The position of equilibrium can be quantified by the equilibrium law. The equilibrium constant for a particular reaction only depends on the temperature.

  • The equilibrium law is used in calculations of reacting concentrations and equilibrium mixtures.
  • Equilibrium calculations can be performed by identifying the initial concentrations, the changes in concentration and the equilibrium concentrations.
  • Some approximations can be used to simplify the calculation when $\mathrm K_c$ is small – When $\mathrm K_c$ is very small, $\rm [Reactant]_{eqm} \approx [Reactant]_{initial}$. (See worked example).

Worked Example

$\rm 0.0500~mol$ of $\rm N_2(g)$ and $\rm 0.100~mol$ of $\rm O_2(g)$ are mixed in a $\rm 1~dm^3$ closed container and react according to the equation.

$\rm N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$ 

Calculate the equilibrium concentrations when $\mathrm K_c = 2.0 \times 10^{–37}$

Solution

As $\mathrm K_c$ $x$ is small this allows us to simplify the calculation.

  $\rm N_2(g) + O_2(g)\rightleftharpoons
2NO(g)$
Initial concentration $0.0500$ $0.100$ $0$
Eqm concentration $= 0.0500 - x$ $0.100 - x$ $2x$
  $\approx 0.0500$ $\approx 0.100$ $2x$

Answer

$\mathrm K_c = \dfrac{4x^2}{(0.0500 \times 0.100)} = 2.0 \times 10^{–37}$
$4x^2 = 2.0 \times10^{–37} \times (0.0500 \times 0.100)$
$x^2 = 0.5 \times 10^{–37} \times (0.0500 \times 0.100)$
$x = 1.58 \times 10^{-20}$

Note the small value of x confirms our assumption.

Entropy, Gibbs free energy and Equilibrium

  • Equilibrium occurs at a maximum value of entropy and a minimum value of Gibbs free energy.

Gibbs free energy, $\rm\Delta G$, and the equilibrium constant, $\mathrm K_c$, can both be used to measure the position of an equilibrium reaction. They are related by the expression $\rm \Delta G= –RT \ln K$.