and H2O(l)\text{H}_2\text{O}(l) are −393.5 kJ mol−1-393.5\ \text{kJ mol}^{-1} and −285.8 kJ mol−1-285.8\ \text{kJ mol}^{-1} respectively, calculate the standard enthalpy of formation of methane.

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The combustion reaction is CH4(g)+2O2(g)→CO2(g)+2H2O(l)\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l), with ΔcH∘=−890.3 kJ mol−1\Delta_cH^\circ = -890.3\ \text{kJ mol}^{-1}. By Hess's law, the enthalpy of this reaction equals the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants (each weighted by stoichiometric coefficient), noting ΔfH∘(O2,g)=0\Delta_fH^\circ(\text{O}_2, g) = 0 since oxygen is an element in its standard state: ΔcH∘=[ΔfH∘(CO2)+2ΔfH∘(H2O)]−[ΔfH∘(CH4)+2(0)]\Delta_cH^\circ = \left[\Delta_fH^\circ(\text{CO}_2) + 2\Delta_fH^\circ(\text{H}_2\text{O})\right] - \left[\Delta_fH^\circ(\text{CH}_4) + 2(0)\right]. Substituting: $-890.3 = \left[-393.5 + 2(-285.8)\right] - \Del …

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