The equilibrium constant and the reaction quotient

The law of mass action

In 1864 Guldberg and Waage found a simple rule. For a reaction at equilibrium, one ratio of concentrations always has the same value at a given temperature. For the general reaction

the equilibrium constant is

Square brackets mean concentrations in mol/L at equilibrium. Products go on top, reactants below, and each coefficient becomes a power.

For hydrogen iodide at 700 K:

It does not matter how much hydrogen or iodine you start with. At 700 K the equilibrium mixture always gives this value. Change the temperature and the value changes.

Like NCERT, we write K without units. Each concentration is divided by the standard concentration, 1 mol/L, so K is a pure number.

How the equation changes K

K belongs to the equation as it is written.

• Reverse the equation: $2\text{HI} \rightleftharpoons \text{H}_2 + \text{I}_2$ has $K' = \dfrac{1}{57.0} = 0.0175$.
• Halve the equation: $\tfrac{1}{2}\text{H}_2 + \tfrac{1}{2}\text{I}_2 \rightleftharpoons \text{HI}$ has $K'' = \sqrt{57.0} = 7.55$.
• Multiply by n: the new K is the old K to the power n.
• Add two equations: multiply their two K values.

Kp for gases

For gases we can use partial pressures instead. For the ammonia equilibrium:

Since $p = cRT$ for each gas, the two constants are linked:

Here $\Delta n$ is the moles of gaseous products minus the moles of gaseous reactants. With pressures in bar, $R = 0.0831$ L bar K$^{-1}$ mol$^{-1}$.

• For ammonia, $\Delta n = 2 - 4 = -2$.
• For $\text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2$, $\Delta n = 2 - 1 = +1$.
• For hydrogen iodide, $\Delta n = 2 - 2 = 0$, so $K_p = K_c$.

Homogeneous and heterogeneous equilibria

In a homogeneous equilibrium everything is in one phase, such as a mixture of gases.

In a heterogeneous equilibrium there is more than one phase. Heat limestone in a closed vessel:

A pure solid or a pure liquid has a constant concentration, however much of it is present. So it is left out of K. At a given temperature, the pressure of carbon dioxide over limestone and lime is fixed.

What the size of K tells

• K above about 1000: products are favoured. The reaction goes nearly to completion.
• K below about 0.001: reactants are favoured. Very little product forms.
• K in between: both reactants and products are present in fair amounts.

The reaction quotient

For any mixture, equilibrium or not, the same expression gives the reaction quotient, $Q_c$. Compare it with $K_c$:

• $Q_c < K_c$: too little product. The reaction goes forward.
• $Q_c > K_c$: too much product. The reaction goes backward.
• $Q_c = K_c$: the mixture is at equilibrium.

Finding the equilibrium mixture

An ICE table keeps track: Initial amounts, the Change, and the amounts at Equilibrium.

Start with 0.100 mol/L each of hydrogen and iodine at 700 K. Let $x$ mol/L of each react.

Then $\dfrac{(2x)^2}{(0.100 - x)^2} = 57.0$. Take the square root: $\dfrac{2x}{0.100 - x} = 7.55$. This gives $x = 0.0791$ mol/L. So $[\text{HI}] = 0.158$ mol/L and $[\text{H}_2] = [\text{I}_2] = 0.0209$ mol/L.

Make Kc very large, then very small. When is almost all of A and B used up? Check that the readout of Qc always equals Kc.

K and Gibbs energy

K is linked to the standard Gibbs energy change of the reaction:

If K is larger than 1, $\Delta G^\circ$ is negative and products are favoured. If K is smaller than 1, $\Delta G^\circ$ is positive. For K = 10 at 298 K, $\Delta G^\circ = -2.303 \times 8.314 \times 298 \times 1 = -5.71$ kJ/mol.

Practise with new numbers

Try it in Python

The program solves the hydrogen iodide ICE table above. It tries values of $x$, halving the gap each time, until $Q_c$ equals $K_c$.

Common mistakes

• Putting reactants on top. Products go in the numerator.
• Forgetting the powers. A coefficient of 2 means a square.
• Putting a solid or a pure liquid into K. Leave it out.
• Using starting amounts in K. Only equilibrium concentrations go into K.
• Using $R = 8.314$ in $K_p = K_c(RT)^{\Delta n}$ with pressures in bar. Use 0.0831 L bar K$^{-1}$ mol$^{-1}$.

Remember these

• $K_c = \dfrac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}$, constant at a given temperature.
• Reverse the equation: $1/K$. Multiply it by n: $K^n$.
• $K_p = K_c(RT)^{\Delta n}$.
• Pure solids and liquids are left out.
• $Q_c < K_c$: forward. $Q_c > K_c$: backward.
• $\Delta G^\circ = -2.303\,RT\log K$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Write $K_c$ for $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$.

2. Write $K_p$ for $\text{CaCO}_3\text{(s)} \rightleftharpoons \text{CaO(s)} + \text{CO}_2\text{(g)}$.

3. A reaction has $K_c = 57.0$. For a mixture of the same reaction, $Q_c = 80$. Which way does it go?

4. For which kind of reaction is $K_p = K_c$?