Confused about underlying reasoning behind reaction quotient/equilibrium constant

Question

I am a high school student taking AP Chemistry. In the unit on chemical equilibrium, there is much emphasis placed on calculating reaction quotients and equilibrium constants using this method: $$aA + bB \rightleftharpoons cC + dD$$ $$Q_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$$ It has been explained that equilibrium constants much larger than 1 indicate that the reaction favors the products, and equilibrium constants much smaller than 1 indicate that the reaction favors the reactants. I am curious why this is.

I am familiar with the notion that the expressions in the numerator and the denominator are proportional to the probability of particle collisions for both respective sides of the reaction. However, I do not wholly understand the reason behind this, or how this relates to the idea that the equilibrium constant can quantify the degree to which the reactants or products are favored.

I also understand that the reaction quotient is another way of expressing the ratio of the rate constants, $\frac{k_f}{k_r}$, but I do not understand how this, too, relates to the reaction constant's ability to quantify reaction favorability.

Edit: Just to clarify, I understand that when the quotient is greater than 1, the numerator is greater, and vice versa. My confusion is about why the concentrations raised to their stoichiometric coefficients quantify reaction favorability.

The fact that "concentration raised to a stochiometric coefficient" is the right form was first observed experimentally, and proposed as a law. Nowadays, we can also derive this from first principle thermodynamical laws. You can read more about it on Wikipedia, but there are also several questions on this StackExchange, like Is there a reason for the mathematical form of the equilibrium constant?. — Domen, Mar 19 '23 at 14:33

score 0 · Answer 1 · answered Mar 19 '23 at 13:23

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Q is a fraction which, like every fraction, could be defined as A/B.

If A > B, Q = A/B is > 1 and the numerator, depending on the final products, is bigger than the denominator, depending on the reactants. The products are more important than the reactants, at the equilibrium.

If A < B, this is the contrary : Q < 1. The reactants are more important than the products.

answered Mar 19 '23 at 13:23

Maurice

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Thank you for your answer. I understand this much about Q. I am confused why the numerator and denominator accurately quantify the "importance" of the reactants or products. Why has the form concentration raised to a stoichiometric coefficient been used to denote "importance"? – Scott Mar 19 '23 at 13:26
In my comment, the word "Importance" is equivalent to the "product of concentrations raised to a stoichiometric coefficient". – Maurice Mar 19 '23 at 19:47

porphyrin · Answer 2 · 2023-03-19T16:29:03.413

The quotient can take on many values, initially suppose that no product is present then its value is zero. Only at equilibrium do we normally know its value, which is the equilibrium constant, unless we solve the differential equations in time and get the changing concentration ratios that way. At equilibrium the concentrations obviously have their equilibrium value and then the quotient becomes the equilibrium constant which is also the ratio of rate constants $K_e = k_f/k_r$ because only here is the rate of forwards and reverse reactions equal.

If $K_e >1$ there are clearly going to be more products that reactants and so the reaction 'favours the products' but in a more general sense when this is the case the expression $-RT\ln(K_e)$ give the 'free energy' and when this is negative the reaction is called 'spontaneous', although that absolutely does not mean that it will occur at all rapidly. Texts on thermodynamics go into this in detail.

Thank you for your response. It has been very helpful to think about the fact that the rate law is the derivative of time vs. concentration. — Scott, Mar 19 '23 at 18:43

Confused about underlying reasoning behind reaction quotient/equilibrium constant

2 Answers2