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

Step 1: Understanding the Question:
We need to calculate the standard Gibbs free energy change (\(\Delta G^{\ominus}\)) using the equilibrium constant \(K_{p}\).
Step 2: Key Formula or Approach:
1. \(\Delta_{r}G^{\ominus} = -RT \ln K_{p}\).
2. \(K_{p} = \frac{(P_{XY_{2}})^{2}}{P_{X_{2}Y_{4}}}\).
Step 3: Detailed Explanation:
Reaction: \(X_{2}Y_{4} \rightleftharpoons 2XY_{2}\).
Initial moles: \(1, 0\).
At equilibrium: \(1 - \alpha, 2\alpha\).
Given \(\alpha = 0.75\).
Equilibrium moles: \(1 - 0.75 = 0.25\) and \(2 \times 0.75 = 1.5\).
Total moles \(= 0.25 + 1.5 = 1.75\).
Partial pressures (\(P_{total} = 1 \text{ atm}\)):
\[ P_{X_{2}Y_{4}} = \frac{0.25}{1.75} \times 1 = \frac{1}{7} \text{ atm} \]
\[ P_{XY_{2}} = \frac{1.5}{1.75} \times 1 = \frac{6}{7} \text{ atm} \]
Calculation of \(K_{p}\):
\[ K_{p} = \frac{(6/7)^{2}}{1/7} = \frac{36/49}{1/7} = \frac{36}{7} \approx 5.14 \]
Calculation of \(\Delta G^{\ominus}\):
\[ \Delta G^{\ominus} = -8.3 \times 600 \times \ln(36/7) \]
\[ \ln(36/7) = 2.3 \log(36/7) = 2.3 [\log 36 - \log 7] \]
\[ = 2.3 [2 \log 2 + 2 \log 3 - \log 7] = 2.3 [2(0.3) + 2(0.48) - 0.84] \]
\[ = 2.3 [0.6 + 0.96 - 0.84] = 2.3 [0.72] = 1.656 \]
\[ \Delta G^{\ominus} = -8.3 \times 600 \times 1.656 = -4980 \times 1.656 \approx -8246 \text{ J/mol} \]
Magnitude in kJ/mol \(\approx 8.246 \rightarrow 8\).
Step 4: Final Answer:
The magnitude is 8 kJ/mol.