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Chapter 5: Q 5.88 (page 216)

Express ∂ΔG°/∂Pin 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

Expert verified

Hence, there is no change in equilibrium constant.

Step by step solution

01

Given information

∂ΔG°/∂P in terms of the volumes of solutions of reactants and products, for a chemical reaction of dilute solutes.

02

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(1)

Change in energy is given by:

dU=SdT-PdV

Substitute this into (1),

dG=SdT-PdV+PdV+VdP-TdS-SdTdG=VdP-TdS

By changing the pressure,

∂G∂P=V

Standard change in Gibbs energy:

∂ΔG°∂P=ΔV

03

Explanation

The equilibrium constant is given as:

K=e-ΔG°/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.

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