Van 't Hoff equation

The van 't Hoff equation (also known as the van 't Hoff isochore) in chemical thermodynamics relates the change in temperature ("T") to the change in the equilibrium constant ("K") given the standard enthalpy change ("ΔH^o") for the process. The equation was first derived by Jacobus Henricus van 't Hoff.

:$frac\{d\; ln\; K\}\{dT\}\; =\; frac\{Delta\; H^ominus\}\{RT^2\}$

This can also be written

:$frac\{d\; ln\; K\}\{d\; \{frac$1 T = -frac{Delta H^ominus}{R} [Atkins, P., De Paula, J.: "Physical Chemistry", page 212. W.H. Freeman and Company, 2006]

If the enthalpy change of reaction is assumed to be constant with temperature, the definite integral of this differential equation between temperatures "T₁" and "T₂" is given by

:$ln\; left(\; \{frac$K_2 K_1 } ight) = frac - Delta H^ominus {R}left( {frac{1}T_2 - frac{1}T_1 } ight)

In this equation "K₁" is the equilibrium constant at absolute temperature "T₁" and "K₂" is the equilibrium constant at absolute temperature "T₂". "ΔH^o" is the standard enthalpy change and "R" is the gas constant.

Since:$Delta\; G^ominus\; =\; Delta\; H^ominus\; -\; TDelta\; S^ominus$and:$Delta\; G^ominus\; =\; -RT\; ln\; K$it follows that:$ln\; K\; =\; -\; frac$Delta H^ominus{RT}+ fracDelta S^ominus {R}

Therefore, a plot of the natural logarithm of the equilibrium constant versus the reciprocal temperature gives a straight line. The slope of the line is equal to minus the standard enthalpy change divided by the gas constant, "ΔH^o/R" and the intercept is equal to the standard entropy change divided by the gas constant, "ΔS^o/R". Differentiation of this expression yields the van 't Hoff equation.

ee also

*Clausius-Clapeyron relation

References