# The equilibrium constant

### The equilibrium constant

#### Lessons

In this lesson, we will learn:

• To write an expression for the equilibrium constant Keq.
• How to interpret the value of Keq and describe the reaction using this value.
• How to use the equilibrium expression with equilibrium concentrations to solve for Keq (and vice versa).
• To write an expression for the reaction quotient, Q, and learn the difference between Q and Keq.

Notes:

• We now know the definition of equilibrium; a chemical process where the forward reaction rate is equal to the reverse rate. Be careful – this tells us nothing about how much product or reactant is there! To find that, we need to use the equilibrium constant expression.

• Using measurements of reactant and product concentrations, it is possible to find what is called the equilibrium constant, Keq, (sometimes Kc) of a given reaction at equilibrium. This is done using the expression:
For the reaction at equilibrium:
$aA + bB \rightleftharpoons cC + dD$

Keq is calculated by:
$\large K_{eq} = \frac{[C]^c[D]^d}{[A]^a[B]^b}$

• The whole equation is the equilibrium expression.
• Keq is the equilibrium constant.
• Keq is the general term – if the equilibrium is measuring concentration (in mol dm-3) it might be called Kc. Kp would be used if it was partial pressures (for gases).

• The equilibrium constant is called a constant because it is not affected by changes in some conditions. Changes to concentration of reactants or products and changes in pressure do not affect the equilibrium constant!
• Remember Le Chatelier’s principle: the system will counteract any change made. If you add reactant, more product will be made to counteract the change. This keeps Keq constant in the long run.
• Changing temperature WILL affect Keq, depending on whether the reaction is endothermic or exothermic.
Because of this, ALWAYS quote Keq with a temperature.

• The equilibrium expression looks complicated, so breaking it down using simple math can help.
• It is a fraction. It has terms for amount of product and reactant.
• You can write fractions as ratios. So…
• It is a ratio of products to reactants in the reaction. Written as a decimal, the value of Keq tells us something important:
• Keq is smaller than 1: There is less product than reactant in the reaction mixture. The smaller the value of Keq, the less product there is.
• Keq is approximately 1: There is roughly the same amount of product as reactant in the reaction mixture.
• Keq is larger than 1: There is more product than reactant in the reaction mixture. The larger Keq is, the more product compared to reactant.

• When writing Keq for heterogeneous systems, where the substances are not all in the same phase, ignore any substances in the solid state. Solid reagents do not affect the equilibrium constant; everything else in the Keq expression is written as normal.
For example, let’s look at the thermal decomposition of calcium carbonate:

• CaCO3 (s) $\, \rightleftharpoons \,$ CaO (s) + CO2 (g)

• If CO2 escapes the reaction vessel because it’s open, equilibrium cannot and will not be established. CO2 alone dictates whether the equilibrium happens, so that is all the Keq expression contains.

• Keq = [ CO2 (g)]

• Remember that Keq is only used for reactions at equilibrium.
The reaction quotient (symbol Q) is used for reactions not at equilibrium to reveal which direction a reaction favours. Practically, Q is a Keq value for reactions that are not yet at equilibrium!
It is calculated using largely the same information as Keq would be.

• For the reaction:

aA + bB $\,$$\,$ cC + dD

We find Q by calculating:

$Q =$ $\large \frac{[C]^{c} [D]^{d}} {[A]^{a} [B]^{b} }$

This would be for reactants and products in the aqueous or gaseous phase, where concentration is measured by partial pressure (see Kp and partial pressure). Like with Keq, pure solids and liquids are not included in this calculation.

For some reactions, Q isn’t very useful. Reactions that go to completion will have an infinitely large Q (all product, no reactant), and reactions that don’t proceed will have a Q value equal to zero. A reaction with Q = 1 is already at equilibrium.

But many aqueous and gaseous reactions can go to equilibrium. Recall Le Chatelier’s principle: Q is useful because comparing Q to Keq lets us predict changes to concentration as the reaction tries to reach equilibrium.
You can think of Q then as a pendulum; Keq is the point at rest and Q is where it currently is:
• If Q is smaller than Keqfor a reaction, the products will be favoured. Q > Keq suggests there are more reactants in the mixture than the ideal equilibrium concentrations, so according to Le Chatelier’s principle there will be a shift back toward the products.
• If Q is equal to Keq, there is no favouring of products or reactants. Q = Keq means the reaction is already at ideal equilibrium concentrations. In this situation, nothing would be expected to change.
• If Q is greater than Keq, the reactants will be favoured. Q < Keq suggests that there is more product in the vessel than the ideal equilibrium concentration, so the equilibrium will shift back toward the reactants to remove this product surplus.

This Q value is larger than the quoted Keq. Which means there is currently more product than there is at ideal equilibrium conditions. Applying Le Chatelier’s principle, we expect that in the current conditions, the reaction will favour the reactants.
• Introduction
What is the equilibrium constant?
a)
The equilibrium constant and equilibrium expression.

b)
Changing the equilibrium CONSTANT.

c)
Keq: a number for "where is the equilibrium?"

d)
Heterogeneous systems and equilibria.

• 1.
Write the expression for the equilibrium constant, Keq and interpret its value.
The equation for the decomposition of compound A, is below:

A (g) $\, \rightleftharpoons \,$ 2B (g) + C (g)

At 298 K, Keq = 4.5*1015
a)
Write an expression for Keq for this reaction.

b)
What does the value of Keq at 298K tell you about the reaction mixture?

• 2.
Calculate the equilibrium constant for the reaction at equilibrium.
The decomposition of PCl5 is shown by the equation below:

PCl5 (g) $\, \rightleftharpoons \,$ PCl3 (g) + Cl2 (g)

This reaction was started at room temperature (298K) by placing 0.5 mol of PCl5 in a 20 L container. When the reaction came to equilibrium, 0.3 mol of Cl2 (g) was detected.
a)
Find the number of moles of each substance in the equilibrium mixture.

b)
Calculate the equilibrium constant, Keq given the equilibrium quantities found in question a).

c)
Using the previous Keq equilibrium constant, what would be the equilibrium amounts of PCl3 and Cl2 if 0.8 mol of PCl5 was present?

d)
This equilibrium mixture is then heated up to 400K. Explain whether or not, and if so how, this will affect the Keq value.