Question

For the following reaction run at 298 K; a mixture of equilibrium contains \(P_{O_2}=0.50\) atm, \(P_{SO_2}=2.0\) atm. Calculate \(P_{SO_3}\) at equilibrium at 298 K?\[\huge{2SO_2(g)+O_2(g)\rightleftharpoons2SO_3(g)}\]\[\large{\Delta G°_{f,SO_3(g)}=-371\frac{kJ}{mol}}\] \[\large{\Delta G°_{f,SO_2(g)}=-300\frac{kJ}{mol}}\]

Answer 1

\[\Delta G°_{rxn}=(\text{moles }SO_3)(\Delta G°_{f,SO_3(g)})-(\text{moles }SO_2)(\Delta G°_{f,SO_2(g)})\]\[\to(2\text{ mol})(\text{-371 kJ/mol})-(2\text{ mol})(\text{-300 kJ/mol})\]\[\to\text{-742 kJ}-(\text{-600 kJ})=\text{142 kJ}\]

Answer 2

As \(\Delta G°_{rxn}=-RT\ln{k}\), we find the equilibrium constant of pressure, \(K_p\):\[\Delta G°_{rxn}=\text{142 kJ}=-(\text{0.00831 kJ/mol}\cdot\text{K})(\text{298 K})\ln K_p\]

Answer 3

Typo correction: The calculated \(\Delta G°_{rxn}\) should be NEGATIVE 142 kilo-Joules.\[\ln K_p=\frac{\text{-142 kJ}}{-(\text{0.00831 kJ/mol}\cdot\text{K})(\text{298 K})}\approx57.3\]\[\therefore K_p=e^{57.3}\approx7.67\times10^{24}\]

Answer 4

And per the following,\[K_p=\frac{P^2_{SO_3}}{P_{O_2}\cdot P^2_{SO_2}}\]will become\[7.67\times10^{24}=\frac{P^2_{SO_3}}{(\text{0.50 atm})(\text{2.0 atm})}\]
... am I doing this right?

Answer 5

Note: I posted this on another site.