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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 $Δ_\text{r}G^{\circ} = f(T)$.

Vincent Thacker
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Anton
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1 Answers1

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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.

Chemomechanics
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  • Doesn't the subscript of standard state denote standard pressure e.g.? ($1$ atm or $1$ bar) – Anton Jul 25 '21 at 18:06
  • Sometimes we use 1 atm/1 bar, sometimes we wish to use the equilibrium pressure at some set of standard conditions, and sometimes (generally when logarithms are involved) we can use any consistent value as long as it has units, so we don't end up taking the logarithm of a unit. Which definition to apply depends on the model/calculation. In the link, it's the last case; in SI, we'd typically use 1 Pa. – Chemomechanics Jul 25 '21 at 20:46