# Calculation of equilibrium constant in a gaseous system A ⇌ B + C

I am supposed to determine whether the following two statements are true or false:

1. For the reaction $$\ce{A (g) <=> B (g) + C(g)}$$, $$K_p = \pu{1 atm}$$. If we start with equal moles of all gases at $$\pu{9 atm}$$ of initial pressure, then at equilibrium the partial pressure of $$\ce{A}$$ increases.

2. The reaction quotient $$Q_p > K_p$$, hence the equilibrium shifts in the backward direction.

I discussed it with my friend, who gave me a weird explanation abut how to find the reaction quotient $$Q$$ at the start. He claimed that it would be equal to the total pressure at the starting point, $$\pu{9 atm}$$. But I believe that my friend is wrong, according to the following calculation:

Let us assume that $$x~\mathrm{mol}$$ of each gas is present. Hence the mole fraction of each gas is $$x/(3x)$$ = $$1/3$$. The partial pressure of $$\ce{A}$$ is thus $$p_\ce{A} = (1/3) \times 9 = 3$$ . Similarly we get $$p_\ce{B} = 3$$ and $$p_\ce{C} = 3$$, which gives

$$Q_p = (p_\ce{C} \times p_\ce{B})/p_\ce{A} = (3 \times 3)/3 = 3$$

Regardless of whether $$Q$$ is 3 or 9, it is still larger than $$K$$, so the eventual answer is not changed. However, I wanted to know what the correct value of $$Q$$ was.

## 1 Answer

Your analysis is correct, except that you don't account for how the the units cancel out in your calculation.

The initial reaction quotient, $$Q_P$$, is indeed 3, as you have calculated. And since $$K_P$$ is 1, the reaction will shift to the left, increasing $$P_A$$. Hence Statement-1 is nearly correct. I say "nearly" because $$K_P$$ is dimensionless, so it should be 1, not 1 atm (each partial pressure should be divided by whatever you define as the standard pressure, 1 atm or 1 bar -- that's what you've left out of your calculation; you should try this yourself as an exercise).

Statement-2 is not quite correct, because it says the "equilibrium shifts in the backward direction." Assuming you are not changing the temperature, the equilibrium does not shift (it's a constant: $$K_P$$ = 1). Rather, the reaction (and thus $$Q_P$$) shifts in the backwards direction until equilibrium is attained.

• How you set up an equilibrium constant is your choice; there is no reason for it to be dimensionless. The worst part about this statement is the use of 'moles' as a synonym for 'amounts of substance'. – Martin - マーチン Aug 24 '19 at 8:29
• @Martin-マーチン Yes, equilibrium constants, depending on how they're expressed and the convention used, can sometimes have dimension. However, I've never seen a modern physical chemistry textbook that expressed $K_P$ as anything other than dimensionless. And the statements here concerned $K_P$. Also, I don't understand your objection to the use of "moles". "Amounts of substance" is ambiguous—it could mean the number of molecules or the masses of the molecules. So they weren't using "moles" as a synonym for the ambiguous "amounts of substance", they were using "moles" to mean "moles". – theorist Aug 25 '19 at 0:21
• I have seen in plenty of textbooks that $K_p$ (not a capital p) is defined in partial pressures, therefore it had a unit. All I wanted to convey was that your boiler plate statement is not correct, and why I down voted your answer attempt. The term 'amount of substance' is not ambiguous; it is clearly defined in the gold book. They were using a unit to refer to a quantity, which is wrong. You don't have to understand my objection, but I think it is necessary to point out, so that users who come here to learn get the chance to learn from correct information. – Martin - マーチン Aug 25 '19 at 10:58
• @Martin-マーチン Is your objection that the statement said "equal moles" instead of "an equal number of moles"? If so, I think you are confusing an English typographical error with a nomenclature error. Both statements are clearly written with awkward English, so I implicitly read "equal moles" to be a bad English version of (not a bad nomenclature version of) "equal numbers of moles". – theorist Aug 25 '19 at 18:41
• That usage is wrong. The quantity is called 'amount of substance'. The unit for this quantity is 'mole'. Don't use the one as a synonym for the other. I cannot explain it in any more simpler terms. It's been an issue, which had been brought up on this platform multiple times, I'm on mobile and I don't have the patience to look up the references. We're also beyond the point where any more commenting serves a purpose to improve your answer attempt. If there were more questions which need to be resolved, maybe chat is more appropriate. – Martin - マーチン Aug 25 '19 at 18:48