JEE Main 2026 — Chemistry Physical Chemistry

The given equilibrium reaction is:

$$\text{X}_2\text{Y}_4(g) \rightleftharpoons 2\text{XY}_2(g)$$X2​Y4​(g)⇌2XY2​(g)

Initial moles: $1$1

Degree of dissociation, $\alpha = 0.75$α=0.75

Moles of $\text{X}_2\text{Y}_4$X2​Y4​ at equilibrium $= 1 - \alpha = 1 - 0.75 = 0.25$=1−α=1−0.75=0.25

Moles of $\text{XY}_2$XY2​ at equilibrium $= 2\alpha = 2 \times 0.75 = 1.5$=2α=2×0.75=1.5

Total moles at equilibrium $= 0.25 + 1.5 = 1.75$=0.25+1.5=1.75

Given total pressure, $P = 1$P=1 atm.

Partial pressure of $\text{X}_2\text{Y}_4$X2​Y4​, $P_{\text{X}_2\text{Y}_4} = \dfrac{0.25}{1.75} \times 1 = \dfrac{1}{7}$PX2​Y4​​=1.750.25​×1=71​ atm

Partial pressure of $\text{XY}_2$XY2​, $P_{\text{XY}_2} = \dfrac{1.5}{1.75} \times 1 = \dfrac{6}{7}$PXY2​​=1.751.5​×1=76​ atm

The equilibrium constant $K_p$Kp​ is given by:

$$K_p = \dfrac{(P_{\text{XY}_2})^2}{P_{\text{X}_2\text{Y}_4}} = \dfrac{(\dfrac{6}{7})^2}{\dfrac{1}{7}} = \dfrac{36}{7}$$Kp​=PX2​Y4​​(PXY2​​)2​=71​(76​)2​=736​

The standard Gibbs free energy change is:

$$\Delta_r G^{\ominus} = -RT \ln K_p = -2.3 RT \log K_p$$Δr​G⊖=−RTlnKp​=−2.3RTlogKp​

Calculating $\log K_p$logKp​:

$$\log \left(\dfrac{36}{7}\right) = \log 36 - \log 7 = \log (2^2 \times 3^2) - \log 7$$log(736​)=log36−log7=log(22×32)−log7

$$\log \left(\dfrac{36}{7}\right) = 2 \log 2 + 2 \log 3 - \log 7$$log(736​)=2log2+2log3−log7

$$\log \left(\dfrac{36}{7}\right) = 2(0.3) + 2(0.48) - 0.84 = 0.6 + 0.96 - 0.84 = 0.72$$log(736​)=2(0.3)+2(0.48)−0.84=0.6+0.96−0.84=0.72

Substituting the values into the $\Delta_r G^{\ominus}$Δr​G⊖ equation:

$$\Delta_r G^{\ominus} = -2.3 \times 8.3 \times 600 \times 0.72$$Δr​G⊖=−2.3×8.3×600×0.72

$$\Delta_r G^{\ominus} = -8246.88 \text{ J mol}^{-1} = -8.24688 \text{ kJ mol}^{-1}$$Δr​G⊖=−8246.88 J mol−1=−8.24688 kJ mol−1

The magnitude of $\Delta_r G^{\ominus}$Δr​G⊖ is $8.24688 \text{ kJ mol}^{-1}$8.24688 kJ mol−1.

Rounding to the nearest integer, we get $8$8.

Answer: $8$8