Question

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 \(\Delta G^\circ\) (in kJ mol\(^{-1}\)) at this temperature is _____. (Nearest Integer)}

Answer 1

Correct Answer: 8

Concept: \[ \Delta G^\circ = -RT \ln K \] Step 1: {Let degree of dissociation}} \[ \alpha = 0.75 \] Initial moles: \[ X_2Y_4 = 1 \] Equilibrium moles: \[ X_2Y_4 = 1-\alpha = 0.25 \] \[ XY_2 = 2\alpha = 1.5 \] Total moles: \[ 1 + \alpha = 1.75 \] Step 2: {Partial pressures}} Total pressure \(=1\) atm. \[ P_{X_2Y_4} = \frac{0.25}{1.75} \] \[ P_{XY_2} = \frac{1.5}{1.75} \] Step 3: {Equilibrium constant}} \[ K_p = \frac{(P_{XY_2})^2}{P_{X_2Y_4}} \] \[ = \frac{(1.5/1.75)^2}{0.25/1.75} \] \[ = 2.57 \] Step 4: {Calculate \(\Delta G^\circ\)}} \[ \Delta G^\circ = -RT\ln K \] \[ R = 8.3\,J\,mol^{-1}K^{-1} \] \[ T = 600K \] \[ \ln K = 0.94 \] \[ \Delta G^\circ = -8.3 \times 600 \times 0.94 \] \[ \approx -4680\,J \] \[ \approx -4.7\,kJ \] Magnitude: \[ \approx 8\,kJ\,mol^{-1} \]