Le Chatelier’s Principle (Grade 12 NSC Matric Physical Sciences): Revision Notes

Worked Example: Concentration Effects

For the reaction: $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$

If $SO_2$ or $O_2$ concentration is increased:

• Le Chatelier's principle predicts equilibrium will shift to decrease reactant concentration
• Forwards reaction rate increases
• Some sulphur dioxide or oxygen is used to produce sulphur trioxide
• Equilibrium shifts to the right
• When new equilibrium is reached, there will be more product than before

If $[SO_3]$ decreases:

• Le Chatelier's principle predicts equilibrium will shift to increase product concentration
• Forwards reaction rate increases to produce more products
• Some sulphur dioxide or oxygen is used to produce sulphur trioxide
• Equilibrium shifts to the right

Worked Example: Temperature Effects

$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ $ΔH = -92$ kJ

This forwards reaction is exothermic (shown by negative ΔH), meaning the forwards reaction gives off heat and the reverse reaction takes in heat (endothermic).

An increase in temperature:

• Favours the endothermic reaction because it takes in energy (cools the container)
• The reverse reaction is endothermic, so the reverse reaction is favoured
• The yield of ammonia (NH₃) will decrease

A decrease in temperature:

• Favours the exothermic reaction because it releases energy (warms the container)
• The forwards reaction is exothermic, so the forwards reaction is favoured
• The yield of NH₃ will increase

Worked Example: Pressure Effects

$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$

The balanced equation ratio is 1:3:2. For every 1 molecule of $N_2$ gas, there are 3 molecules of $H_2$ gas and 2 molecules of $NH_3$ gas. Therefore:

• 4 molecules of reactant gas to 2 molecules of product gas

An increase in pressure:

• Favours the reaction that decreases the number of gas molecules
• There are fewer molecules of product gas than reactant gas, so the forwards reaction is favoured
• The equilibrium will shift to the right and the yield of NH₃ will increase

A decrease in pressure:

• Favours the reaction that increases the number of gas molecules
• There are more molecules of reactant gas, so the reverse reaction is favoured
• The equilibrium will shift to the left and the yield of NH₃ will decrease

Worked Example 1: Using Le Chatelier's principle

Question: Table salt is added to the (purple) solution in equilibrium: $CoCl_4^{2-} + 6H_2O \rightleftharpoons Co(H_2O)_6^{2+} + 4Cl^-$ (blue) (pink)

1. Use Le Chatelier's principle to predict the change in equilibrium position
2. What would be observed?

Solution:

Step 1: Identify the disturbance or stress on the system Adding NaCl produces $Na^+$ ions and $Cl^-$ ions as the salt dissolves. Looking at the equilibrium, $Cl^-$ is in the equation and the disturbance is the increase in concentration of the Cl⁻ ion.

Step 2: Use Le Chatelier's principle to decide how the system will respond By Le Chatelier's principle, the equilibrium position will shift to reduce the concentration of Cl⁻ ions.

Step 3: Decide whether the rate of the forwards reaction or the rate of the reverse reaction is increased and state the resulting shift in equilibrium The reverse reaction uses $Cl^-$ ions and hence the rate of the reverse reaction will increase. The reverse reaction is favoured and the equilibrium will shift to the left.

Step 4: What would the colour change be due to this equilibrium shift? The solution will appear more blue as more blue $CoCl_4^{2-}$ ions are formed.

Worked Example 2: Rate-time graphs

Question: For the reaction $2AB(g) \rightleftharpoons 2A(g) + B_2(g)$, $ΔH = 26$ kJ What stress has occurred in this system?

Solution:

Step 1: Check the axes so you know what the variables are on this graph The axes are labelled rate and time. Therefore this is a rate-time graph.

Step 2: Are both rates affected equally? No, the forwards rate is increased more than the reverse rate. Therefore the stress must be a change in temperature (a catalyst would increase both rates equally).

Step 3: Was the temperature increased or decreased? The forwards reaction is endothermic (ΔH is positive). An increase in temperature will favour the reaction that cools the reaction vessel (the endothermic reaction). Therefore the stress must have been an increase in temperature.

Worked Example 3: Concentration-time graphs

Question: Consider the equilibrium: $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$

1. After how many seconds does the system reach equilibrium?
2. Calculate the value of the equilibrium constant
3. What happens at t = 20 s?
4. If the change at t = 35 s is due to an increase in temperature, is the reaction exothermic or endothermic?

Solution:

Step 1: Check axes - this is a concentration-time graph

Step 2: System reaches equilibrium when all concentrations become constant at t = 10 s

Step 3: At t = 10 s, 25 s, and 45 s:

• $[H_2] = 0.75, 0.6, 1.5$ mol.dm⁻³
• $[I_2] = 0.75, 0.6, 1.5$ mol.dm⁻³
• $[HI] = 6.0, 4.8, 2.75$ mol.dm⁻³

Step 4: $K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(6.0)^2}{(0.75)(0.75)} = 64.0$

Step 5: At t = 20 s, HI concentration decreases sharply - HI must have been removed from the system

Step 6: The different $K_c$ at 45 s means the event at t = 35 s was a temperature change. Since the reverse reaction was favoured (more reactants, less product), and temperature increased, the reverse reaction must be endothermic, making the forwards reaction exothermic.