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Chemistry Question on Thermodynamics terms

The combustion enthalpies of carbon, hydrogen and methane are −395.5 kJ mol−1,−284.8 kJ mol1-395.5\, kJ\, mol ^{-1},-284.8\, kJ\, mol ^{1} and −890.4 kJ mol−1-890.4\, kJ \,mol ^{-1} respectively at 25∘C25^{\circ} C. The value of standard formation enthalpies of methane at that temperature is

A

890.4 kJ mol−1890.4\, kJ\, mol ^{-1}

B

−298.8 kJ mol−1-298.8\, kJ\, mol ^{-1}

C

−74.7 kJ mol−1-74.7\, kJ\, mol ^{-1}

D

−107.7 kJ mol−1-107.7\, kJ\, mol ^{-1}

Answer

−74.7 kJ mol−1-74.7\, kJ\, mol ^{-1}

Explanation

Solution

Combustion of carbon; C(s)+O2(s)→CO2(g);ΔH1=395.5C_{\left(s\right)}+O_{2\left(s\right)} \rightarrow CO_{2\left(g\right);} \Delta H_{1}=395.5 kJ/mol Combustion of hydrogen; H2(g)+12O2(s)→H2O(g);ΔH2=−284.8H_{2\left(g\right)}+\frac{1}{2}O_{2\left(s\right)} \rightarrow H_{2}O_{\left(g\right)}; \Delta H_{2}=-284.8 kJ/mol Combustion of methane; CH4(g)+2O2(g)→CO2+2H2O;ΔH3=−890.4CH_{4\left(g\right)}+2O_{2\left(g\right)} \rightarrow CO_{2}+2H_{2}O; \Delta H_{3}=-890.4 kJ/mol Formation of methane; C(s)+2H2(g)→CH2(g);ΔH=?C_{\left(s\right)}+2H_{2\left(g\right)} \rightarrow CH_{2\left(g\right)}; \Delta H=? From the above equations, ΔH=ΔH1+2ΔH2−ΔH3\Delta H=\Delta H_{1}+2\Delta H_{2}-\Delta H_{3} −395.5−2??284.8+890.4- 395.5 - 2 ??284.8 + 890.4 kJ/mol =−74.7= - 74.7 kJ/mol