Topic 9 · page 13 of 15

Temperature, catalysts and writing the justification

Raising the temperature shifts the position in the endothermic direction, and lowering it shifts it in the exothermic direction. A catalyst does not shift the position at all. Then bring all the changes together in a full justification.

13 min read9.10

Read first: Le Chatelier — concentration and pressure

Temperature 9.10

Every reversible reaction is exothermic in one direction and endothermic in the other, with the same size of enthalpy change. For example:

NX2OX4(g)⇌2 NOX2(g)ΔH=+57 kJ mol−1 \ce{N2O4(g) <=> 2NO2(g)} \qquad \Delta H = +57\ \text{kJ mol}^{-1}

The forward reaction is endothermic (ΔH positive) and the backward reaction is exothermic, at −57 kJ mol⁻¹.

The system opposes a temperature change by absorbing or releasing heat:

Forward reaction is Raise the temperature Lower the temperature
exothermic (ΔH negative) shifts left, less product shifts right, more product
endothermic (ΔH positive) shifts right, more product shifts left, less product

So for N₂O₄ ⇌ 2NO₂ with ΔH = +57 kJ mol⁻¹, heating moves the position to the right, which makes more brown NO₂: the tube darkens, exactly as in the sealed-tube picture on page 11.

Temperature also changes the rate, always upwards when it is raised. That is a separate effect, so a high temperature can speed up a reaction while moving the position the wrong way. That conflict is what the next page is about.

Catalysts 9.10

A catalyst has no effect on the position of equilibrium. It speeds up the forward and backward reactions by the same factor, because it lowers the activation energy of both by the same amount, as you saw on page 8 (a fall of 60 kJ mol⁻¹ in both directions). The two rates stay equal, so the equilibrium mixture is unchanged.

What it does change is how quickly the equilibrium is reached.

All the changes together 9.10

Change Effect on the position Effect on the rate
Add a reactant shifts right forward rate rises at first
Remove a product shifts right backward rate falls at first
Raise the pressure (gases) towards the side with fewer gas moles rises
Raise the temperature towards the endothermic direction rises
Add a catalyst no shift both rates rise equally

Worked exampleJustifying the conditions for the Contact process

2 SOX2(g)+OX2(g)⇌2 SOX3(g)ΔH=−196 kJ mol−1 \ce{2SO2(g) + O2(g) <=> 2SO3(g)} \qquad \Delta H = -196\ \text{kJ mol}^{-1}

Predict and justify the effect on the equilibrium yield of SO₃ of (a) raising the temperature, (b) raising the pressure, (c) adding a catalyst.

(a) Temperature.

  1. The temperature is raised, so the system opposes the change by absorbing heat.
  2. The forward reaction is exothermic (ΔH = −196), so the position shifts in the endothermic direction, to the left.
  3. The yield of SO₃ falls.

(b) Pressure.

  1. The pressure is raised, so the system opposes the change by reducing the pressure.
  2. There are 3 mol of gas on the left (2 + 1) and 2 mol on the right, so the position shifts to the right.
  3. The yield of SO₃ rises.

(c) Catalyst.

  1. The catalyst lowers the activation energy of the forward and backward reactions equally, so the position does not change.
  2. The yield is the same, but it is reached faster.

Check yourself

  1. State the effect of raising the temperature on the position of equilibrium of an exothermic reaction, and explain it.
  2. The reaction N₂O₄(g) ⇌ 2NO₂(g) has ΔH = +57 kJ mol⁻¹. Predict and justify the colour change when a sealed tube of the mixture is cooled.
  3. For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ mol⁻¹. Predict the effect on the equilibrium yield of ammonia of (a) raising the temperature, (b) raising the pressure.
  4. Explain why a catalyst does not change the position of equilibrium.
  5. For the Contact process in the worked example, justify the effect on the equilibrium yield of SO₃ of adding more O₂ to the mixture.
  6. A student writes: "Heating a reversible reaction always moves the equilibrium to the products, since heat speeds up the forward reaction." Give two errors.

Answers cover these until you have tried the questions

  1. The position shifts to the left, so the yield of products falls. The system opposes the rise in temperature by absorbing heat, which favours the endothermic direction, and the backward reaction is the endothermic one.
  2. The temperature falls, so the system opposes the change by releasing heat. It shifts in the exothermic direction. Since the forward reaction is endothermic (+57), the exothermic direction is the backward one, to the left, giving more N₂O₄. The tube becomes paler (less brown NO₂).
  3. (a) The forward reaction is exothermic, so a higher temperature shifts the position left (endothermic direction): the yield falls. (b) 4 mol of gas on the left and 2 mol on the right, so a higher pressure shifts it right: the yield rises.
  4. A catalyst lowers the activation energy of the forward and backward reactions by the same amount, so it increases both rates equally. The rates remain equal, so the equilibrium composition is the same. It only makes equilibrium arrive sooner.
  5. Adding O₂ increases its concentration, so the system opposes the change by using up some of the O₂. The position shifts to the right, and the yield of SO₃ (based on SO₂) rises.
  6. (1) The position of equilibrium depends on the sign of ΔH: heating favours the endothermic direction, which is the products only if the forward reaction is endothermic. (2) The rate and the position are different effects: heating speeds up both the forward and the backward reactions.