Effect of Conditions on Equilibrium Constants

Spec Mapping — OCR H432 Module 5.1.2 — How far? The factors that affect the value of the equilibrium constant ($K_c$Kc​ or $K_p$Kp​): only temperature changes $K$K. Concentration changes, pressure changes (of gases) and catalysts shift the position of equilibrium but do not change the value of $K$K. The dependence of $K$K on the sign of $\Delta H$ΔH for the forward reaction: exothermic-forward → $K$K decreases as $T$T rises; endothermic-forward → $K$K increases as $T$T rises (refer to the official OCR H432 specification document for exact wording).

This closing lesson of OCR Module 5.1.2 establishes the most-tested single fact about equilibrium constants: only temperature changes the value of $K$K. The intuitive temptation — "I added more reactant, so equilibrium has shifted, so $K$K must have changed" — is wrong. Le Chatelier's principle tells us the position responds to concentration, pressure and catalyst changes, but the equilibrium constant itself is fixed at a given temperature. The lesson develops the physical reasoning via the relationship $K = k_{\text{fwd}}/k_{\text{rev}}$K=kfwd​/krev​ and the Arrhenius equation — temperature is the only variable that changes the two rate constants by different factors (because $E_{a,\text{fwd}}$Ea,fwd​ and $E_{a,\text{rev}}$Ea,rev​ differ by $\Delta H$ΔH). Three worked examples explore: identifying the sign of $\Delta H$ΔH from how $K$K changes with $T$T; predicting whether $K$K rises or falls on heating; and revisiting the Haber and Contact compromise conditions. The lesson closes with the van't Hoff equation — formally beyond OCR but conceptually beautiful — which makes the $K$K-vs-$T$T relationship quantitative.

Key Equation: the relationship between equilibrium constant and temperature $\boxed{K = \frac{k_{\text{fwd}}}{k_{\text{rev}}}}$K=krev​kfwd​​​ with both $k_{\text{fwd}}$kfwd​ and $k_{\text{rev}}$krev​ obeying the Arrhenius equation $k = A e^{-E_a/(RT)}$k=Ae−Ea​/(RT). Because $E_{a,\text{fwd}} - E_{a,\text{rev}} = \Delta H$Ea,fwd​−Ea,rev​=ΔH, temperature shifts $k_{\text{fwd}}$kfwd​ and $k_{\text{rev}}$krev​ by different multiplicative factors, so their ratio $K$K changes. For an exothermic forward reaction, raising $T$T decreases $K$K; for an endothermic forward reaction, raising $T$T increases $K$K. Concentration, pressure and catalyst changes affect $k_{\text{fwd}}$kfwd​ and $k_{\text{rev}}$krev​ identically (or not at all) — so $K$K is unchanged.

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The key idea

A crucial take-home message of OCR Year 2 equilibrium: only temperature changes the value of $K$K. Concentration, pressure and catalysts affect rate and/or position, but they do not change the numerical value of $K_c$Kc​ or $K_p$Kp​.

Change	Affects rate?	Affects position?	Affects $K$K value?
Concentration	Yes (if of reactant/product)	Yes	No
Pressure (gas)	Yes	Yes (if $\Delta n_{\text{gas}} \neq 0$Δngas​=0)	No
Temperature	Yes	Yes	Yes
Catalyst	Yes (faster both ways)	No	No

The result is so important that OCR examiners ask it in some form every paper. Memorise the table; understand the mechanism.

Why temperature changes $K$K

Temperature changes $K$K because it changes the ratio of forward to reverse rate constants by different multiplicative factors.

• Both $k_{\text{fwd}}$kfwd​ and $k_{\text{rev}}$krev​ obey Arrhenius: $k = A e^{-E_a/(RT)}$k=Ae−Ea​/(RT) (Lesson 9). Raising $T$T increases both.
• The forward and reverse activation energies differ: $E_{a,\text{fwd}} - E_{a,\text{rev}} = \Delta H$Ea,fwd​−Ea,rev​=ΔH for the reaction.
• The exponential term is more sensitive to changes in $T$T for the step with the larger $E_a$Ea​. This means $k_{\text{fwd}}$kfwd​ and $k_{\text{rev}}$krev​ change at different rates with $T$T.

Since at equilibrium $K = k_{\text{fwd}}/k_{\text{rev}}$K=kfwd​/krev​, this ratio changes as $T$T changes — and $K$K along with it.

Exothermic forward reaction ($\Delta H < 0$ΔH<0)

• $E_{a,\text{rev}} > E_{a,\text{fwd}}$Ea,rev​>Ea,fwd​ — the reverse barrier is higher.
• Raising $T$T speeds up $k_{\text{rev}}$krev​ more than $k_{\text{fwd}}$kfwd​.
• Ratio $k_{\text{fwd}}/k_{\text{rev}}$kfwd​/krev​ decreases — $K$K decreases.
• Equilibrium shifts backwards (Le Chatelier: heat is a "product" of the exothermic forward direction; adding heat opposes the forward reaction).

