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MHT CET · Chemistry · Thermodynamics (C)

Calculate the standard enthalpy change of following reaction.
\(\begin{aligned}
& \mathrm{CH}_{4(\mathrm{~g})}+2 \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{2(\mathrm{~g})}+2 \mathrm{H}_2 \mathrm{O}_{(\ell)} \\
& \text { if } \quad \Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(C H_4\right)=-75 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(C \mathrm{O}_2\right)=-390 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{H}_2 \mathrm{O}\right)=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}\)

  1. A -887.00 kJ
  2. B -1325.00 kJ
  3. C -1035.00 kJ
  4. D -1770.00 kJ
Verified Solution

Answer & Solution

Correct Answer

(A) -887.00 kJ

Step-by-step Solution

Detailed explanation

\( \Delta_{\mathrm{r}} \mathrm{H}^{\circ} = [1 \times (-390) + 2 \times (-286)] - [1 \times (-75) + 2 \times (0)] \) \( \Delta_{\mathrm{r}} \mathrm{H}^{\circ} = -390 - 572 + 75 \)