MCQ 11 Mark
The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, $-890.3 kJ mol ^{-1},-393.5 kJ mol ^{-1}$ and $-285.8 kJ mol ^{-1}$ respectively. Enthalpy of formation of $CH _4(g)$ will be
- A$+74.8 kJ mol ^{-1}$
- B$+52.26 kJ mol ^{-1}$
- C$-74.8 kJ mol ^{-1}$
- D$-52.27 kJ mol ^{-1}$
Answer
$\begin{aligned} & \text { (c) }-74.8 kJ mol ^{-1} \\ & \text { Explanation: } CH _4+2 O _2 \rightarrow CO _2+2 H _2 O \Delta H _1=-890.3 KJ / mol \ldots(1) \\ & C ( s )+ O _2 \rightarrow CO _2 \Delta H _2=-393.5 KJ / mol \ldots(2) \\ & H _2+0.5 O _2 \rightarrow H _2 O \Delta H _3=-285.8 KJ / mol \ldots(3) \\ & C ( s )+2 H _2 \rightarrow CH _4 \Delta H =\Delta H _2+2\left(\Delta H _3\right)-\Delta H _1 \\ & \Delta H =-393.5+2(-285.8)-(-890.3) \\ & =-74.8 kJ / mol \end{aligned}$
View full question & answer→$\begin{aligned} & \text { (c) }-74.8 kJ mol ^{-1} \\ & \text { Explanation: } CH _4+2 O _2 \rightarrow CO _2+2 H _2 O \Delta H _1=-890.3 KJ / mol \ldots(1) \\ & C ( s )+ O _2 \rightarrow CO _2 \Delta H _2=-393.5 KJ / mol \ldots(2) \\ & H _2+0.5 O _2 \rightarrow H _2 O \Delta H _3=-285.8 KJ / mol \ldots(3) \\ & C ( s )+2 H _2 \rightarrow CH _4 \Delta H =\Delta H _2+2\left(\Delta H _3\right)-\Delta H _1 \\ & \Delta H =-393.5+2(-285.8)-(-890.3) \\ & =-74.8 kJ / mol \end{aligned}$