MHT CET · Chemistry · Thermodynamics (C)
Calculate the standard enthalpy change of following reaction
\(\begin{aligned} & \mathrm{CH}_{4(\mathrm{~g})}+2 \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{2(\mathrm{~g})}+2 \mathrm{H}_2 \mathrm{O}_{(\ell)} \\ & \text { If } \quad \Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{CH}_4\right)=-75 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{CO}_2\right)=-394 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{H}_2 \mathrm{O}\right)=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}\)
- A \(-891 \mathrm{~kJ}\)
- B \(-1041 \mathrm{~kJ}\)
- C \(-966 \mathrm{~kJ}\)
- D \(-1782 \mathrm{~kJ}\)
Answer & Solution
Correct Answer
(A) \(-891 \mathrm{~kJ}\)
Step-by-step Solution
Detailed explanation
\(\Delta_{\mathrm{r}} \mathrm{H}^{\circ} = \left[1 \cdot \Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{CO}_2) + 2 \cdot \Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{H}_2 \mathrm{O})\right] - \left[1 \cdot \Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{CH}_4) + 2 \cdot \Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{O}_2)\right]\) \(\Delta_{\mathrm{r}} \mathrm{H}^{\circ} = [1 \cdot (-394) + 2 \cdot (-286)] - [1 \cdot (-75) + 2 \cdot (0)]\)
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