I was wondering what is the difference between the Clausius-Clapeyron equation and the Van't Hoff equation. They appear to have the exact same physical meaning and are often used interchangeably.


1 Answer 1


They're two forms of the same equation, but Clausius-Clapeyron uses vapor pressure ($p^*$) where Van 't Hoff uses the reaction equilibrium constant ($K$).

Why does this work out? Well, think of vaporization as a chemical reaction:

$$ \text{X} (l) \longrightarrow \text{X} (\text{g}) $$

The equilibrium constant is defined in terms of activity (a):

$$ K=\left.\frac{a_\text{X(g)}}{a_{\text{X}(l)}}\right|_\text{equil} $$

For an ideal liquid solution at modest pressure, a is just the mole fraction x. And for an ideal gas, a is just its partial pressure in bar:

$$ K=\left.\frac px\,\right|_\text{equil}$$

One equilibrium condition is $x=1\ $ and $p=p^*,\ $ so...

$$ K=p^* $$

Or $p^*$ is the equilibrium coefficient for vaporization. Now since K is characteristic of a reaction at a given temperature, this would imply that if we change x, then the new partial pressure at equilibrium would change according to

$$ K=p^*=\frac{p}{x} \qquad\Rightarrow\qquad p=xp^*$$

And that's exactly what happens. Neat, huh?

  • 1
    $\begingroup$ Please, take care of the units! K and a are dimensionless, which is correct, but p is [bar] and x is also dimensionless, so K = p/x can't be true in this form. Instead of p, the molar fraction of X in the gas phase, pX/ptotal should be used. Why is x=1 a requirement for equilibrium? Can't mixtures be in equilibrium? $\endgroup$ Commented Nov 20, 2019 at 2:40

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