Question

The standard free energy change for the reaction, \( {SO}_2 + \frac{1}{2} {O}_2 \rightleftharpoons {SO}_3 \) at equilibrium is given by \( \Delta G^\circ = -94600 + 89.37T \), where \( T \) is in Kelvin and \( \Delta G^\circ \) is in Joules. The equilibrium constant (\( K_p \)) at 1050 K is (rounded off to two decimal places) ...........
Given: Universal gas constant (R) $= 8.314\ {J} \cdot {K}^{-1} \cdot {mol}^{-1}$

Hint:

When calculating the equilibrium constant from the standard free energy change, remember to use the correct units for \( R \) and \( \Delta G^\circ \), and solve the equation for \( K_p \) carefully.

Answer 1

To calculate the equilibrium constant \( K_p \) from the standard free energy change \( \Delta G^\circ \), we use the equation:
\[ \Delta G^\circ = -RT \ln K_p \] where:
- \( \Delta G^\circ = -94600 + 89.37 \times T \),
- \( R = 8.314 \, {J/mol·K} \) is the universal gas constant,
- \( T = 1050 \, {K} \) is the temperature.
Step 1: Calculate \( \Delta G^\circ \) at \( T = 1050 \, {K} \)
Substituting the value of \( T = 1050 \, {K} \) into the expression for \( \Delta G^\circ \): \[ \Delta G^\circ = -94600 + 89.37 \times 1050 \] \[ \Delta G^\circ = -94600 + 93838.5 = -761.5 \, {J/mol}. \] Step 2: Use the equation to find \( K_p \)
Now, substitute \( \Delta G^\circ = -761.5 \, {J/mol} \) into the equation: \[ -761.5 = -8.314 \times 1050 \ln K_p \] Solving for \( \ln K_p \): \[ \ln K_p = \frac{-761.5}{-8.314 \times 1050} \approx 0.0906 \] Exponentiating both sides: \[ K_p = e^{0.0906} \approx 1.05. \] Thus, the equilibrium constant \( K_p \) at 1050 K is approximately \( 1.05 \).