Q24 / 75 · JEE Main 2026 (06 April Shift 2) · Chemistry
In a closed flask at $600$ K, one mole of X$_2$Y$_4$(g) attains equilibrium as given below :
$\text{X}_2\text{Y}_4(g) \rightleftharpoons 2\text{XY}_2(g)$
At equilibrium, $75\%$ X$_2$Y$_4$(g) was dissociated and the total pressure is $1$ atm. The magnitude of $\Delta_r G^{\ominus}$ (in kJ mol$^{-1}$) at this temperature is __________. (Nearest Integer) (Given : R $= 8.3$ J mol$^{-1}$ K$^{-1}$; $\ln 10 = 2.3$, $\log 2 = 0.3$, $\log 3 = 0.48$, $\log 5 = 0.69$, $\log 7 = 0.84$)
$\text{X}_2\text{Y}_4(g) \rightleftharpoons 2\text{XY}_2(g)$
At equilibrium, $75\%$ X$_2$Y$_4$(g) was dissociated and the total pressure is $1$ atm. The magnitude of $\Delta_r G^{\ominus}$ (in kJ mol$^{-1}$) at this temperature is __________. (Nearest Integer) (Given : R $= 8.3$ J mol$^{-1}$ K$^{-1}$; $\ln 10 = 2.3$, $\log 2 = 0.3$, $\log 3 = 0.48$, $\log 5 = 0.69$, $\log 7 = 0.84$)
Correct answer: 8
Solution
The given equilibrium reaction is:
$\text{X}_2\text{Y}_4(g) \rightleftharpoons 2\text{XY}_2(g)$
Initial moles: $1$
Degree of dissociation, $\alpha = 0.75$
Moles of $\text{X}_2\text{Y}_4$ at equilibrium $= 1 - \alpha = 1 - 0.75 = 0.25$
Moles of $\text{XY}_2$ at equilibrium $= 2\alpha = 2 \times 0.75 = 1.5$
Total moles at equilibrium $= 0.25 + 1.5 = 1.75$
Given total pressure, $P = 1$ atm.
Partial pressure of $\text{X}_2\text{Y}_4$, $P_{\text{X}_2\text{Y}_4} = \dfrac{0.25}{1.75} \times 1 = \dfrac{1}{7}$ atm
Partial pressure of $\text{XY}_2$, $P_{\text{XY}_2} = \dfrac{1.5}{1.75} \times 1 = \dfrac{6}{7}$ atm
The equilibrium constant $K_p$ is given by:
$K_p = \dfrac{(P_{\text{XY}_2})^2}{P_{\text{X}_2\text{Y}_4}} = \dfrac{(\dfrac{6}{7})^2}{\dfrac{1}{7}} = \dfrac{36}{7}$
The standard Gibbs free energy change is:
$\Delta_r G^{\ominus} = -RT \ln K_p = -2.3 RT \log K_p$
Calculating $\log K_p$:
$\log \left(\dfrac{36}{7}\right) = \log 36 - \log 7 = \log (2^2 \times 3^2) - \log 7$
$\log \left(\dfrac{36}{7}\right) = 2 \log 2 + 2 \log 3 - \log 7$
$\log \left(\dfrac{36}{7}\right) = 2(0.3) + 2(0.48) - 0.84 = 0.6 + 0.96 - 0.84 = 0.72$
Substituting the values into the $\Delta_r G^{\ominus}$ equation:
$\Delta_r G^{\ominus} = -2.3 \times 8.3 \times 600 \times 0.72$
$\Delta_r G^{\ominus} = -8246.88 \text{ J mol}^{-1} = -8.24688 \text{ kJ mol}^{-1}$
The magnitude of $\Delta_r G^{\ominus}$ is $8.24688 \text{ kJ mol}^{-1}$.
Rounding to the nearest integer, we get $8$.
Answer: $8$
$\text{X}_2\text{Y}_4(g) \rightleftharpoons 2\text{XY}_2(g)$
Initial moles: $1$
Degree of dissociation, $\alpha = 0.75$
Moles of $\text{X}_2\text{Y}_4$ at equilibrium $= 1 - \alpha = 1 - 0.75 = 0.25$
Moles of $\text{XY}_2$ at equilibrium $= 2\alpha = 2 \times 0.75 = 1.5$
Total moles at equilibrium $= 0.25 + 1.5 = 1.75$
Given total pressure, $P = 1$ atm.
Partial pressure of $\text{X}_2\text{Y}_4$, $P_{\text{X}_2\text{Y}_4} = \dfrac{0.25}{1.75} \times 1 = \dfrac{1}{7}$ atm
Partial pressure of $\text{XY}_2$, $P_{\text{XY}_2} = \dfrac{1.5}{1.75} \times 1 = \dfrac{6}{7}$ atm
The equilibrium constant $K_p$ is given by:
$K_p = \dfrac{(P_{\text{XY}_2})^2}{P_{\text{X}_2\text{Y}_4}} = \dfrac{(\dfrac{6}{7})^2}{\dfrac{1}{7}} = \dfrac{36}{7}$
The standard Gibbs free energy change is:
$\Delta_r G^{\ominus} = -RT \ln K_p = -2.3 RT \log K_p$
Calculating $\log K_p$:
$\log \left(\dfrac{36}{7}\right) = \log 36 - \log 7 = \log (2^2 \times 3^2) - \log 7$
$\log \left(\dfrac{36}{7}\right) = 2 \log 2 + 2 \log 3 - \log 7$
$\log \left(\dfrac{36}{7}\right) = 2(0.3) + 2(0.48) - 0.84 = 0.6 + 0.96 - 0.84 = 0.72$
Substituting the values into the $\Delta_r G^{\ominus}$ equation:
$\Delta_r G^{\ominus} = -2.3 \times 8.3 \times 600 \times 0.72$
$\Delta_r G^{\ominus} = -8246.88 \text{ J mol}^{-1} = -8.24688 \text{ kJ mol}^{-1}$
The magnitude of $\Delta_r G^{\ominus}$ is $8.24688 \text{ kJ mol}^{-1}$.
Rounding to the nearest integer, we get $8$.
Answer: $8$
Hence, the correct answer is 8.