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MHT CET · Chemistry · Thermodynamics (C)

Calculate standard enthalpy change of reaction
\(\begin{aligned} & \mathrm{C}_2 \mathrm{H}_{2(\mathrm{~g})}+\frac{5}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_2 \mathrm{O}_{(\ell)} \text { if } \\ & \Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{CO}_2\right)=-393 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{H}_2 \mathrm{O}\right)=-286 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{C}_2 \mathrm{H}_2\right)=227 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}\)

  1. A \(-650 \mathrm{~kJ}\)
  2. B \(-1950 \mathrm{~kJ}\)
  3. C \(-1299 \mathrm{~kJ}\)
  4. D \(-2598 \mathrm{~kJ}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(-1299 \mathrm{~kJ}\)

Step-by-step Solution

Detailed explanation

\( \Delta_{\mathrm{r}} \mathrm{H}^{\circ} = \sum n \Delta_{\mathrm{f}} \mathrm{H}^{\circ} \text{(products)} - \sum m \Delta_{\mathrm{f}} \mathrm{H}^{\circ} \text{(reactants)} \) \( \Delta_{\mathrm{r}} \mathrm{H}^{\circ} = [2 \Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{CO}_2) + 1 \Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{H}_2 \mathrm{O})] - [1 \Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{C}_2 \mathrm{H}_2) + \frac{5}{2} \Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{O}_2)] \)