JEE Chemistry 2026 Question with Solution

Given: $A_2(g)+B_2(g) \rightleftharpoons 2AB(g)$ at $500 \, \text{K}$. Also, $\log K = 2.2$, $S^\circ(AB)=222 \, \text{J K}^{-1} \text{mol}^{-1}$, $S^\circ(A_2)=146 \, \text{J K}^{-1} \text{mol}^{-1}$, $S^\circ(B_2)=280 \, \text{J K}^{-1} \text{mol}^{-1}$, $\Delta_f H^\circ(AB)=32 \, \text{kJ mol}^{-1}$, $\Delta_f H^\circ(A_2)=6 \, \text{kJ mol}^{-1}$, and $\Delta_f H^\circ(B_2)=x$.

Find: The value of $x$.

From the solution data,

$$\Delta G^\circ = -2.303RT \log K$$ $$\Delta G^\circ = -2.303 \times 8.3 \times 500 \times 2.2$$ $$\Delta G^\circ \approx -21027 \, \text{J/mol} \approx -21 \, \text{kJ/mol}$$

Now calculate the standard entropy change of reaction:

$$\Delta S^\circ_{\text{rxn}} = 2S^\circ(AB) - \left[S^\circ(A_2) + S^\circ(B_2)\right]$$ $$\Delta S^\circ_{\text{rxn}} = 2(222) - (146 + 280) = 444 - 426 = 18 \, \text{J K}^{-1} \text{mol}^{-1}$$

Using

$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$ $$-21000 = \Delta H^\circ - 500(18)$$ $$-21000 = \Delta H^\circ - 9000$$ $$\Delta H^\circ = -12000 \, \text{J/mol} = -12 \, \text{kJ/mol}$$

For the reaction enthalpy,

$$\Delta H^\circ_{\text{rxn}} = 2\Delta H(AB) - \left[\Delta H(A_2) + H_{B_2}\right]$$ $$-12 = 2(32) - (6 + H_{B_2})$$ $$-12 = 64 - 6 - H_{B_2}$$ $$-12 = 58 - H_{B_2}$$ $$H_{B_2} = 70 \, \text{kJ/mol}$$

Therefore, the value of $x$ is $70$.

The answer key shows $26$, but the extracted solution working concludes $70$, so the solution is taken as authoritative.

Given: Equilibrium data at $500 \, \text{K}$ and thermodynamic table values.

Find: Missing enthalpy value $x$ for $B_2(g)$.

1. First convert equilibrium information into free energy change:

$$\Delta G^\circ = -2.303RT \log K$$

Substituting the given values:

$$\Delta G^\circ = -2.303 \times 8.3 \times 500 \times 2.2 \approx -21 \, \text{kJ/mol}$$

2. Next compute reaction entropy:

$$\Delta S^\circ_{\text{rxn}} = 2(222) - (146+280) = 18 \, \text{J K}^{-1} \text{mol}^{-1}$$

This is

$$18 \, \text{J K}^{-1} \text{mol}^{-1} = 0.018 \, \text{kJ K}^{-1} \text{mol}^{-1}$$

3. Use the relation

$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$

So,

$$\Delta H^\circ = \Delta G^\circ + T\Delta S^\circ$$ $$\Delta H^\circ = -21 + 500(0.018) = -21 + 9 = -12 \, \text{kJ/mol}$$

4. Express reaction enthalpy in terms of formation enthalpies:

$$\Delta H^\circ_{\text{rxn}} = 2\Delta_f H^\circ(AB) - \left[\Delta_f H^\circ(A_2) + \Delta_f H^\circ(B_2)\right]$$ $$-12 = 2(32) - (6 + x)$$ $$-12 = 64 - 6 - x = 58 - x$$ $$x = 70$$

Therefore, the nearest integer value is $70$.