Endothermic forward reaction ($\Delta H > 0$ΔH>0)

• $E_{a,\text{fwd}} > E_{a,\text{rev}}$Ea,fwd​>Ea,rev​ — the forward barrier is higher.
• Raising $T$T speeds up $k_{\text{fwd}}$kfwd​ more than $k_{\text{rev}}$krev​.
• Ratio $k_{\text{fwd}}/k_{\text{rev}}$kfwd​/krev​ increases — $K$K increases.
• Equilibrium shifts forwards (heat is a "reactant"; adding heat drives the forward reaction).

Summary table

Forward $\Delta H$ΔH	Effect of raising $T$T	Effect on $K$K
Exothermic ($\Delta H < 0$ΔH<0)	Shift backwards	$K$K decreases
Endothermic ($\Delta H > 0$ΔH>0)	Shift forwards	$K$K increases

Cooling has the opposite effect in each case.

Why concentration does NOT change $K$K

Imagine a system at equilibrium with $[\text{A}] = [\text{B}] = 1.0\,\text{mol dm}^{-3}$[A]=[B]=1.0mol dm−3 and $K_c = 1.0$Kc​=1.0. Now add more A so $[\text{A}] = 2.0$[A]=2.0:

• Reaction quotient $Q = [\text{B}]/[\text{A}] = 1.0/2.0 = 0.5 < K_c$Q=[B]/[A]=1.0/2.0=0.5<Kc​.
• System responds by converting A to B until $Q$Q returns to $K_c$Kc​.
• Eventually: $[\text{A}] = 1.5$[A]=1.5, $[\text{B}] = 1.5$[B]=1.5, so $Q = 1.0 = K_c$Q=1.0=Kc​ once more.

The ratio defining $K_c$Kc​ is restored — but the individual concentrations are now different. $K_c$Kc​ itself is unchanged. The equilibrium position has shifted towards products, but $K_c$Kc​ remains 1.0.

The reaction quotient $Q$Q (same expression as $K$K but with current, possibly non-equilibrium concentrations) is the key diagnostic: if $Q < K$Q<K, the forward reaction proceeds; if $Q > K$Q>K, the reverse reaction proceeds; at $Q = K$Q=K, the system is at equilibrium.

Why pressure does NOT change $K_p$Kp​

Pressure changes the partial pressures of all gases proportionally, but $K_p$Kp​ depends only on temperature. Increasing total pressure (by compressing) causes the system to shift to the side with fewer moles of gas — but the value of $K_p$Kp​ does not change.

Example — Haber under compression

$\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$N2​+3H2​⇌2NH3​, $K_p$Kp​ fixed at fixed $T$T.

• Compress the mixture (raise $P_{\text{total}}$Ptotal​). All four $p_i$pi​ double initially. Substituting into the $K_p$Kp​ expression: $p_{\text{NH}_3}^2$pNH3​2​ scales by $2^2 = 4$22=4; $p_{\text{N}_2} \cdot p_{\text{H}_2}^3$pN2​​⋅pH2​3​ scales by $2 \times 2^3 = 16$2×23=16. Ratio $Q = K_p \times 4/16 = K_p/4$Q=Kp​×4/16=Kp​/4. So $Q < K_p$Q<Kp​.
• System shifts forward until partial pressures re-satisfy $K_p = p_{\text{NH}_3}^2/(p_{\text{N}_2} \cdot p_{\text{H}_2}^3)$Kp​=pNH3​2​/(pN2​​⋅pH2​3​).
• Final $K_p$Kp​ is the same number; individual partial pressures are different.

Pressure shifts the position (towards fewer gas moles) but leaves $K_p$Kp​ unchanged.

Why catalysts do NOT change $K$K or position

A catalyst lowers the activation energies of both the forward and reverse reactions by the same amount (because they share a common transition state). Therefore:

• $k_{\text{fwd}}$kfwd​ increases by some factor.
• $k_{\text{rev}}$krev​ increases by the same factor.
• Their ratio $k_{\text{fwd}}/k_{\text{rev}} = K$kfwd​/krev​=K is unchanged.
• The equilibrium position is unchanged.
• The rate of reaching equilibrium is faster (in both directions equally).

The classic OCR exam-room phrase: "A catalyst provides an alternative pathway with a lower activation energy, increasing the rate of both forward and reverse reactions equally, so equilibrium is reached more quickly but $K$K and the position are unchanged."

Mermaid — catalyst effect

graph LR
  A[Catalyst lowers Ea,fwd by some amount delta] --> B[k_fwd increases by exp delta/RT]
  A --> C[Catalyst lowers Ea,rev by SAME delta]
  C --> D[k_rev increases by SAME factor exp delta/RT]
  B --> E[Ratio k_fwd/k_rev = K UNCHANGED]
  D --> E
  E --> F[Position unchanged, equilibrium reached faster]

Worked Example 1 — calculating $K$K at a new temperature qualitatively

$K_c$Kc​ for $\text{A} \rightleftharpoons \text{B}$A⇌B is 4.0 at 300 K. The reaction is exothermic in the forward direction ($\Delta H < 0$ΔH<0). Will $K_c$Kc​ at 350 K be larger or smaller?

