MHT CET · Chemistry · Thermodynamics (C)
Calculate standard enthalpy change of reaction
\(\begin{aligned} & \mathrm{C}_2 \mathrm{H}_{2(\mathrm{~g})}+\frac{5}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_2 \mathrm{O}_{(\ell)} \text { if } \\ & \Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{CO}_2\right)=-393 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{H}_2 \mathrm{O}\right)=-286 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{C}_2 \mathrm{H}_2\right)=227 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}\)
- A \(-650 \mathrm{~kJ}\)
- B \(-1950 \mathrm{~kJ}\)
- C \(-1299 \mathrm{~kJ}\)
- D \(-2598 \mathrm{~kJ}\)
Answer & Solution
Correct Answer
(C) \(-1299 \mathrm{~kJ}\)
Step-by-step Solution
Detailed explanation
\( \Delta_{\mathrm{r}} \mathrm{H}^{\circ} = \sum n \Delta_{\mathrm{f}} \mathrm{H}^{\circ} \text{(products)} - \sum m \Delta_{\mathrm{f}} \mathrm{H}^{\circ} \text{(reactants)} \) \( \Delta_{\mathrm{r}} \mathrm{H}^{\circ} = [2 \Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{CO}_2) + 1 \Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{H}_2 \mathrm{O})] - [1 \Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{C}_2 \mathrm{H}_2) + \frac{5}{2} \Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{O}_2)] \)
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