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

Question

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.

Answer 1

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.