Controlling the Position of Equilibrium Revision Notes for OCR A-Level Chemistry A

Controlling the Position of Equilibrium

Understanding equilibrium position shifts

When a chemical system at equilibrium experiences a change in conditions, the equilibrium position responds to counteract that change. This is explained by Le Chatelier's principle, which you learned about in earlier topics. But why does this happen? The answer lies in understanding equilibrium constants.

There are three key rules for predicting equilibrium shifts:

Concentration changes: If the concentration of a species increases, the equilibrium position shifts in the direction that reduces that concentration
Pressure changes: If pressure increases, the equilibrium position shifts towards the side with fewer gaseous moles
Temperature changes: If temperature increases, the equilibrium position shifts in the endothermic direction

infoNote

The reason Le Chatelier's principle works is because the equilibrium constant ( $K$ ) controls the relative amounts of reactants and products present at equilibrium.

The meaning of equilibrium constants

The magnitude of an equilibrium constant tells us the extent of a chemical equilibrium. Understanding what different values of $K$ mean is crucial:

$K = 1$ indicates an equilibrium halfway between reactants and products
$K = 100$ indicates an equilibrium well in favour of the products
$K = 1 \times 10^{-2}$ indicates an equilibrium well in favour of the reactants

The value of $K$ gives the exact position of equilibrium. When reactants or products of a reversible reaction are mixed together, the reaction proceeds until the concentrations (or partial pressures) of the equilibrium species match the value of $K$ when placed into the equilibrium constant expression.

What changes equilibrium constants?

chatImportant

Here's a crucial point that many students find confusing: at a set temperature, $K$ is constant and does not change despite any modifications to concentration, pressure, or the presence of a catalyst.

However, $K$ does change if the temperature is changed. Temperature change is the only condition that will cause $K$ to change its value.

Effect of temperature on equilibrium constants

The value of $K$ changes when you alter the temperature. Whether $K$ gets larger or smaller depends on whether the forward reaction is exothermic or endothermic.

Exothermic reactions

For an exothermic forward reaction (one that gives out energy):

The equilibrium constant decreases with increasing temperature
Raising the temperature decreases the equilibrium yield of products

Let's look at a specific example: the sulfur dioxide/oxygen/sulfur trioxide equilibrium.

$2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g) \quad \Delta H^{\circ} = -197 \text{ kJ mol}^{-1}$

This is an exothermic reaction. The table below shows how $K_p$ changes with temperature:

As temperature increases from 500 K to 1100 K, $K_p$ decreases dramatically from $2.5 \times 10^{10}$ to $1.3 \times 10^{-1}$ atm $^{-1}$ . The equilibrium position shifts to the left (towards reactants).

lightbulbExample

Worked Example: Understanding the Temperature Effect on an Exothermic Equilibrium

At equilibrium at 500 K: $K_p = \frac{p(\text{SO}_3)^2}{p(\text{SO}_2)^2 \times p(\text{O}_2)}$

Step 1: Identify the initial equilibrium

At 500 K: $K_p = 2.5 \times 10^{10}$ atm $^{-1}$
The system is at equilibrium with this $K_p$ value

Step 2: Analyze what happens when temperature increases to 700 K

$K_p$ decreases from $2.5 \times 10^{10}$ to $3.0 \times 10^4$ atm $^{-1}$
The system is no longer in equilibrium
The ratio $\frac{p(\text{SO}_3)^2}{p(\text{SO}_2)^2 \times p(\text{O}_2)}$ is now greater than the new $K_p$

Step 3: Determine how the system responds

To restore equilibrium with the new, smaller $K_p$ value, the partial pressures must adjust:

The partial pressure of product SO $_3$ must decrease
The partial pressures of reactants SO $_2$ and O $_2$ must increase
The position of equilibrium shifts towards the reactants (to the left)

A new equilibrium is reached where the ratio equals the new $K_p$ value.

Endothermic reactions

For an endothermic forward reaction (one that takes in energy):

The equilibrium constant increases with increasing temperature
Raising the temperature increases the equilibrium yield of products

Consider the nitrogen/oxygen/nitrogen monoxide equilibrium:

$\text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}(g) \quad \Delta H^{\circ} = +180 \text{ kJ mol}^{-1}$

This is an endothermic reaction. As temperature increases, $K_p$ increases significantly, and the equilibrium position shifts to the right (towards products).

lightbulbExample

Worked Example: Understanding the Temperature Effect on an Endothermic Equilibrium

At equilibrium at 700 K: $K_p = \frac{p(\text{NO})^2}{p(\text{N}_2) \times p(\text{O}_2)}$

Step 1: Identify the initial equilibrium

At 700 K: $K_p = 5 \times 10^{-13}$ atm $^{-1}$
The system is at equilibrium with this $K_p$ value

Step 2: Analyze what happens when temperature increases to 1100 K

$K_p$ increases from $5 \times 10^{-13}$ to $4 \times 10^{-8}$ atm $^{-1}$
The system is no longer in equilibrium
The ratio $\frac{p(\text{NO})^2}{p(\text{N}_2) \times p(\text{O}_2)}$ is now less than the new $K_p$

Step 3: Determine how the system responds

To match the new, larger $K_p$ value:

The partial pressure of product NO must increase
The partial pressures of reactants N $_2$ and O $_2$ must decrease
The position of equilibrium shifts towards the products (to the right)

Summary of temperature effects

Effect of concentration changes on equilibrium

infoNote

The value of an equilibrium constant $K$ is unaffected by changes in concentration or pressure. This seems strange at first because you know from Le Chatelier's principle that the equilibrium position can be shifted by changing concentration or pressure. The key is understanding that the equilibrium shift results from the fact that the equilibrium constant does not change.

