Question

The standard free energy change for the reaction \[ SO_2 + \tfrac{1}{2}O_2 \rightleftharpoons SO_3 \] is given by \(\Delta G^\circ = -94600 + 89.37T\) (J). At \(T=1050\,K\), the equilibrium constant \(K_p\) is (rounded off to two decimal places). Given: Universal gas constant \(R=8.314\,J\,mol^{-1}K^{-1}\).

Hint:

Remember: A slightly negative \(\Delta G^\circ\) leads to \(K_p\) slightly greater than 1.

Answer 1

Correct Answer: 0.9

Step 1: Calculate \(\Delta G^\circ\) at 1050 K.
\[ \Delta G^\circ = -94600 + 89.37(1050) \] \[ = -94600 + 93928.5 = -671.5 \, J \]

Step 2: Relation between \(\Delta G^\circ\) and equilibrium constant.
\[ \Delta G^\circ = -RT \ln K_p \] \[ \ln K_p = -\frac{\Delta G^\circ}{RT} \]

Step 3: Substitute values.
\[ \ln K_p = -\frac{-671.5}{(8.314)(1050)} = \frac{671.5}{8730} \approx 0.0769 \]

Step 4: Exponentiate.
\[ K_p = e^{0.0769} \approx 1.08 \] Recheck values with precision: answer closer to \(\boxed{1.08}\).