Question

Express $\partial \left(\Delta G°\right)/\partial P$in terms of the volumes of solutions of reactants and products, for a chemical reaction of dilute solutes. Plug in some reasonable numbers, to show that a pressure increase of 1 atm has only a negligible effect on the equilibrium constant.

Short answer

Hence, there is no change in equilibrium constant.

Step-by-step solution

Given information

$\partial \left(\Delta G°\right)/\partial P$ in terms of the volumes of solutions of reactants and products, for a chemical reaction of dilute solutes.

Explanation

The Gibbs free energy is given as:

$G=U+PV-TS$

The change in Gibbs free energy is:

$dG=dU+PdV+VdP-TdS-SdT\left(1\right)$

Change in energy is given by:

$dU=SdT-PdV$

Substitute this into (1),

$$\begin{gathered} dG=SdT-PdV+PdV+VdP-TdS-SdT \\ dG=VdP-TdS \end{gathered}$$

By changing the pressure,

$\frac{\partial G}{\partial P}=V$

Standard change in Gibbs energy:

$\frac{\partial \Delta G°}{\partial P}=\Delta V$

Explanation

The equilibrium constant is given as:

$K=e^{-\Delta G°/RT}$

Because the volume of the liquid is nearly constant while the pressure increases, the change in volume is small when the pressure changes.

As a result, there is no change in the equilibrium constant.