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Thermodynamics question

2025 · 22 Jan · Shift 2 · Q21
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Thermodynamics question

2025 · 22 Jan · Shift 2 · Q21

JEE MainChemistryThermodynamicsNumerical+4 / −1
Consider the following cases of standard enthalpy of reaction (ΔHr∘\left(\Delta \mathrm{H}_{\mathrm{r}}^{\circ}\right.(ΔHr∘​ in kJmol−1)\left.\mathrm{kJ} \mathrm{mol}^{-1}\right)kJmol−1) 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}) \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}) \Delta \mathrm{H}_3^{\circ}=-286 \end{aligned}​C2​H6​( g)+27​O2​( g)→2CO2​( g)+3H2​O(l)ΔH1∘​=−1550C( graphite )+O2​( g)→CO2​( g)ΔH2∘​=−393.5H2​( g)+21​O2​( g)→H2​O(l)ΔH3∘​=−286​ The magnitude of ΔHfC2H6( g)∘\Delta \mathrm{H}_{f \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})}^{\circ}ΔHfC2​H6​( g)∘​ is ‾\underline{\hspace{2cm}}​kJmol−1\mathrm{kJ} \mathrm{mol}^{-1}kJmol−1 (Nearest integer).
Numerical answer
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