Equilibrium applies to **reversible reactions**, meaning as products are formed, some are also converting back to the original reactants.
At equilbrium, the **rate of forward reaction = rate of backwards reaction**.

Initially the rate of forward reaction is much higher than the rate of backwards reaction, as the concentration of reactants is very high, and reduces over time as more products form.

In an reversible reaction **not all reactants end up as products**, there is never 100% conversion.

## Dynamic equilbrium

- Equilibrium may only be obtained in a closed system
- rate of forward reaction = rate of backward reaction
- Catalysts have no effect on the equilibrium point
- Observable or physical properties should remain constant

## Equilibrium constants

$K_c$ - the equlibrium values expressed as concentration in $mol\ dm^{-3}$

$K_p$ - the equilbrium values expressed as partial pressures (See Equilibrium 2)

## Equilibrium calculations

for an equilbrium reaction in the form: \begin{equation}aA+bB\ \rightleftharpoons\ cC+dD\end{equation} and at a constant temperature Kc is shown by: \begin{align}\frac{\left[C\right]^c\ast\left[D\right]^d}{\left[A\right]^a\ast\left[B\right]^b}=\ K_c\end{align} For example: \begin{equation}K_c=\left[\frac{\left[NH_3\right]^2}{\left[N_2\right]^1\ast\left[H_2\right]^3}\right]\end{equation} The most important part of these calculations are units, as they vary based in the ratios of the reactants. To work them out: \begin{equation}\frac{\left(mol\ dm^{-3}\right)^2}{\left(mol\ dm^{-3}\right)^1\ast\left(mol\ dm^{-3}\right)^3}=\left(mol\ dm^{-3}\right)^{2-1-3}=\left(mol^{-2}dm^6\right)\end{equation}