Listen to the Podcast Episode Here
What is the Equilibrium Constant?
In an earlier article we defined reversible reactions and chemical equilibrium as well as considering how to use Le Chatelier’s Principle to predict the way that an equilibrium position shifts when certain conditions are changed.
Those changes we considered were qualitative, not quantitative. In other words, we considered how the equilibrium changes, but not how much it changes nor the original position of equilibrium.
In your exam you may be be asked to quantify the position of equilibrium. In fact it’s almost guaranteed.
And the way we quantify equilibrium is by using the equilibrium constant, Kc
We need the equilibrium concentration of reactants and products when we calculate equilibrium constant. The basic form of the equation can be shown as:
I find the above equation a good one to remember. It’s a visual of the “first principles” when I need to think about constructing the equation for a particular reaction. But it is too basic to use as it is, so we next need to consider a generic reaction. Let’s consider this reaction between A and B to produce C and D:
We can write a more specific version of the equilibrium constant for this reaction, and note that we need to include the values number of moles of each reactants and products (a, b, c and d).
Because of the structure of the equations (with products over reactants) we can make some statements based upon the value of Kc.
If the value is well over 1 it suggests that there are more products than reactants at equilibrium, and that the forward reaction is favoured.
If the value is much less than 1 it suggests that there are more reactants than products at equilibrium, and that the reverse reaction is favoured.
When comparing the Kc values for two temperatures, the higher Kc value indicates the temperature that favours the forward reaction.
The “Expression” for Kc
You may see this phrase in exam questions from past papers, and you will likely see it in your actual exam too. But what does it mean?
Usually, you see a question related to a particular reversible reaction and are asked something like “write an expression for the equilibrium constant” for the reaction.
This just means to write the equation for Kc for the specific reaction.
To illustrate this we need to consider a specific reaction. We’ll use an example that comes up regularly – the reaction between ethanol and ethanoic acid. The products are ethyl ethanoate and water. The chemical equation is:
So, how do we write the equation for the equilibrium constant for this reaction? It’s simply a case of taking the general equation for Kc and putting the details from the chemical equation. If we do that we will end up with the following equation:
This is the expression for the equilibrium constant, Kc, for this reaction.
How Does Kc Change When We Change the Conditions?
In our article Reversible Reactions and Chemical Equilibrium we discussed how to use Le Chatelier’s Principle to predict the direction of change of an equilibrium position when we alter certain conditions, but we did not quantify this.
The equilibrium constant quantifies the position of equilibrium, and now we need to consider how it changes, if at all, when we disturb the temperature, a concentration of a substance, or introduce a catalyst.
How Does Kc Change When We Change the Temperature?
We already discussed that raising the temperature of a reaction at equilibrium will drive the reaction in the endothermic direction, moving the equilibrium position in that direction. If the endothermic reaction is the forward reaction the value of Kc will increase. As you would expect, if the reverse reaction is the endothermic reaction the value of Kc will decrease.
In fact, changing temperature is the only condition that will change the value of Kc.
How Does Kc Change When We Change the Reaction Composition?
If we change the concentration of one of the substances in a reaction at equilibrium, the equilibrium moves to minimise the effect of the disturbance. As you will remember, this follows what Le Chatelier’s Principle tells us.
The equilibrium will re-establish and the Kc value will be unchanged. Only the concentrations are different.
How Does Kc Change When We Introduce a Catalyst?
You’ve probably learnt elsewhere that a catalyst increases the rate of a reaction. So, what does it for a reversible reaction?
It increases the rate of both the forward and the reverse reactions, to the same degree. Therefore, it does not affect the position of equilibrium, nor Kc .
The only change in conditions that changes the equilibrium constant Kc is a change in temperature.
Do you have a question for the podcast?
Just submit your question here and your suggestion may be included as a future episode.