'a' moles of ethanoic acid and 'b' moles of ethanol were reacted together at 50°C until equilibrium was established. At equilibrium, x mol of the ester were found to be present. Write an expression for the equilibrium constant, Kc, for the reaction in terms of a, b and x.

The reaction is between ethanoic acid and ethanol to form ethyl ethanoate (ester) and water.

$$\text{CH}_3\text{COOH(l)} + \text{CH}_3\text{CH}_2\text{OH(l)} \rightleftharpoons \text{CH}_3\text{COOCH}_2\text{CH}_3\text{(l)} + \text{H}_2\text{O(l)}$$

Initial moles: Ethanoic acid = $a$ mol, Ethanol = $b$ mol, Ester = 0 mol, Water = 0 mol.
At equilibrium, $x$ mol of the ester are present. According to the stoichiometry (1:1:1:1), if $x$ moles of ester are formed, then $x$ moles of water are also formed, and $x$ moles of ethanoic acid and $x$ moles of ethanol must have reacted.

Species	Initial (mol)	Change (mol)	Equilibrium (mol)
$\text{CH}_3\text{COOH}$	$a$	$-x$	$a - x$
$\text{CH}_3\text{CH}_2\text{OH}$	$b$	$-x$	$b - x$
$\text{CH}_3\text{COOCH}_2\text{CH}_3$	0	$+x$	$x$
$\text{H}_2\text{O}$	0	$+x$	$x$

Let the volume of the reaction vessel be $V \text{ dm}^3$. The equilibrium concentration of each species is its equilibrium moles divided by the volume $V$.

$$[\text{CH}_3\text{COOH}] = \frac{a-x}{V}$$ $$[\text{CH}_3\text{CH}_2\text{OH}] = \frac{b-x}{V}$$ $$[\text{CH}_3\text{COOCH}_2\text{CH}_3] = \frac{x}{V}$$ $$[\text{H}_2\text{O}] = \frac{x}{V}$$

The equilibrium constant $K_c$ is given by:

$$K_c = \frac{[\text{CH}_3\text{COOCH}_2\text{CH}_3][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{CH}_3\text{CH}_2\text{OH}]}$$

$$K_c = \frac{\left(\frac{x}{V}\right)\left(\frac{x}{V}\right)}{\left(\frac{a-x}{V}\right)\left(\frac{b-x}{V}\right)}$$ $$K_c = \frac{\frac{x^2}{V^2}}{\frac{(a-x)(b-x)}{V^2}}$$

The $V^2$ terms in the numerator and denominator cancel out.

$$K_c = \frac{x^2}{(a-x)(b-x)}$$

The expression for the equilibrium constant $K_c$ is $\boxed{K_c = \frac{x^2}{(a-x)(b-x)}}$.