JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
Consider the following cases of standard enthalpy of reaction \(\left(\Delta \mathrm{H}_{\mathrm{r}}^{\circ}\right.\) in \(\left.\mathrm{kJ} \mathrm{mol}{ }^{-1}\right)\)
\(\mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})+\frac{7}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \Delta \mathrm{H}_1^{\circ}=-1550 \)
\( \mathrm{C} \text { (graphite) }+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) \quad \Delta \mathrm{H}_2^{\circ}=-393.5 \)
\( \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \quad \Delta \mathrm{H}_3^{\circ}=-286\)
The magnitude of \(\Delta \mathrm{H}_{f \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})}^{\circ}\) is _______ \(\mathrm{kJ} \mathrm{mol}{ }^{-1}\) (Nearest integer).
- A 90
- B 100
- C 95
- D 97
Answer & Solution
Correct Answer
(C) 95
Step-by-step Solution
Detailed explanation
\begin{aligned} & 2 \mathrm{C}_{\text {(graphite) }}+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g}) \quad \Delta \mathrm{H}_{\mathrm{f}}=? \\ & \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})+\frac{7}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 2…
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