Question

Find $\Delta G$ of the reaction in calorie at 300 K for A(g) $\rightleftharpoons$ 2B (g) ; $K_p = 10$ at 300 K when partial pressures of A(g) & B(g) are 100 kPa & 1000 kPa respectively Given : $\ln 10 = 2.3$ ; $R = 2$ cal/mol-K

• A. 1380 cal/mol
• B. 0 cal/mol
• C. 600 cal/mol
• D. 2300 cal/mol

Answer 1

Answer: (A)

1380 cal/mol

Answer 2

The Gibbs Free Energy change ($\Delta G$) is related to the reaction quotient ($Q$) by the equation: $\Delta G = RT \ln \left(\frac{Q}{K_p}\right)$ For the reaction A(g) $\rightleftharpoons$ 2B(g), the reaction quotient ($Q_p$) is expressed in terms of partial pressures relative to a standard pressure ($P^\circ = 1$ bar = 100 kPa): $Q_p = \frac{(P_B/P^\circ)^2}{(P_A/P^\circ)}$ Given:

• Temperature, $T = 300$ K
• Equilibrium constant, $K_p = 10$
• Partial pressure of A, $P_A = 100$ kPa
• Partial pressure of B, $P_B = 1000$ kPa
• Gas constant, $R = 2$ cal/mol-K
• Standard pressure, $P^\circ = 100$ kPa

Calculate the partial pressures relative to the standard pressure:

• $P_A/P^\circ = \frac{100 \text{ kPa}}{100 \text{ kPa}} = 1$
• $P_B/P^\circ = \frac{1000 \text{ kPa}}{100 \text{ kPa}} = 10$

Calculate the reaction quotient ($Q_p$): $Q_p = \frac{(10)^2}{1} = 100$

Now, calculate $\Delta G$: $\Delta G = RT \ln \left(\frac{Q_p}{K_p}\right)$ $\Delta G = (2 \text{ cal/mol-K}) \times (300 \text{ K}) \times \ln \left(\frac{100}{10}\right)$ $\Delta G = 600 \text{ cal/mol} \times \ln(10)$ Using $\ln 10 = 2.3$: $\Delta G = 600 \text{ cal/mol} \times 2.3$ $\Delta G = 1380 \text{ cal/mol}$