Answer: since the forward is exothermic, raising $T$T decreases $K$K. So $K_c$Kc​ at 350 K will be smaller than 4.0.

If, say, $K_c = 1.5$Kc​=1.5 at 350 K, the equilibrium position has shifted towards A: more reactant remains at higher temperature. Consistent with Le Chatelier — raising $T$T pushes equilibrium in the endothermic direction, which for this system is the reverse direction.

Worked Example 2 — identifying $\Delta H$ΔH sign from $K$K vs $T$T

A reaction $\text{X} \rightleftharpoons \text{Y}$X⇌Y has $K_p = 12$Kp​=12 at 400 K and $K_p = 45$Kp​=45 at 500 K. Is the forward reaction exothermic or endothermic?

$K_p$Kp​ has increased with $T$T. This means the forward reaction is endothermic.

(Reasoning: raising $T$T shifts equilibrium in the endothermic direction. Since equilibrium shifted forwards — more Y at the higher $T$T, larger $K_p$Kp​ — the forward direction must be the endothermic direction.)

Worked Example 3 — Haber compromise conditions revisited

The Haber process: $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$N2​+3H2​⇌2NH3​, $\Delta H = -92\,\text{kJ mol}^{-1}$ΔH=−92kJ mol−1.

• Low $T$T would give high $K_p$Kp​ (since forward is exothermic, cooling favours products). But at low $T$T, the Boltzmann factor $e^{-E_a/(RT)}$e−Ea​/(RT) is small, so $k$k is small and the rate is impracticably slow.
• High $T$T gives low $K_p$Kp​ (low yield) but a high rate.
• Compromise: 450 °C ($\sim 723$∼723 K) gives a workable $K_p$Kp​ (around $1.6 \times 10^{-4}$1.6×10−4 mol$^{-2}$−2 dm$^{6}$6) and a workable rate.

Note the conceptual distinction: choosing $T$T involves a trade-off between $K_p$Kp​ (which depends on $T$T alone, via Le Chatelier) and rate (which depends on $T$T via Arrhenius). The 450 °C is not "optimal for yield" — that would be much lower — but rather the temperature at which the rate is fast enough that the reaction reaches equilibrium in industrially-feasible residence times.

High pressure (200 atm) improves the equilibrium position towards NH$_3$3​ (fewer gas moles on the right: $2 < 4$2<4), but does not change $K_p$Kp​. The 200 atm is the compromise between yield-from-Le-Chatelier and capital cost of pressure-rated steel reactors.

Iron catalyst does not change $K_p$Kp​ or position — it simply accelerates the attainment of equilibrium. Without it, the synthesis would require much higher temperatures (where rate is fast but $K_p$Kp​ is uneconomic) or much longer residence times.

The van't Hoff equation (extension — beyond OCR)

The quantitative relationship between $K$K and $T$T for a given $\Delta H$ΔH is the van't Hoff equation:

$\ln\!\left(\frac{K_2}{K_1}\right) = -\frac{\Delta H}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$ln(K1​K2​​)=−RΔH​(T2​1​−T1​1​)

This mirrors the Arrhenius equation in form. A plot of $\ln K$lnK against $1/T$1/T is a straight line with gradient $-\Delta H/R$−ΔH/R — exactly analogous to the Arrhenius plot of $\ln k$lnk against $1/T$1/T with gradient $-E_a/R$−Ea​/R. OCR does not require this equation for examination but it makes the qualitative results above quantitative: knowing $K$K at one temperature and $\Delta H$ΔH for the forward reaction, you can compute $K$K at any other temperature.

Synoptic Links

Synoptic Links — Connects to:

• ocr-alevel-chemistry-quantitative-rates-equilibrium / the-arrhenius-equation (Lesson 9 — the Arrhenius equation underpins the temperature dependence of $K = k_{\text{fwd}}/k_{\text{rev}}$K=kfwd​/krev​).
• ocr-alevel-chemistry-quantitative-rates-equilibrium / kc-and-kp-calculations (Lesson 11 — quantitative calculation of $K_c$Kc​, $K_p$Kp​).
• ocr-alevel-chemistry-enthalpy-rates-equilibrium / dynamic-equilibrium-le-chatelier-kc (Lesson 10 of OCR Module 3.2 — Le Chatelier's principle and compromise conditions; this lesson is the quantitative deepening).
• ocr-alevel-chemistry-enthalpy-rates-equilibrium / enthalpy-changes-and-standard-conditions (the sign of $\Delta H$ΔH determines the direction $K$K moves with $T$T).
• ocr-alevel-chemistry-enthalpy-rates-equilibrium / catalysts (the explicit shared-transition-state argument for why catalysts cannot change $K$K).