Chemistry · Thermodynamics & Thermochemistry

JEE Main 2025 — 22 January, Evening Shift — Question 46

Consider the following cases of standard enthalpy of reaction (ΔHro\left(\Delta \mathrm{H}_{\mathrm{r}}^{\mathrm{o}}\right. in kJmol−1)\left.\mathrm{kJ} \mathrm{mol}^{-1}\right)

C2H6( g)+72O2( g)→2CO2( g)+3H2O(ℓ)ΔH1o=−1550\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}(\ell) \Delta \mathrm{H}_{1}^{\mathrm{o}}=-1550

C (graphite) +O2( g)→CO2( g)ΔH2o=−393.5+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}) \Delta \mathrm{H}_{2}^{\mathrm{o}}=-393.5

H2( g)+12O2( g)→H2O(ℓ)ΔH3o=−286\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\ell) \Delta \mathrm{H}_{3}^{\mathrm{o}}=-286

The magnitude of ΔHf2H6( g)o\Delta \mathrm{H}_{\mathrm{f}_{2} \mathrm{H}_{6}(\mathrm{~g})}^{\mathrm{o}} is____ kJmol−1\mathrm{kJ} \mathrm{mol}^{-1} (Nearest integer).

Answer: 95

Numerical answer — enter this value.

Step-by-step solution

2C_{(graphite)} + 3H_2(g) &\rightarrow C_2H_6(g) \quad \Delta H_f = ? \\ C_2H_6(g) + \frac{7}{2}O_2(g) &\rightarrow 2CO_2(g) + 3H_2O(l) \quad \Delta H_1 = -1550 \\ C_{(graphite)} + O_2(g) &\rightarrow CO_2(g) \quad \Delta H_2 = -393.5 \\ H_2(g) + \frac{1}{2}O_2(g) &\rightarrow H_2O(l) \quad \Delta H_3 = -286 \\ \Delta H_f &= 2\Delta H_2 + 3\Delta H_3 - \Delta H_1 \\ &= 2(-393.5) + 3(-286) - (-1550) \\ &= -787 - 858 + 1550 \\ &= 95 \text{ kJ/mole} \end{aligned}$$ \end{document}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Chemistry
Chapter
Thermodynamics & Thermochemistry
Topic
Thermochemistry and Enthalpy Changes
Consider the following cases of standard enthalpy of reaction (Δ H r… | JEE Main 2025 PYQ with Solution · DhiX AI