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JEE Main 202522 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
22 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2025 · 22 January · Evening Shift · Chemistry · Thermodynamics (C)

Consider the following cases of standard enthalpy of reaction (Δ H_...

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) C2H6(g)+72O2(g)→2CO2(g)+3H2O(l)ΔH1∘=−1550C (graphite) +O2(g)→CO2(g)ΔH2∘=−393.5H2(g)+12O2(g)→H2O(l)ΔH3∘=−286\begin{aligned} & \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 \end{aligned}ThemagnitudeofThe magnitude of\Delta \mathrm{H}_{f \mathrm{C}_2 \mathrm{H}_6(\mathrm{ g})}^{\circ} is _______ kJmol−1\mathrm{kJ} \mathrm{mol}{ }^{-1} (Nearest integer).

Correct answer

0

Step-by-step solution

2C(graphite) +3H2(g)→C2H6(g)ΔHf=?C2H6(g)+72O2(g)→2CO2(g)+3H2O(l)ΔH1=−1550C(graphite) +O2(g)→CO2(g)ΔH2=−393.5H2(g)+12O2(g)→H2O(/)ΔH3=−286ΔHf=2ΔH2+3ΔH3−ΔH1=95kJ/mole2\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 \mathrm{CO}_2(\mathrm{ g})+3 \mathrm{H}_2 \mathrm{O}(l) \\ & \quad \Delta \mathrm{H}_1=-1550 \\ & \mathrm{C}_{\text{(graphite) }}+\mathrm{O}_2(\mathrm{ g}) \rightarrow \mathrm{CO}_2(\mathrm{ g}) \Delta \mathrm{H}_2=-393.5 \\ & \mathrm{H}_2(\mathrm{ g})+\frac{1}{2} \mathrm{O}_2(\mathrm{ g}) \rightarrow \mathrm{H}_2 \mathrm{O}(/) \Delta \mathrm{H}_3=-286 \\ & \Delta \mathrm{H}_{\mathrm{f}}=2 \Delta \mathrm{H}_2+3 \Delta \mathrm{H}_3-\Delta \mathrm{H}_1 \\ & \quad=95 \mathrm{ kJ} / \mathrm{mole}^2\end{aligned}
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