Free Energy and Equilibrium
Introduced| English | Meaning |
|---|---|
| free energy/friː ˈenədʒi/ | In this chemical context, Gibbs energy: enthalpy minus absolute temperature times entropy. |
Linking the verdict to the balance
- Free energy predicts direction; the constant $K$ describes the balance.
- These two ideas turn out to be tightly linked.
- A negative free energy means products win at equilibrium.
- One equation ties them together for good.
Free energy and K
- The link is $\Delta G° = -RT \ln K$.
- A negative $\Delta G°$ gives $K > 1$, so products are favoured.
- A positive $\Delta G°$ gives $K < 1$, so reactants are favoured.
When a reaction is feasible from the signs of enthalpy and entropy change
A reaction with $\Delta G° < 0$ has an equilibrium constant that is...
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Negative $\Delta G°$ favours products, so $K > 1$.
The equation linking free energy and K is $\Delta G° = -RT \ln$ ____.
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$\Delta G° = -RT \ln K$.
Reading the connection
- Favourable ($\Delta G° < 0$) matches product-favoured ($K > 1$).
- Unfavourable ($\Delta G° > 0$) matches reactant-favoured ($K < 1$).
- $\Delta G° = 0$ matches $K = 1$.
If $\Delta G° > 0$, then...
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A positive $\Delta G°$ means reactants dominate, so $K < 1$.
If $\Delta G° = 0$, the equilibrium constant $K$ equals...
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$\ln K = 0$ means $K = 1$.
Zero at equilibrium
- As a reaction proceeds, its actual free energy 自由能 heads toward zero.
- At equilibrium, $\Delta G = 0$ and the reaction stops shifting.
- The standard $\Delta G°$ still fixes where that balance lies.
Free energy and K
Link the sign of ΔG-standard to the size of K.
At equilibrium, the actual $\Delta G$ equals...
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The reaction stops shifting when $\Delta G = 0$.
A reaction has $\Delta G° < 0$. Is $K$ greater or less than 1?
- A negative $\Delta G°$ means products are favoured.
- So $K > 1$.
The standard $\Delta G°$ sets $K$, while the actual $\Delta G$ changes as the reaction runs.
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They are different quantities; only the actual $\Delta G$ reaches zero.
$\Delta G°$ (standard) fixes $K$, while the actual $\Delta G$ changes as the reaction proceeds and hits zero at equilibrium — do not confuse them. A negative $\Delta G°$ means $K > 1$, not $K = 0$. And the temperature in $\Delta G° = -RT \ln K$ is in kelvin.
Standard free energy and the equilibrium constant are linked by $\Delta G° = -RT \ln K$: a negative $\Delta G°$ gives $K > 1$ (product-favoured), a positive one gives $K < 1$. As a reaction runs, its actual $\Delta G$ falls to zero at equilibrium.