MHT CET · Chemistry · Thermodynamics (C)
Calculate the standard enthalpy change of following reaction.
\(\mathrm{C}_2 \mathrm{H}_{4(\mathrm{~g})}+3 \mathrm{O}_{2(\mathrm{~g})} \longrightarrow 2 \mathrm{CO}_{2(\mathrm{~g})}+2 \mathrm{H}_2 \mathrm{O}_{(\ell)}\)
If \(\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{C}_2 \mathrm{H}_4\right)=-52 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
\(\begin{aligned}
\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{CO}_2\right) & =-390 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{H}_2 \mathrm{O}\right) & =-286 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}\)
- A \(-650 \mathrm{~kJ}\)
- B \(-1300 \mathrm{~kJ}\)
- C \(-1950 \mathrm{~kJ}\)
- D \(-1500 \mathrm{~kJ}\)
Answer & Solution
Correct Answer
(B) \(-1300 \mathrm{~kJ}\)
Step-by-step Solution
Detailed explanation
\( \Delta_{\mathrm{r}} \mathrm{H}^{\circ} = [2 \Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{CO}_2) + 2 \Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{H}_2 \mathrm{O})] - [\Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{C}_2 \mathrm{H}_4) + 3 \Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{O}_2)] \) \( \Delta_{\mathrm{r}} \mathrm{H}^{\circ} = [2(-390) + 2(-286)] - [-52 + 3(0)] \)
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