Textbook Problem 5.57

[Meghna Sehrawat 3A]

In this textbook problem, I got the right answer, however the explanation on the solution manual confused me. https://drive.google.com/file/d/1sGeOQf2ZMyT7MzCsy4MC7tfol_GJrq-R/view. This is the problem, and on part B I just did the equilibrium with the moles given even though it was gas. However, the solution manual says this: " Since there are two moles of gas on both sides of the equation, the volume cancels out of the equilibrium constant expression and it is possible to use moles directly for each component."I understand that you can't do concentration values for the gases, but I'm having trouble understanding what the book is saying exactly.

[Francis Le 2K]

Since there are two moles of gases on each side of the equation, the equilibrium constant expression would be K=[mol SO2/L][mol NO2/L] / [mol SO3/L][mol NO/L]. Thus, the units would be [(mol SO2)(mol NO2)/L^2] / [(mol SO3)(mol NO)/L^2]. Since both the numerator and denominator have L^2, it cancels out, leaving only moles. This allows you to use moles directly instead of concentrations.

However, if, for example, a reaction has 2 moles of product and 1 moles of reactants, the equilibrium constant would have units [(mol Product 1)(mol Product 2)/L^2] / [(mol Reactant)/L]. The numerator would have L^2 while the denominator would have L, so the volumes do not cancel out completely, preventing you from using moles directly.