Van 't Hoff Equation (Reaction Isobar)

Use the van 't Hoff equation in its integrated form: $\ln{}(\frac{K_{2}}{K_{1}}) = -\Delta{}H$°/R·(1/T₂ − 1/T₁). Insert the standard reaction enthalpy ΔH° in $\frac{J}{mol}$, the gas constant $R =\frac{ 8.314 J}{(mol\cdot{}K)}$ and both temperatures in kelvin. From the calculated logarithm you obtain the ratio $\frac{K_{2}}{K_{1}}$ by exponentiating. Example: for the ammonia synthesis with $\Delta H^\circ=-92\,\mathrm{kJ/mol}$ on heating from 298 K to 350 K, $\ln{}(\frac{K_{2}}{K_{1}}) \approx{} -5.52$, so $\frac{K_{2}}{K_{1}} \approx{} 0.004$; the equilibrium constant collapses strongly. Make sure to convert ΔH° to $\frac{J}{mol}$ so the units match R, and to insert the temperatures in the correct order, because the sign depends on it.

For an exothermic reaction ΔH° is negative. Inserting this into the van 't Hoff equation makes the expression $\ln{}(\frac{K_{2}}{K_{1}})$ negative on heating, so the equilibrium constant K decreases with rising temperature. The equilibrium shifts to the reactant side, less product forms. This is exactly the quantitative form of Le Chatelier's principle: for an exothermic reaction heat is treated like a product, and supplying heat shifts the equilibrium back. A practical example is the ammonia synthesis, which is exothermic. Therefore a low temperature would be favourable for high yield; in industry one nevertheless chooses moderate temperatures, because otherwise the reaction would run far too slowly.

Plot the natural logarithm of the equilibrium constant ln K against the reciprocal of the absolute temperature $\frac{1}{T}$. According to $\ln{} K = -\Delta{}H$°/(R·T) + ΔS°/R this gives a straight line with slope $-\frac{\Delta H^\circ}{R}$. From the measured slope m the standard reaction enthalpy follows as ΔH° = −m·R. If the slope is negative, the reaction is endothermic, because then K rises with temperature; a positive slope belongs to an exothermic reaction. The intercept additionally yields ΔS°/R and thus the reaction entropy. You need at least two K values at different temperatures, more points make the evaluation more accurate. This graphical method is the standard way to obtain thermodynamic quantities from equilibrium measurements.

Both equations have a similar form with an exponential temperature term but describe different things. The van 't Hoff equation describes how the equilibrium constant K depends on temperature and contains the reaction enthalpy ΔH°. It belongs to thermodynamics and tells you about the position of the equilibrium. The Arrhenius equation, by contrast, describes how the rate constant k depends on temperature and contains the activation energy E_A. It belongs to kinetics and tells you about the rate, not the equilibrium. You must not confuse the two: K determines how far a reaction proceeds, k determines how fast. Both complement each other into a complete picture of a reaction.

In the van 't Hoff equation the temperature appears as the reciprocal $\frac{1}{T}$ in the exponent, and this expression is very sensitive to the zero point of the scale. The Kelvin scale begins at absolute zero, which is thermodynamically the only meaningful reference temperature. If you accidentally insert Celsius, the reciprocal $\frac{1}{T}$ is completely wrong, because for example 0 °C is by no means an infinitely large or near-zero quantity but 273.15 K. Even small errors in the zero point lead to large errors in the result, because the exponential term amplifies them. Therefore you consistently calculate with absolute temperatures in kelvin and convert beforehand: T in kelvin equals T in Celsius plus 273.15. This holds for all thermodynamic equations with T in the denominator.