MHT CET · Chemistry · Thermodynamics (C)
Calculate the standard enthalpy change for reaction,
\(\begin{aligned}
& \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(\ell)}+3 \mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{CO}_{2(\mathrm{~g})}+3 \mathrm{H}_2 \mathrm{O}_{(\ell)} \\
& \text {If,}\\
& \text {} \begin{aligned}
\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right) & =-280 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{CO}_2\right) & =-390 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{H}_2 \mathrm{O}\right) & =-285 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
\end{aligned}\)
- A \(-678.00 \mathrm{~kJ}\)
- B \(-2033 \cdot 00 \mathrm{~kJ}\)
- C \(-1355 \cdot 00 \mathrm{~kJ}\)
- D \(-1016 \cdot 00 \mathrm{~kJ}\)
Answer & Solution
Correct Answer
(C) \(-1355 \cdot 00 \mathrm{~kJ}\)
Step-by-step Solution
Detailed explanation
\(\Delta_{\mathrm{r}} \mathrm{H}^{\circ} = [2 \times \Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{CO}_2) + 3 \times \Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{H}_2 \mathrm{O})] - [\Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}) + 3 \times \Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{O}_2)]\) \(\Delta_{\mathrm{r}} \mathrm{H}^{\circ} = [2 \times (-390 \mathrm{~kJ}) + 3 \times (-285 \mathrm{~kJ})] - [-280 \mathrm{~kJ} + 3 \times 0 \mathrm{~kJ}]\)
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