Unit 7: Topic 10 - Reaction Quotient and Le Châtelier’s Principle

Question 1

How does Q shift when a system is disturbed out of equilibrium?

Answer 1

Let’s use an example reaction : A (aq) + B (aq) -> C (aq) + D (aq)

When the reaction is at equilibrium : Q = K = [C][D] / ([A][B]) **

Note : Q is calculated in the same way as K, we just use Q when the reaction is not at equilibrium

However, what if we increased the concentration of A?
A is a reactant so we divide by [A] in the expression for Q. By increasing [A], we decrease Q.

K has not been changed so now, Q < K.

We already know that according to Le Châtelier’s principle, the reaction will shift towards products. Therefore, [C] and [D] will increase until Q=K once again (equilibrium has been reachieved).

If we redo this scenario but increase the concentration of [C] or [D] as opposed to [A], then Q > K and the reaction will shift towards reactants to decrease Q until equilibrium is achieved

Question 2

What is the effect of temperature on Q and K?

Answer 2

The effect of temperature on K depends on the enthalpy of the reaction. We know that endothermic reactions have heat as a “reactant” while exothermic reactions have heat as a “product.” Increasing/decreasing temperature will shift K just like how increasing/decreasing concentrations of reactants will shift Q.

One possible scenario is that we raise the temperature of an exothermic reaction :
In this case, heat is a product so by raising the temperature, the reactant side becomes more favored and K decreases. Q has not changed so Q>K.

The concentration of reactants will increase until Q increases enough for Q=K.

Question 3

What is the effect of changes in pressure and volume on Q and K?

Answer 3

If pressure increases, then the side of the reaction with fewer gaseous molecules will be favored. If the reactant side is favored, then K decreases. If the product side is favored, then K increases.

An example could be : A (g) + 3B (l) -> 2C (g) + D (g)
If pressure decreases, then the product side is more favored because it has 3 moles of gas compared to 1 on the reactant side. Therefore, K increases.

Once again, Q has not changed and so Q