So say I have a reaction where there are x moles of gas on the reactant side, and y moles of gas on the product side, where x>y. The volume decreases, favoring the products side. The textbook I am using says that the equilibrium concentration of the reactants would decrease, but wouldn't it end up increasing due to the lowered volume?


1 Answer 1


Suppose you have the following generic reaction:

$\ce{xA_{(g)} <=> yB_{(g)}}$

With the condition, $x > y$. The reaction quotient, written in terms of concentration would be:

$Q = \frac{(\frac{n_B}{V})^y}{(\frac{n_A}{V})^x}$

Where $n_A$ and $n_B$ are moles of the respective species and $V$ is volume. With some algebraic manipulation, this can be rewritten as follows (note here I'm explicitly treating $Q$ as dimensionless, q.v., the relevant Wikipedia page on equilibrium constant units, which applies to the reaction quotient as well):

$Q = \frac{n_B^y}{n_A^x} \cdot V^{x-y}$

Written in that form, it should be apparent that, if $x > y$, then $x - y > 0$, and therefore $Q$, the ratio of products to reactants, varies directly with $V$. In other words, as volume grows, $Q$ grows, and as volume decreases, $Q$ decreases. Suppose that, prior to some volume change, the reaction is at equilibrium, i.e., $Q = K$. If volume instantaneously begins to increase, $Q$ becomes instantaneously greater than $K$, and therefore the rate of the reverse reaction becomes greater than the rate of the forward reaction, favoring generation of $A$ until equilibrium is reestablished. Conversely, if volume instantaneously begins to decrease, the opposite happens and production of $B$ is favored. This is precisely what one would expect based on an intuitive understanding of Le Chatelier's principle: if volume decreases, equilibrium shifts to favor production of fewer moles of gas to counteract the increasing pressure; if volume increases, equilibrium shifts to favor production of more moles of gas to counteract the decrease in pressure.

Edit: I misread part of your question initially. In short, if the volume decreases, generation of product $B$ becomes favored, so the concentration of reactant $A$ must decrease relative to $B$ (since $x$ moles of $A$ are consumed in the production of $y$ moles of $B$, and the reaction quotient approaches the equilibrium constant over time).

Just to clarify this further, it's true that, in an absolute sense, if volume decreases in an instantaneous fashion, then the concentrations of the reactant and product, considered separately, would both have to increase. However, in relative terms, the concentration of $B$ must increase relative to $A$ as the reaction proceeds. The ratio between the two must grow to compensate for the diminished volume, otherwise $Q$ will remain perpetually smaller than $K$, when in fact $Q$ always converges to $K$, for a given set of conditions, over time.

  • $\begingroup$ Yeah, I knew that relatively [A] would have to decrease with respect to [B], but I wasn't sure if absolutely they would both increase or not. My book only asks "Which of the following is increased by decreasing the volume of the reaction system", and then lists "Equilibrium concentration of reactants" as one of the options, so it reads as if it means absolutely, but the answer says that it meant relatively. I guess it's just a case of bad book writing (seeing as it's an SAT review book, I wouldn't expect the greatest quality though). Thank you for your response. $\endgroup$
    – jsmith
    Jul 30, 2013 at 15:39

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