Effect of conditions on KEdexcel A-Level Chemistry: Revision notes

Section 1

What can change K

The equilibrium constants Kc and Kp have a fixed value for a given reaction at a given temperature.

Changing the temperature changes the value of K.
Changing the concentration or pressure, or adding a catalyst, does not change K.

The position of equilibrium can still move, because the composition adjusts so that the ratio in the K expression returns to the same value.

Key termsequilibrium constantposition of equilibrium

Section 2

Temperature: exothermic and endothermic reactions

Exothermic forward reaction (ΔH negative): K decreases as the temperature rises, and increases as it falls.
Endothermic forward reaction (ΔH positive): K increases as the temperature rises, and decreases as it falls.

Examples: N₂ + 3H₂ ⇌ 2NH₃ (ΔH = −92 kJ mol⁻¹): Kp falls as T rises, so less ammonia. N₂O₄ ⇌ 2NO₂ (ΔH = +57 kJ mol⁻¹): Kp rises as T rises, so more NO₂.

This agrees with Le Chatelier's principle: heating shifts the position towards the endothermic direction.

Key termsexothermicendothermic

Common mistake

Do not say K changes because the rate changes. K changes because the enthalpy change makes one direction favoured at the new temperature.

Section 3

Temperature explains the shift in position

A change of temperature moves the position of equilibrium because it changes the value of K.

Worked explanation for an endothermic reaction, A ⇌ B, heated:

K increases because the forward reaction is endothermic.
The ratio [B] ÷ [A] at equilibrium must equal the larger K.
So the position shifts to the right: more B, less A.

For an exothermic reaction the same steps give a smaller K and a shift to the left.

Key termsratioshift

Exam tip

Link the three ideas in order: sign of ΔH, direction K changes, shift in position and amounts.

Section 4

Concentration and pressure: K unchanged

Changing a concentration or pressure disturbs the equilibrium: the ratio in the K expression no longer equals K. The system then changes composition until the ratio equals the same K again.

Example. $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$ , $K_p = \frac{p_{\text{NO}_2}^2}{p_{\text{N}_2\text{O}_4}}$ . Doubling the total pressure doubles each partial pressure, so the ratio becomes $2K_p$ (too large). The position shifts to the left, to N₂O₄ (fewer moles of gas), until the ratio is Kp again.

If there are the same number of moles of gas on each side, a pressure change alters the top and bottom of the expression equally, so the position does not change.

Key termspartial pressuremoles of gas

Common mistake

The position can move when concentration or pressure changes, but K itself does not.

Section 5

Catalysts and summary

A catalyst increases the rates of the forward and reverse reactions equally, so it does not change the position of equilibrium or the value of K. Equilibrium is reached sooner.

Raise temperature: position shifts to the endothermic direction; K changes
Change concentration: position shifts to oppose the change; K unchanged
Change pressure: position shifts to the side with fewer moles of gas; K unchanged
Add a catalyst: no change to the position or K

Key termscatalyst

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Effect of conditions on K

The Haber process makes ammonia: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ mol⁻¹. A chemical engineer studies how the equilibrium constant Kp, and the composition of the equilibrium mixture, change when the conditions are altered.
The iron catalyst is added to the reactor. Explain why this does not change the value of Kp.2 marks
Hydrogen is made on a large scale by steam reforming of methane: CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g), ΔH = +206 kJ mol⁻¹. The reformers run at about 1100 K.
Explain why the percentage of hydrogen in the equilibrium mixture is higher at 1100 K than at 900 K.2 marks
In hydrogen plants carbon monoxide reacts with steam in the water–gas shift reaction: CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g), ΔH = −41 kJ mol⁻¹. A first reactor runs at about 650 K and a second reactor at about 470 K.
Explain the effect on Kp, and on the position of equilibrium and the yield of hydrogen, of lowering the temperature from 650 K to 470 K.3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).