Question

2 SO2(g) + O2(g) equilibrium reaction arrow 2 SO3(g) Calculate the equilibrium partial pressures of SO2, O2, and SO3 produced from an initial mixture in which the partial pressures of SO2 and O2 = 0.52 atm and the partial pressure of SO3 = 0 (exactly).

Answer 1

Answer:

$p_{SO_2} \approx 0 atm$

$p_{O_2} = 0.52 - x = 0.26 atm$

$p_{SO_3} = 2x = 0.52 atm$

Explanation:

The equilibrium system is defined by:

$2 SO_2 (g) + O_2 (g)\rightleftharpoons 2 SO_3 (g)$

First of all, we need to be given the equilibrium constant for this reaction. Equilibrium constant is temperature-dependent. For the purpose of this problem and illustration of the thought process, we'll assume room temperature for which the equilibrium constant of this reaction is defined as:

$K_{eq} = 4.3\cdot 10^6$

Define the equilibrium constant in terms of equilibrium molarities:

$K_{eq} = \frac{p_{SO_3}^2}{p_{SO_2}^2p_{O_2}}$

Let's say that x atm of oxygen react, then, according to stoichiometry, 2x atm of sulfur dioxide react and 2x atm of sulfur trioxide are formed. The equilibrium amounts are then:

$p_{SO_2} = 0.52 - 2x$

$p_{O_2} = 0.52 - x$

$p_{SO_3} = 2x$

Substitute into the K expression:

$K_{eq} = 4.3\cdot 10^6 = \frac{(2x)^2}{(0.52 - 2x)^2(0.52 - x)}$

For such a huge equilibrium constant, we may assume that this reaction nearly goes to completion, so we'd have a limiting reactant, in which:

$0.52 - 2x\approx 0$

This means:

$x = 0.26 atm$

And the equilibrium partial pressures become:

$p_{SO_2} \approx 0 atm$

$p_{O_2} = 0.52 - x = 0.26 atm$

$p_{SO_3} = 2x = 0.52 atm$