In a closed flask at 600 K, one mole of $$X_2Y_4(g)$$ attains equilibrium as given below : $$$X_2Y_4(g) \rightleftharpoons 2XY_2(g)$$$ At equilibrium, 75% $$X_2Y_4(g)$$ 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)

For the reaction

$$X_2Y_4(g)\rightleftharpoons2XY_2(g),$$

initially, (1) mole of ($$X_{2}Y_4$$) is present and (75%) of it dissociates at equilibrium.

Hence, the equilibrium moles are:

• ($$X_{2}Y_4$$ = 1-0.75 = 0.25)
• ($$XY_2$$ = 2$$\times$$0.75 = 1.5)

The total number of moles at equilibrium is

$$0.25+1.5=1.75.$$

Since the total pressure is (1) atm, the mole fractions are

$$x_{X_2Y_4}=\frac{0.25}{1.75}=\frac{1}{7},$$

and

$$x_{XY_2}=\frac{1.5}{1.75}=\frac{6}{7}.$$

Therefore, the partial pressures are

$$P_{X_2Y_4}=\frac{1}{7}\text{ atm}$$

and

$$P_{XY_2}=\frac{6}{7}\text{ atm}.$$

The equilibrium constant is

$$K_p=\frac{(P_{XY_2})^2}{P_{X_2Y_4}}=\frac{\left(\frac{6}{7}\right)^2}{\frac{1}{7}}=\frac{36}{49}\times7=\frac{36}{7}\approx5.14.$$

The standard Gibbs free energy change is given by

$$\Delta_rG^\ominus=-RT\ln K_p.$$

Using

$$R=8.3\ \text{J mol}^{-1}\text{K}^{-1},\qquad T=600\ \text{K},$$

and

$$\ln K_p=\ln\left(\frac{36}{7}\right)=2\ln6-\ln7,$$

with

$$\ln x=2.3\log x,$$

we obtain

$$\ln K_p=2.3,[2(\log2+\log3)-\log7]$$

$$=2.3,[2(0.3+0.48)-0.84]$$

$$=2.3,(1.56-0.84)$$

$$=2.3\times0.72$$

$$=1.656.$$

Substituting into the Gibbs free energy equation,

$$\Delta_rG^\ominus=-(8.3)\times600\times1.656$$

$$\approx-8251\ \text{J mol}^{-1}$$

$$\approx-8.25\ \text{kJ mol}^{-1}.$$

Rounding to the nearest integer, the magnitude of the standard Gibbs free energy change is ($$8\ \text{kJ mol}^{-1}$$).