Question

For the reaction SO2(g) + 1/2 O2(g) SO3(g) H°298 = –98.32 kJ/mole, S°298 = –95.0 J/K/mole. Find the Kp for this reaction at 298 K

• A. 1.30 x 10^12
• B. 1.30 x 10^-12
• C. 6.50 x 10^11
• D. 6.50 x 10^-11

Answer 1

Answer: (A)

1.30 x 10^12

Answer 2

The relationship between the standard Gibbs free energy change ($\Delta G^\circ$) and the equilibrium constant ($K_p$) is given by: $\Delta G^\circ = -RT \ln K_p$ The standard Gibbs free energy change can be calculated from the standard enthalpy and entropy changes using the Gibbs-Helmholtz equation: $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$

Step 1: Convert units to be consistent. Given values: $\Delta H^\circ_{298} = -98.32 \text{ kJ/mole}$ $\Delta S^\circ_{298} = -95.0 \text{ J/K/mole}$ $T = 298 \text{ K}$ Ideal gas constant $R = 8.314 \text{ J/K/mole}$.

Convert $\Delta H^\circ$ to Joules: $\Delta H^\circ_{298} = -98.32 \text{ kJ/mole} \times 1000 \text{ J/kJ} = -98320 \text{ J/mole}$

Step 2: Calculate the standard Gibbs free energy change ($\Delta G^\circ$). $\Delta G^\circ_{298} = \Delta H^\circ_{298} - T\Delta S^\circ_{298}$ $\Delta G^\circ_{298} = (-98320 \text{ J/mole}) - (298 \text{ K}) \times (-95.0 \text{ J/K/mole})$ $\Delta G^\circ_{298} = -98320 \text{ J/mole} + 28310 \text{ J/mole}$ $\Delta G^\circ_{298} = -70010 \text{ J/mole}$

Step 3: Calculate the equilibrium constant ($K_p$). $\ln K_p = \frac{-\Delta G^\circ}{RT}$ $\ln K_p = \frac{-(-70010 \text{ J/mole})}{8.314 \text{ J/K/mole} \times 298 \text{ K}}$ $\ln K_p = \frac{70010 \text{ J/mole}}{2477.572 \text{ J/mole}}$ $\ln K_p \approx 28.2570$

To find $K_p$: $K_p = e^{28.2570}$ $K_p \approx 1.301 \times 10^{12}$

Rounding to three significant figures, $K_p \approx 1.30 \times 10^{12}$.