Gibbs energy and equilibrium

Delta G and delta G standard

$\Delta G^\circ$, with the small circle, is the Gibbs energy change when every reactant and product is in its standard state: $1$ bar for gases, $1\ \text{mol/L}$ for solutes. It is one fixed number for a reaction at a given temperature.

$\Delta G$, without the circle, is the change for the mixture as it is at the moment. It changes as the reaction goes on. A reaction moves in the direction that makes $\Delta G$ smaller. It stops changing when $\Delta G$ reaches zero.

• As the reaction proceeds, the composition changes, and so does $\Delta G$.
• At equilibrium, $\Delta G = 0$. Then the mixture has the composition given by the equilibrium constant $K$.

The link to K

Thermodynamics shows that at equilibrium

• $\Delta G^\circ = -RT \ln K$

Changing to common logarithms, with $\ln K = 2.303 \log K$:

• $\Delta G^\circ = -2.303\, RT \log K$

Here $R = 8.314\ \text{J/(K mol)}$, $T$ is in kelvin and $\Delta G^\circ$ is in J/mol. Turned round:

• $\log K = \dfrac{-\Delta G^\circ}{2.303\, RT}$

What the sign says

A negative $\Delta G^\circ$ makes $\log K$ positive, and a positive one makes it negative.

At $298\ \text{K}$, $2.303\, RT = 2.303 \times 8.314 \times 298 = 5706\ \text{J/mol}$, about $5.71\ \text{kJ/mol}$. So every $5.71\ \text{kJ/mol}$ of $\Delta G^\circ$ changes $K$ by a factor of $10$.

Example. A reaction has $\Delta G^\circ = -11.4\ \text{kJ/mol}$ at $298\ \text{K}$.

• $\log K = \dfrac{11\,400}{5706} = 2.00$, so $K = 10^{2.00} = 100$.

A modest $\Delta G^\circ$ gives a large $K$. A $\Delta G^\circ$ of $-57\ \text{kJ/mol}$ gives $K = 10^{10}$: the reaction goes almost to completion.

K at another temperature

$\Delta G^\circ$ itself depends on temperature, through $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$. So to find $K$ at a new temperature:

1. Work out $\Delta G^\circ$ at that temperature from $\Delta H^\circ$ and $\Delta S^\circ$.
2. Put it into $\log K = -\Delta G^\circ / (2.303\, RT)$, with the same $T$.

For an endothermic reaction with $\Delta S^\circ > 0$, $\Delta G^\circ$ falls as $T$ rises, so $K$ grows. Heating favours the products, as Le Chatelier's principle also says.

Practise with new numbers

Try it in Python

The program works out $\log K$ and $K$ for five values of $\Delta G^\circ$ at $298\ \text{K}$.

Common mistakes

• Mixing up $\Delta G$ and $\Delta G^\circ$. At equilibrium $\Delta G = 0$, but $\Delta G^\circ$ is usually not zero.
• Using kJ/mol with $R$ in J/(K mol). Change $\Delta G^\circ$ to J/mol first.
• Dropping the minus sign. A negative $\Delta G^\circ$ gives a positive $\log K$ and a $K$ above $1$.
• Using $\ln$ with the factor $2.303$. Use $\ln K$ with $RT$, or $\log K$ with $2.303\, RT$, not a mix.

Remember these

• $\Delta G^\circ = -RT \ln K = -2.303\, RT \log K$.
• $\Delta G^\circ < 0$: $K > 1$, products favoured. $\Delta G^\circ > 0$: $K < 1$, reactants favoured.
• At $298\ \text{K}$, $5.71\ \text{kJ/mol}$ of $\Delta G^\circ$ changes $K$ tenfold.
• At equilibrium, $\Delta G = 0$.

Check yourself

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

1. What is $K$ for a reaction with $\Delta G^\circ = 0$?

2. A reaction has $\Delta G^\circ = +5.71\ \text{kJ/mol}$ at $298\ \text{K}$. What is $K$?

3. Why is $\Delta G$ zero at equilibrium, although $\Delta G^\circ$ is not?

4. Heating an endothermic reaction with $\Delta S^\circ > 0$: does $K$ rise or fall?