Question

Consider the following reaction at 298 K.

Answer 1

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:

$Kp = \frac{P^{2}_{SO_{3} } }{P_{SO_{2}} ^{2}P_{O_{2}} }$

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:

$P_{SO_{2} } = \sqrt{\frac{P_{SO_{3} } ^{2} }{Kp P_{O_{2} } } } = \sqrt{\frac{(2.6 atm)^{2} }{(7 x 10^{24)}(0.43 atm) } } = 1.5 x 10^{-12} atm$

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