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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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Most popular questions from this chapter

Most pasta recipes instruct you to add a teaspoon of salt to a pot

of boiling water. Does this have a significant effect on the boiling temperature?

Justify your answer with a rough numerical estimate.

How can diamond ever be more stable than graphite, when it has

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can increase the total entropy of the carbon plus its environment.

Sketch qualitatively accurate graphs of Gvs.Tfor the three phases ofH2O(ice, water, and steam) at atmospheric pressure. Put all three graphs on the same set of axes, and label the temperatures0°Cand 100°C. How would the graphs differ at a pressure of0.001bar?

When plotting graphs and performing numerical calculations, it is convenient to work in terms of reduced variables, Rewrite the van der Waals equation in terms of these variables, and notice that the constants a and b disappear.

The enthalpy and Gibbs free energy, as defined in this section, give special treatment to mechanical (compression-expansion) work, -PdV. Analogous quantities can be defined for other kinds of work, for instance, magnetic work." Consider the situation shown in Figure 5.7, where a long solenoid ( Nturns, total length N) surrounds a magnetic specimen (perhaps a paramagnetic solid). If the magnetic field inside the specimen is B→and its total magnetic moment is M→, then we define an auxilliary field H→(often called simply the magnetic field) by the relation

H→≡1μ0B→-M→V,

where μ0is the "permeability of free space," 4π×10-7N/A2. Assuming cylindrical symmetry, all vectors must point either left or right, so we can drop the -→symbols and agree that rightward is positive, leftward negative. From Ampere's law, one can also show that when the current in the wire is I, the Hfield inside the solenoid is NI/L, whether or not the specimen is present.

(a) Imagine making an infinitesimal change in the current in the wire, resulting in infinitesimal changes in B, M, and H. Use Faraday's law to show that the work required (from the power supply) to accomplish this change is Wtotal=VHdB. (Neglect the resistance of the wire.)

(b) Rewrite the result of part (a) in terms of Hand M, then subtract off the work that would be required even if the specimen were not present. If we define W, the work done on the system, †to be what's left, show that W=μ0HdM.

(c) What is the thermodynamic identity for this system? (Include magnetic work but not mechanical work or particle flow.)

(d) How would you define analogues of the enthalpy and Gibbs free energy for a magnetic system? (The Helmholtz free energy is defined in the same way as for a mechanical system.) Derive the thermodynamic identities for each of these quantities, and discuss their interpretations.
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