Chemistry · Thermodynamics & Thermochemistry

JEE Main 2024 — 4 April, Shift 1 — Question 74

The magnitude of enthalpy of formation of $\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)$ from ethylene by addition of hydrogen where the bondenergies of $\mathrm{C}-\mathrm{H}, \mathrm{C}-\mathrm{C}, \mathrm{H}-\mathrm{H}$ are $414 \mathrm{~kJ}, 347 \mathrm{~kJ}$ , 615 kJ and 435 kJ respectively is - $\qquad$ kJ

Answer: 125

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})$

$\Delta \mathrm{H}=\mathrm{BE}(\mathrm{C}=\mathrm{C})+4 \mathrm{BE}(\mathrm{C}-\mathrm{H})+\mathrm{BE}(\mathrm{H}-\mathrm{H})$

$-\mathrm{BE}(\mathrm{C}-\mathrm{C})-6 \mathrm{BE}(\mathrm{C}-\mathrm{H})$ $\Delta \mathrm{H}=\mathrm{BE}(\mathrm{C}=\mathrm{C})+\mathrm{BE}(\mathrm{H}-\mathrm{H})-\mathrm{BE}(\mathrm{C}-\mathrm{C})$

$-2 \mathrm{BE}(\mathrm{C}-\mathrm{H})$ $=615+435-347-2 \times 414$ $=-125 \mathrm{~kJ}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 1
Subject: Chemistry
Chapter: Thermodynamics & Thermochemistry
Topic: Thermochemistry and Enthalpy Changes