# How to show that when the equilibrium constant is larger than one, some real-life equilibria are still reactant favoured?

My instructor assigned us with this question:

When real examples of chemical equilibrium are used, the conclusions about reactant- or product-favoured versus the value of $K$ must be modified (e.g. real equilibria can have $K > 1$ but still be reactant-favoured). Discuss this idea, using examples of real chemical equilibria and assumed values of equilibrium concentration.

I was taught that when $K > 1$, it will be product favoured, assuming it’s already at equilibrium ($Q = K$).

How would I go about answering this question? Do I approach it with regards to initial concentrations in the system, their initial states (mixed vs. unmixed), pressure and volume changes?

• This question can't be answered without a more precise definition of "product-favored" and "reactant-favored." – Zhe Mar 18 '17 at 20:32
• @Zhe Doesn't "reactant-favored" imply $k_\mathrm{b} > k_\mathrm{f}$? Or are there other definitions? – Gaurang Tandon Jun 15 '18 at 5:20
• @GaurangTandon In a case of $\ce{A->B}$, sure, but what about an arbitrarily complex reaction? What if the system is very concentrated versus very dilute? – Zhe Jun 15 '18 at 14:00

The terms reactant-favored and product-favored

As discussed in the comments and in YB609's answer, these are not technical terms with precise definitions.

First possible definition One straight-forward definition would be to call reactant-favored reactions those for which $$K$$ is smaller than one, and product-favored those for which $$K$$ is larger than one. You would have to specify the reaction (with phase information), the temperature and the standard state you are using. With this definition, you would be correct to say that such a reaction, if initiated at standard state, would go in the net forward direction to reach equilibrium if $$K$$ is greater one (and therefor, according to this definition, it is product-favored). Of course, if you take this definition, the question does not make any sense.

Second possible definition You could make a definition based on whether there is "more" reactant or product at equilibrium. This is problematic for multiple reasons. First, you could have a lot of one product, little of another product, and something in between for the reactant. So is there more reactant or more product? Second, if you use amounts of substance (moles) as the criteria for "more" and have a heterogeneous reaction, the relative amounts depend on the volumes (compare oxygen in equilibrium between hemoglobin and the dioxygen in a test tube vs in the entire atmosphere).

Third possible definition If you look at the concentrations to define "more", it still is confusing under some circumstances, even for a homogeneous reaction (everything in the same phase). If the sum of coefficients of reactants is not equal to that of products, diluting the system will disturb an equilibrium, so reactant: product ratios will depend on the dilution of the system.

You would start by asking for the definition of reactant-favored and product-favored.

Do I approach it with regards to initial concentrations in the system, their initial states (mixed vs. unmixed), pressure and volume changes?

Once the terms are clearly defined, it would be sufficient to look at equilibrium states (and you need to know the definition of the standard state, otherwise the value of $$K$$ has no meaning).

Actually you should specify correctly what exactly product favoured and reactant favoured mean according to you. Some text use product favoured or reactant favoured term to notate the direction of reaction and some text use product favoured or reactant favoured term to notate the domination of concentration.

1. If product favoured term is used to notate direction of reaction:

If $$K < 1$$ that means reactant concentration is greater than product concentration. Now at equilibrium concentration of reactant can be greater than that of product or vice-versa. Say for a reaction $$\ce{A -> B}$$, we get $$K = \frac{\ce{[B]}}{\ce{[A]}}$$ and let's say this $$K$$ is less than 1. Now this ratio will always remain constant. Now if you add some more reactant, then it will get consumed and yield product so as to keep $$K = \frac{\ce{[B]}}{\ce{[A]}}$$ constant. Now adding reactant made the reaction move towards right i.e product favoured (i.e reaction moves towards the product) but still $$K < 1$$ is being maintained.

2. If product favoured is used to notate domination of concentration in the reaction i.e. if by the term product favoured you mean to say at equilibrium concentration dominates product concentration, then according to this notation:

Again taking the example of $$\ce{A -> B}$$, we get $$K = \frac{\ce{[B]}}{\ce{[A]}}$$ and let's say this K is less than 1. That means at equilibrium reactant concentration is greater than product concentration. So according to the notation we are using now, we should say it to be reactant favoured. Now $$K < 1$$ will always be reactant favoured and can never be product favour if you are using the term reactant favoured or product favour to notate domination of concentration in the reaction.

If you are using the terms product favoured and reactant favoured to notate the direction of reaction, then saying K < 1 and saying product being favoured makes sense. According to your professor I think product favoured and reactant favoured mean the direction of reaction. So, it depends how you use those terms.

• By only discussing $\ce{A -> B}$ examples, you miss a lot of the subtleties. – Karsten Theis Aug 3 '20 at 17:49