What changes the equilibrium constant?

10 minAS Level7.1.9

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can say, for any change to a system at equilibrium, whether it alters the value of the equilibrium constant — and explain why in one sentence.

Two different questions

There are two separate questions you can ask about a change:

What happens to the position of equilibrium — that is, the amounts in the mixture?
What happens to the value of the constant?

Le Chatelier's principle answers the first. The second has a much shorter answer.

Definition

What $K$ depends on

The value of $K_c$ or $K_p$ depends on temperature only.

Changing a concentration does not change $K$ .
Changing a pressure (or volume) does not change $K$ .
Adding a catalyst does not change $K$ .
Changing the temperature does change $K$ .

Everything else changes the mixture; only temperature changes the constant.

Why temperature is the exception

$K$ is the ratio of the rate constant for the forward reaction to the rate constant for the reverse reaction:

$K = \frac{k_{\text{forward}}}{k_{\text{reverse}}}$

Rate constants depend on temperature — that is the central idea of the Arrhenius equation. They do not depend on concentration or pressure.

So $K$ cannot change unless the temperature changes.

How temperature changes it

The direction is set by the sign of $\Delta H$ :

Forward reaction exothermic: raising the temperature decreases $K$ .
Forward reaction endothermic: raising the temperature increases $K$ .

This is consistent with Le Chatelier: if heating shifts the equilibrium left, the product side of the $K$ expression must have become relatively smaller, so $K$ has fallen.

But the position still moves without $K$ changing

Here is the point students find hardest. Raising the pressure on the Haber equilibrium shifts it right — the mixture contains proportionally more ammonia — yet $K_p$ is unchanged.

Both statements are true because the shift restores the same ratio. Every partial pressure in the expression increases when the gas is compressed, but they increase by amounts that leave the ratio at its old value.

So the composition changes while the constant does not. The constant is the destination; the composition is where you are on the way there.

The examinable summary

Change	Effect on position	Effect on $K$
Add a reactant	shifts right	none
Remove a product	shifts right	none
Increase pressure (fewer gas moles on the right)	shifts right	none
Add a catalyst	no change	none
Raise temperature (forward exothermic)	shifts left	decreases
Raise temperature (forward endothermic)	shifts right	increases

Read the last column from top to bottom: only the temperature rows move.

Worked example: Worked example: reading a table of $K$ values

For an equilibrium, $K_c$ was measured at several temperatures. The data are illustrative:

$T\ /\ \text{K}$	$K_c$
$400$	$1.2 \times 10^{-1}$
$500$	$4.0 \times 10^{-3}$
$600$	$2.5 \times 10^{-4}$

(a) Is the forward reaction exothermic or endothermic? (b) State the effect of raising the temperature on the equilibrium yield of the products.

Have a go first?

A useful one-line test

If a change can be made without touching the thermostat, $K$ does not move.

That single sentence answers a surprising number of exam questions — and it is exactly why pressure and concentration questions can be answered with Le Chatelier while the constant stays put.

Common mistakes

Increasing the pressure increases $K_p$ because the yield rises.

The yield can rise while $K_p$ stays constant. The new mixture satisfies the same constant.

In the exam

Answer the question that was asked: "effect on the value of $K$ " is not the same as "effect on the yield".
Use the phrase "K depends only on temperature"; it is the mark scheme's own wording.
If a table of $K$ values is given, compare two rows explicitly: " $K_c$ falls from ... to ... as $T$ rises".
For pressure changes, expect two marks: one for the shift, one for " $K$ unchanged".
Never write that a catalyst changes $K$ ; it changes only the time taken to reach equilibrium.

Checking the idea

Multiple choiceQuestion 1 · Recall

Multiple choiceQuestion 2 · Apply

Answer the 4 questions above to continue.