Chemistry · Thermodynamics & Thermochemistry
JEE Main 2025 — 22 January, Evening Shift — Question 46
Consider the following cases of standard enthalpy of reaction in
C (graphite)
The magnitude of is____ (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