Relationship Between \(K\) and \(\Delta_{r}G^{\circ}\)
This section explains why the standard Gibbs energy change \(\Delta_{r}G^{\circ}\) fixes the
equilibrium constant \(K\) at a given temperature, and how to convert between them in practice.
1) Core thermodynamic identities
The fundamental link between free energy and composition is
\[
\Delta_{r}G \;=\; \Delta_{r}G^{\circ} \;+\; R\,T\,\ln Q
\]
At equilibrium, \(\Delta_{r}G = 0\) and the reaction quotient equals the equilibrium constant \((Q=K)\):
\[
0 \;=\; \Delta_{r}G^{\circ} \;+\; R\,T\,\ln K
\Longrightarrow
\Delta_{r}G^{\circ} \;=\; -\,R\,T\,\ln K
\]
Algebraic rearrangements you will use in the calculator:
\[
K \;=\; \exp\!\Big(-\dfrac{\Delta_{r}G^{\circ}}{R\,T}\Big)
\]
\[
\Delta_{r}G^{\circ} \;=\; -\,R\,T\,\ln K
\]
\[
T \;=\; -\,\dfrac{\Delta_{r}G^{\circ}}{R\,\ln K}
\]
2) Non-standard conditions and spontaneity
When the current composition is not at equilibrium \((Q\neq K)\), the driving force is
\[
\Delta_{r}G \;=\; R\,T\,\ln\!\Big(\dfrac{Q}{K}\Big)
\]
| Condition | Sign of \(\Delta_{r}G\) | Spontaneous direction |
|---|
\(Q| \(<0\) | Forward (left \(\to\) right) | |
| \(Q>K\) | \(>0\) | Reverse (right \(\to\) left) |
| \(Q=K\) | \(=0\) | Equilibrium (no net change) |
3) Units, constants, and practical notes
- Always use absolute temperature: \(T\) in kelvin (K). If you start in °C, convert using \(T(\mathrm{K}) = T({}^{\circ}\mathrm{C}) + 273.15\).
- Use \(R = 8.314462618\ \mathrm{J\,mol^{-1}\,K^{-1}}\). This forces \(\Delta_{r}G^{\circ}\) to be in J·mol⁻¹. If you are given kJ·mol⁻¹, multiply by \(10^{3}\).
- \(K\) is dimensionless (activities). When you compute with concentrations/partial pressures, you are using their numerical values relative to standard states.
- Rule-of-thumb at \(298\ \mathrm{K}\): \(\Delta_{r}G^{\circ}(\mathrm{kJ\,mol^{-1}})\approx -5.708\,\log_{10}K\). Thus \(|\Delta_{r}G^{\circ}|\gtrsim 20\ \mathrm{kJ\,mol^{-1}}\) implies \(K\) is very large/small.
- Temperature dependence: \( \dfrac{\mathrm{d}\ln K}{\mathrm{d}T}=\dfrac{\Delta_{r}H^{\circ}}{R\,T^{2}}\) (van ’t Hoff). Our calculator assumes a single specified \(T\).
4) Worked example
Given: \(\Delta_{r}G^{\circ}=-12.0\ \mathrm{kJ\,mol^{-1}}\) at \(T=298.15\ \mathrm{K}\). Find: \(K\).
\[
\begin{aligned}
\Delta_{r}G^{\circ} &= -12.0\ \mathrm{kJ\,mol^{-1}} \;=\; -1.200\times 10^{4}\ \mathrm{J\,mol^{-1}} \\[6pt]
RT &= (8.314462618\ \mathrm{J\,mol^{-1}\,K^{-1}})\cdot 298.15\ \mathrm{K}
\;=\; 2.478\times 10^{3}\ \mathrm{J\,mol^{-1}} \\[6pt]
\ln K &= -\dfrac{\Delta_{r}G^{\circ}}{RT}
\;=\; -\,\dfrac{-1.200\times 10^{4}}{2.478\times 10^{3}}
\;=\; 4.84 \\[6pt]
K &= \exp(4.84)
\;\approx\; 1.27\times 10^{2}
\end{aligned}
\]
Interpretation: \(K\gg 1\) ⇒ products are favored at equilibrium.
5) How this connects to the calculator
- Solve for \(K\): enter \(\Delta_{r}G^{\circ}\) (J·mol⁻¹ or kJ·mol⁻¹) and \(T\) (K or °C). The tool converts units, computes \(K=\exp(-\Delta_{r}G^{\circ}/RT)\), and shows scientific notation.
- Solve for \(\Delta_{r}G^{\circ}\): enter \(K\) and \(T\). It evaluates \(-RT\ln K\) and returns J·mol⁻¹ and kJ·mol⁻¹.
- Solve for \(T\): enter \(K\) and \(\Delta_{r}G^{\circ}\) and use \(T=-\Delta_{r}G^{\circ}/(R\ln K)\).
- Optional \(Q\): if you supply \(Q\), the tool reports \(\Delta_{r}G=RT\ln(Q/K)\) and the spontaneous direction according to the table above.
6) Common pitfalls
- Forgetting to convert °C to K or kJ·mol⁻¹ to J·mol⁻¹ before applying the formulas.
- Using \(K\le 0\) (undefined \(\ln K\)) or \(K=1\) when solving for \(T\) (division by zero in \(\ln K\)).
- Confusing \(\Delta_{r}G\) (depends on composition via \(Q\)) with \(\Delta_{r}G^{\circ}\) (a standard-state property at the given \(T\)).
- Comparing \(Q\) and \(K\) at different temperatures—both must refer to the same \(T\).
