Gibbs Energy of Reaction at Nonstandard Conditions
For a process at constant temperature \(T\) and (usually) constant pressure,
the Gibbs energy of reaction relates the actual composition (through the reaction
quotient \(Q\)) to the standard Gibbs energy change:
\[
\begin{aligned}
\Delta_{\mathrm r}G &= \Delta_{\mathrm r}G^{\circ} + RT\,\ln Q
\end{aligned}
\]
Reaction quotient \(Q\)
The quotient is defined from activities:
\[
\begin{aligned}
Q &= \prod_i a_i^{\,\nu_i}
\end{aligned}
\]
where \(a_i\) is the activity of species \(i\) and \(\nu_i\) is its stoichiometric
coefficient (positive for products, negative for reactants). In common idealized cases:
-
Ideal gases (using partial pressures \(P_i\)):
\[
Q_p \;=\; \prod_i \left(\frac{P_i}{P^{\circ}}\right)^{\nu_i}, \qquad P^{\circ}=1\ \text{bar}
\]
-
Dilute solutions (using molar concentration \(c_i\)):
\[
Q_c \;=\; \prod_i \left(\frac{c_i}{c^{\circ}}\right)^{\nu_i}, \qquad c^{\circ}=1\ \text{mol L}^{-1}
\]
-
Pure solids and pure liquids have \(a \approx 1\) (they usually do not appear in \(Q\)).
\(Q\) is dimensionless because each intensive variable is normalized by its standard-state value.
Connection to equilibrium
At equilibrium \(\Delta_{\mathrm r}G=0\) and \(Q=K\) (the equilibrium constant), so:
\[
\begin{aligned}
0 &= \Delta_{\mathrm r}G^{\circ} + RT\,\ln K
\\
\Rightarrow\quad
K &= \exp\!\left[-\frac{\Delta_{\mathrm r}G^{\circ}}{RT}\right]
\end{aligned}
\]
Eliminating \(\Delta_{\mathrm r}G^{\circ}\) gives a useful form for directionality:
\[
\begin{aligned}
\Delta_{\mathrm r}G
&= RT\,\ln\!\left(\frac{Q}{K}\right)
\end{aligned}
\]
Spontaneity criterion (at the specified \(T\) and composition)
- \(\Delta_{\mathrm r}G < 0\): spontaneous left \(\to\) right (forward).
- \(\Delta_{\mathrm r}G > 0\): spontaneous right \(\to\) left (reverse).
- \(\Delta_{\mathrm r}G = 0\): at equilibrium.
Units and constants
- Use \(R = 8.314\,462\,618\ \text{J mol}^{-1}\text{K}^{-1}\).
- Temperatures must be in kelvin (K).
- If \(\Delta_{\mathrm r}G^{\circ}\) is given in kJ·mol⁻¹, convert to J·mol⁻¹ before using \(RT\ln Q\).
Worked example (numbers as in many texts)
Given \(\Delta_{\mathrm r}G^{\circ} = -32.90\ \text{kJ mol}^{-1}\),
\(T=298.15\ \text{K}\), and \(Q=100\),
\[
\begin{aligned}
\Delta_{\mathrm r}G
&= \Delta_{\mathrm r}G^{\circ} + RT\,\ln Q
\\
&= -32.90\ \text{kJ mol}^{-1}
+ \big(8.314\times10^{-3}\ \text{kJ mol}^{-1}\text{K}^{-1}\big)
(298.15\ \text{K})\,\ln(100)
\\
&= -32.90\ \text{kJ mol}^{-1} + 11.42\ \text{kJ mol}^{-1}
\\
&= -21.48\ \text{kJ mol}^{-1}
\end{aligned}
\]
Since \(\Delta_{\mathrm r}G<0\), the forward direction is spontaneous for these conditions.
How to use the calculator
- Choose what to solve for: \(\Delta_{\mathrm r}G\), \(Q\), or \(T\).
- Enter \(\Delta_{\mathrm r}G^{\circ}\), \(T\), and the dimensionless \(Q\).
The calculator does not compute \(Q\) from amounts; you must supply it.
- Press Calculate. When solving for \(Q\) or \(T\), you may provide a target \(\Delta_{\mathrm r}G\);
the special case \(\Delta_{\mathrm r}G=0\) yields \(K\) or the equilibrium temperature for the specified \(Q\).
Guidance mirrors the figures and tables commonly shown in textbooks (e.g., direction rules based on
\(\Delta_{\mathrm r}G\) and the \(\Delta_{\mathrm r}G\)–\(\xi\) slope interpretation).
