0

I am reading thermodynamics and I bumped into the following equation:

$$ \left( \frac{\partial \ln{K}}{\partial P}\right)_T = -\frac{Δ_\text{r}V^{\circ}}{RT} $$

Why is it valid to differentiate with respect to pressure when the equilibrium constant is function of temperature only? I mean $Δ_\text{r}G^{\circ} = f(T)$.

Anton
  • 271

1 Answers1

0

The equilibrium constant is generally not a function of temperature only. (For some idealized cases, however, such as the ideal gas, the dependence is on temperature only.) The equilibrium constant typically has a strong dependence on temperature and a weak dependence on pressure, and the latter aspect is often ignored as negligible.

Here is an example calculation deriving the (slight) influence of atmospheric pressure on the equilibrium vapor pressure of condensed matter; as with the equation you show above, there's a volume dependence. The influence is slight because the molar volume of condensed matter is small.