Consider the following reaction at 298 K. 2 SO2(g) + O2(g) → 2 SO3(g)
An equilibrium mixture contains O2(g) and SO3(g) at partial pressures of 0.43 atm and 2.6 atm, respectively. Using data from Appendix 4, determine the equilibrium partial pressure of SO2 in the mixture.
atm.

Answer and Explanation:

The reaction is in the gas phase, so the equilibrium constant is expressed in terms of the partial pressures (P) of the products and reactants, as follows:

We have the following data:

P(SO₃) = 2.6 atm

P(O₂) = 0.43 atm

We need Kp for this reaction. We can assume that in Appendix 4 we found that Kp = 7 x 10²⁴.

Then, we introduce the data in the equilibrium constant expression to calculate the partial pressure f SO₂ (PSO₂), as follows:

Therefore, the partial pressure of SO₂ is 1.5 x 10⁻¹² atm (for the given Kp).