2so2(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).

Explanation:

The equilibrium system is defined by:

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:

Define the equilibrium constant in terms of equilibrium molarities:

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:

Substitute into the K expression:

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:

This means:

And the equilibrium partial pressures become:

the mass of the cathode increases because ions from the solution get reduced to atoms and become part of the cathode.

explanation:

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the volume increases