How concentration changes work

Let's examine a specific equilibrium to understand this properly:

$\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$

lightbulbExample

Worked Example: Effect of Concentration Change on Equilibrium Position

At a constant temperature, suppose the equilibrium concentrations are:

$[\text{NO}_2(g)] = 0.400 \text{ mol dm}^{-3}$
$[\text{N}_2\text{O}_4(g)] = 0.010 \text{ mol dm}^{-3}$

Step 1: Calculate the equilibrium constant $K_c$

$K_c = \frac{[\text{NO}_2(g)]^2}{[\text{N}_2\text{O}_4(g)]} = \frac{0.400^2}{0.010} = 16.0 \text{ mol dm}^{-3}$

Step 2: Analyze what happens if we increase $[\text{N}_2\text{O}_4]$ to $0.020 \text{ mol dm}^{-3}$

Calculate the new ratio: $\frac{[\text{NO}_2(g)]^2}{[\text{N}_2\text{O}_4(g)]} = \frac{0.400^2}{0.020} = 8.0 \text{ mol dm}^{-3}$

This ratio is now less than $K_c$ (which remains 16.0), so the system is no longer in equilibrium.

Step 3: Determine the direction of shift

To restore equilibrium and return the ratio to its $K_c$ value of 16.0 mol dm $^{-3}$ , the concentrations must change:

The concentration of product NO $_2$ must increase
The concentration of reactant N $_2$ O $_4$ must decrease
A new equilibrium is established where $\frac{[\text{NO}_2(g)]^2}{[\text{N}_2\text{O}_4(g)]}$ is restored to its $K_c$ value

The equilibrium position shifts to the right (towards products).

This is exactly what Le Chatelier's principle predicts! By decreasing [N $_2$ O $_4$ ], the equilibrium shifts from left to right. But now you understand why: it's because $K_c$ controls the relative concentrations of reactants and products present at equilibrium, and $K_c$ itself doesn't change with concentration changes.

infoNote

The same principle applies when using $K_p$ for equilibrium constants involving partial pressures instead of concentrations.

Effect of pressure changes on equilibrium

For gaseous equilibria, pressure changes also don't change the value of $K$ , but they do shift the equilibrium position.

How pressure changes work

Using the same equilibrium as before:

$\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$

lightbulbExample

Worked Example: Effect of Pressure Change on Equilibrium Position

Suppose at constant temperature, the equilibrium partial pressures are:

$p(\text{NO}_2) = 9.6$ atm
$p(\text{N}_2\text{O}_4) = 0.24$ atm

Step 1: Calculate the equilibrium constant $K_p$

$K_p = \frac{p(\text{NO}_2)^2}{p(\text{N}_2\text{O}_4)} = \frac{9.6^2}{0.24} = 384 \text{ atm}$

Step 2: Analyze what happens if we double the total pressure

Doubling the total pressure doubles the partial pressures of both species:

$p(\text{NO}_2)$ increases to 19.2 atm
$p(\text{N}_2\text{O}_4)$ increases to 0.48 atm

Calculate the new ratio: $\frac{p(\text{NO}_2)^2}{p(\text{N}_2\text{O}_4)} = \frac{19.2^2}{0.48} = 768 \text{ atm}$

This ratio is greater than $K_p$ (which remains 384 atm), so the system is no longer in equilibrium.

Step 3: Determine the direction of shift

To return to equilibrium with the $K_p$ value of 384 atm:

The partial pressure of product NO $_2$ must decrease
The partial pressure of reactant N $_2$ O $_4$ must increase
The equilibrium position shifts to the left (towards reactants)

Using Le Chatelier's principle, you'd predict a shift from right to left to the side with fewer gaseous moles - exactly what we've just shown! The shift is directed by the value of $K_p$ being restored.

General rules for pressure effects

The table below summarizes how increasing pressure affects equilibrium position:

infoNote

Key Principle for Pressure Effects:

If there are fewer moles of gaseous products, the ratio is less than $K$ , and equilibrium shifts right
If there are more moles of gaseous products, the ratio is greater than $K$ , and equilibrium shifts left
If there are equal moles on both sides, the ratio equals $K$ , and there's no effect

Effect of catalysts on equilibrium constants

Equilibrium constants are unaffected by the presence of a catalyst.

infoNote

Catalysts affect the rate of a chemical reaction but not the position of equilibrium. Here's what catalysts do:

They speed up both the forward and reverse reactions by the same factor
Equilibrium is reached more quickly
The equilibrium position itself is not changed

This is important to remember: adding a catalyst gets you to equilibrium faster, but doesn't give you more product once equilibrium is established.

bookmarkSummary

Key Points to Remember:

Temperature is the ONLY factor that changes the value of K. Concentration, pressure, and catalysts do not affect $K$ itself.
For exothermic reactions: Increasing temperature decreases $K$ and shifts equilibrium left (towards reactants). Decreasing temperature increases $K$ and shifts equilibrium right (towards products).
For endothermic reactions: Increasing temperature increases $K$ and shifts equilibrium right (towards products). Decreasing temperature decreases $K$ and shifts equilibrium left (towards reactants).
Le Chatelier's principle works because $K$ controls the equilibrium position. When you change concentration or pressure, the system adjusts to restore the $K$ value by shifting the equilibrium position.
Catalysts speed up the rate of reaching equilibrium but don't change the equilibrium position or the value of $K$ .
Exam tip: When answering questions about equilibrium shifts, always explain your reasoning in terms of how $K$ changes (for temperature) or how the ratio compares to $K$ (for concentration/pressure changes).

Controlling the Position of Equilibrium (OCR A-Level Chemistry A): Revision Notes