van ’t Hoff relation (two-point)
This page summarizes the theory behind the two-point van ’t Hoff relation used by the calculator
to connect the equilibrium constant \(K\) with the absolute temperature \(T\) when the standard
reaction enthalpy \(\Delta_{r}H^{\circ}\) is taken as temperature-independent over the range of interest.
1) From thermodynamics to van ’t Hoff
Start from the standard relation between the Gibbs energy and the equilibrium constant:
\[
\Delta_{r}G^{\circ} \;=\; -\,R\,T\,\ln K .
\]
Combine with \(\Delta_{r}G^{\circ}=\Delta_{r}H^{\circ}-T\,\Delta_{r}S^{\circ}\) and differentiate at constant
pressure. Eliminating \(\Delta_{r}S^{\circ}\) gives the differential van ’t Hoff equation
\[
\frac{\mathrm{d}\ln K}{\mathrm{d}T} \;=\; \frac{\Delta_{r}H^{\circ}}{R\,T^{2}}
\quad\Longleftrightarrow\quad
\frac{\mathrm{d}\ln K}{\mathrm{d}(1/T)} \;=\; -\,\frac{\Delta_{r}H^{\circ}}{R}.
\]
2) Integrated (two-point) form
Assuming \(\Delta_{r}H^{\circ}\) is approximately constant in the temperature window,
integration between \((K_1,T_1)\) and \((K_2,T_2)\) yields
\[
\ln\!\left(\frac{K_2}{K_1}\right)
\;=\; -\,\frac{\Delta_{r}H^{\circ}}{R}\!\left(\frac{1}{T_2}-\frac{1}{T_1}\right).
\]
- If \(K_1,T_1,K_2,T_2\) are known, you can solve for \(\Delta_{r}H^{\circ}\):
\(\displaystyle \Delta_{r}H^{\circ} = -\,R\,\dfrac{\ln(K_2/K_1)}{(1/T_2-1/T_1)}\).
- Given \(K_1,T_1,T_2,\Delta_{r}H^{\circ}\), you can predict \(K_2\):
\(\displaystyle K_2 = K_1 \exp\!\Big[-\dfrac{\Delta_{r}H^{\circ}}{R}\Big(\dfrac{1}{T_2}-\dfrac{1}{T_1}\Big)\Big]\).
- Likewise one can isolate \(T_1\) or \(T_2\) when the other quantities are provided.
3) Linear plot and physical meaning
The equation is linear in \(x=1/T\) and \(y=\ln K\):
\[
y \;=\; \left(-\frac{\Delta_{r}H^{\circ}}{R}\right)\,x \;+\; \frac{\Delta_{r}S^{\circ}}{R}.
\]
- Slope: \(m=-\Delta_{r}H^{\circ}/R\). A more positive slope implies a more negative
\(\Delta_{r}H^{\circ}\) (exothermic). A negative slope corresponds to \(\Delta_{r}H^{\circ}>0\) (endothermic).
- Intercept: \(b=\Delta_{r}S^{\circ}/R\).
- Two experimental points \((1/T_1,\ln K_1)\) and \((1/T_2,\ln K_2)\) define the straight line and
therefore \(\Delta_{r}H^{\circ}\) and \(\Delta_{r}S^{\circ}\) (if desired).
4) Units and input preparation
- Temperatures must be in Kelvin (\(T>0\)).
- \(\Delta_{r}H^{\circ}\) is handled internally in J·mol⁻¹; the calculator accepts kJ·mol⁻¹ or J·mol⁻¹.
- \(K\) must be dimensionless (use standard states: 1 bar for gases, 1 M for solutes).
- Gas constant \(R=8.314462618\ \mathrm{J\,mol^{-1}\,K^{-1}}\).
5) Interpreting how \(K\) varies with \(T\)
- Endothermic reaction (\(\Delta_{r}H^{\circ}>0\)): increasing \(T\) increases \(K\).
- Exothermic reaction (\(\Delta_{r}H^{\circ}<0\)): increasing \(T\) decreases \(K\).
6) Validity and limitations
- The two-point form assumes \(\Delta_{r}H^{\circ}\) does not change with \(T\).
Over wide ranges, temperature-dependence arising from heat-capacity differences
\(\Delta C_{p}^{\circ}\) can matter.
- More complete treatments use
\(\displaystyle \Delta_{r}H^{\circ}(T)=\Delta_{r}H^{\circ}(T_{\mathrm{ref}})+\int_{T_{\mathrm{ref}}}^{T}\Delta C_{p}^{\circ}\,\mathrm{d}T\)
and yield curvature in a plot of \(\ln K\) vs \(1/T\).
- Non-ideality (activities \(\neq\) concentrations/partial pressures) may be important at high pressures
or high ionic strengths; then replace concentrations with activities (or use activity coefficients).
7) What the calculator does
- Normalizes units: converts all temperatures to K and \(\Delta_{r}H^{\circ}\) to J·mol⁻¹.
- Applies the two-point equation to solve for the requested unknown.
- Reports the result and displays the straight-line plot of \(\ln K\) versus \(1/T\) through the two points.
