Writing $K_c$ expressions

10 minAS Level7.1.4

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By the end of this lesson

By the end you can write a $K_c$ expression from any balanced equation, explain why each concentration is raised to its coefficient, and decide whether a large or small $K_c$ means products or reactants dominate.

One number for the whole mixture

At a given temperature, an equilibrium mixture always settles at the same ratio of concentrations, whatever amounts you started with.

That ratio is the equilibrium constant, $K_c$ . It turns "the position of equilibrium" into a number you can calculate with.

Definition

Equilibrium constant, $K_c$

For the general equilibrium

$a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$

the equilibrium constant in terms of concentrations is

$K_c = \frac{[\text{C}]^c\,[\text{D}]^d}{[\text{A}]^a\,[\text{B}]^b}$

The square brackets mean equilibrium concentration in $\text{mol dm}^{-3}$ — not the starting concentration.

Where the powers come from

Each concentration is raised to the coefficient of that substance in the balanced equation. That is not a convention: it comes from the way the rates depend on concentrations.

For the forward reaction the rate is proportional to $[\text{A}]^a[\text{B}]^b$ ; for the reverse it is proportional to $[\text{C}]^c[\text{D}]^d$ .

At equilibrium the two rates are equal, so

$k_{\text{forward}}[\text{A}]^a[\text{B}]^b = k_{\text{reverse}}[\text{C}]^c[\text{D}]^d$

Rearranging gives the ratio above, with $K_c = k_{\text{forward}}/k_{\text{reverse}}$ . The expression is a statement about rates, which is why it always matches the equation.

Worked example: Worked example: an esterification

Ethanoic acid reacts with ethanol:

$\text{CH}_3\text{COOH}(aq) + \text{C}_2\text{H}_5\text{OH}(aq) \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5(aq) + \text{H}_2\text{O}(l)$

Write the $K_c$ expression.

Have a go first?

Leaving out solids and pure liquids

The "concentration" of a pure solid or a pure liquid cannot change: its particles are packed together in the same way however much is present.

So it is a constant, and constants are absorbed into $K_c$ rather than written out.

A useful check: write the expression, then cross out any substance marked $(s)$ or $(l)$ — but keep anything marked $(aq)$ or $(g)$ .

Worked example: Worked example: a heterogeneous equilibrium

Calcium carbonate decomposes in a closed vessel:

$\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$

Write the $K_c$ expression, and explain what it tells you about the amount of carbon dioxide at equilibrium.

Have a go first?

What the size of $K_c$ tells you

Because products sit on the top, $K_c$ is a quick read on the position of equilibrium.

$K_c \gg 1$ : products dominate — the equilibrium lies far to the right.
$K_c \ll 1$ : reactants dominate — the equilibrium lies far to the left.
$K_c \approx 1$ : substantial amounts of everything.

$K_c$ is unitless in practice at this level: Cambridge usually quotes it without units, and you are not required to work them out for $K_c$ .

Why a bigger constant means more product

The expression is a fraction. To make the fraction large you need a large numerator, which means plenty of product.

The same mixture described by the same equation always gives the same $K_c$ , which is what makes the constant useful: it is a fingerprint of the reaction at that temperature.

Common mistakes

$K_c$ is calculated from the starting concentrations.

Only equilibrium concentrations go into $K_c$ . Using the starting values gives a meaningless number.

In the exam

Always check the equation is balanced before writing the expression; every coefficient becomes a power.
Write square brackets with the state in mind: omit $(s)$ and $(l)$ .
If the question says "state the $K_c$ expression", no units and no numbers are needed.
For "what does the value tell you", say "the equilibrium lies to the right/left" rather than "the reaction is fast/slow".
If the equation is reversed in a later part, say $K_c$ becomes $\dfrac{1}{K_c}$ .

Checking the idea

Multiple choiceQuestion 1 · Apply

Answer the 3 questions above to continue.