JEE Chemistry 2025 Question with Solution

Given: Thermochemical data for $\text{CH}_2=\text{CH}_2$ and its bonds are provided.

Find: The value of $\Delta H_{C=C}^0$ for $\text{CH}_2=\text{CH}_2$.

Using the relation between atomization energies and bond enthalpies for ethene:

$$2C(\text{graphite}) + 2H_2(\text{g}) \rightarrow C_2H_4(\text{g})$$

Energy required to atomize the reactants:

$$2 \times 710 + 2 \times 436 = 1420 + 872 = 2292 \, \text{kJ mol}^{-1}$$

For $C_2H_4$, the total bond energy is:

$$4(C-H) + 1(C=C)$$

the solution gives:

$$4 \times 414 + 611 = 1656 + 611 = 2267 \, \text{kJ mol}^{-1}$$

Hence,

$$\Delta H_f^\Theta = \text{Energy required} - \text{Energy released} = 2292 - 2267 = 25 \, \text{kJ mol}^{-1}$$

The extracted solution concludes that the required nearest integer value is $25$.

Given:

• $\Delta_\text{sub}[C(\text{graphite})] = 710 \, \text{kJ mol}^{-1}$
• $\Delta_{H-H} = 436 \, \text{kJ mol}^{-1}$
• Bond data used in the extracted solution for ethene

Find: The nearest integer value asked in the problem.

From the second extracted approach:

$$[\Delta H_f^0]_{C_2H_4(\text{g})} = (2 \times 710) + (2 \times 436) - 611 - 4 \times 414$$ $$[\Delta H_f^0]_{C_2H_4(\text{g})} = 1420 + 872 - 611 - 1656$$ $$[\Delta H_f^0]_{C_2H_4(\text{g})} = 2292 - 2267 = 25 \, \text{kJ mol}^{-1}$$

Therefore, the final answer is $25$.