Topic 13 · page 10 of 12

Temperature changes K

Heating lowers K for an exothermic reaction and raises it for an endothermic one. The position of equilibrium moves because K has changed, so always explain a temperature effect through K.

14 min read13.413.613.7

Read first: What changes K, Pressure and catalysts

Temperature changes K 13.6

Concentration, pressure and catalysts leave $K$ alone. Temperature does not. For a reaction at a different temperature, $K$ has a different value.

The table shows two reactions. The Haber process is exothermic. The dissociation of N₂O₄ is endothermic. Both are calculated from the entropy relationship on page 11, with $\Delta H$ and $\Delta S_\text{system}$ taken as constant, so they are model values. Measured values differ somewhat, but the trends are real.

T / K	$\ce{N2 + 3H2 <=> 2NH3}$ , $\Delta H$ = −92 kJ mol⁻¹, $K_\text{p}$ / atm⁻²	$\ce{N2O4 <=> 2NO2}$ , $\Delta H$ = +57 kJ mol⁻¹, $K_\text{p}$ / atm
298	5.6 × 10⁵	0.16
400	43	56
500	0.17	1.7 × 10³
700	3.0 × 10⁻⁴	8.8 × 10⁴

A graph of log base 10 of K against temperature from 290 to 710 K. For the exothermic Haber reaction the line falls from about 5.7 at 298 K to about minus 3.5 at 700 K. For the endothermic N2O4 dissociation the line rises from about minus 0.8 at 298 K to about 4.9 at 700 K. A horizontal line marks K equals 1. — K changes by many powers of ten over a few hundred kelvin. The exothermic reaction falls, and the endothermic reaction rises.

The trend agrees with the Unit 2 rule that heating favours the endothermic direction. The new idea is the reason for it.

The position follows K 13.713.4

Compare with a concentration change. When you add a reactant, $K$ stays the same and $Q$ is knocked away from it. When you change the temperature, the opposite happens: the mixture has not had time to change yet, so $Q$ is at first the same as before, and $K$ is what moves.

Worked exampleHeating the Haber mixture

$\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$ , $\Delta H$ = −92 kJ mol⁻¹. The temperature is raised from 500 K to 700 K. Explain the effect on the amount of ammonia.

The reaction is exothermic, so $K_\text{p}$ falls, from 0.17 atm⁻² to 3.0 × 10⁻⁴ atm⁻².
At that instant the mixture has not changed, so the quotient $Q_\text{p}$ still has its old value, 0.17 atm⁻². That is larger than the new $K_\text{p}$ .
The quotient is too large, which means too many products. The mixture moves left until $Q_\text{p}$ falls to the new $K_\text{p}$ .
There is less NH₃ at equilibrium at 700 K.

Worked exampleHeating and cooling N₂O₄

$\ce{N2O4(g) <=> 2NO2(g)}$ , $\Delta H$ = +57 kJ mol⁻¹. Colourless N₂O₄ and brown NO₂ share a sealed tube.

Heat. The reaction is endothermic, so $K_\text{p}$ rises. The quotient is now smaller than the new $K_\text{p}$ (too few products), so the mixture moves right: more NO₂ and a darker brown.
Cool in ice. $K_\text{p}$ falls. The quotient is now larger than the new $K_\text{p}$ , so the mixture moves left: more N₂O₄ and a paler colour.
Swap the tubes and $K_\text{p}$ changes back, and so does the colour.

The same sealed mixture at two temperatures. Hot water raises K, so more NO2. Ice lowers K, so more N2O4.

The reverse reaction behaves in the opposite way. Its constant is $1/K$ , so when a forward reaction is exothermic, the reverse reaction is endothermic, and the constant for the reverse reaction rises with temperature.

Check yourself

State how $K_\text{p}$ changes when the temperature is raised for each reaction. (a) $\ce{2SO2(g) + O2(g) <=> 2SO3(g)}$ , $\Delta H$ = −196 kJ mol⁻¹ (b) $\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$ , $\Delta H$ = +178 kJ mol⁻¹ (c) $\ce{N2(g) + O2(g) <=> 2NO(g)}$ , $\Delta H$ = +180 kJ mol⁻¹
Explain, in terms of $K_\text{p}$ , why the yield of SO₃ falls when the temperature of the equilibrium mixture in 1(a) is raised.
$\ce{H2(g) + CO2(g) <=> CO(g) + H2O(g)}$ has $\Delta H$ = +41 kJ mol⁻¹. State and explain the effect on $K_\text{p}$ and on the yield of CO of (a) raising the temperature, and (b) raising the pressure.
A reaction has $K_\text{c}$ = 0.020 at 400 K and 0.31 at 600 K (illustrative). Is it exothermic or endothermic? Explain.
Explain, in terms of $K_\text{p}$ , why the N₂O₄ and NO₂ mixture is paler in ice water than at room temperature.
The forward reaction is exothermic. What happens to the equilibrium constant for the reverse reaction when the temperature is raised?

Answers cover these until you have tried the questions

(a) Exothermic, so $K_\text{p}$ decreases. (b) Endothermic, so $K_\text{p}$ increases. (c) Endothermic, so $K_\text{p}$ increases.
The reaction is exothermic, so $K_\text{p}$ decreases when the temperature rises. The quotient is then larger than the new $K_\text{p}$ , so the mixture moves left until the quotient falls to the new $K_\text{p}$ . There is less SO₃.
(a) The reaction is endothermic, so $K_\text{p}$ increases. The quotient is then smaller than the new $K_\text{p}$ , so the mixture moves right and the yield of CO increases. (b) There are 2 moles of gas on each side, so $P$ cancels from $K_\text{p}$ . Raising the pressure has no effect on $K_\text{p}$ or on the yield.
$K_\text{c}$ rises with temperature, so the forward reaction is endothermic. (An exothermic reaction has a $K_\text{c}$ that falls as the temperature rises.)
The forward reaction (N₂O₄ → 2NO₂) is endothermic, so $K_\text{p}$ falls when the mixture is cooled. The quotient is then larger than the new $K_\text{p}$ , so the mixture moves left towards the colourless N₂O₄, and the brown colour fades.
The constant for the reverse reaction is $1/K$ for the forward reaction. $K$ for the forward (exothermic) reaction falls, so its reciprocal rises: the reverse reaction is endothermic, and its constant increases with temperature.