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Text pages 205 3.3Pressure, Equilibrium and Gibbs Free Energy Dependence of Free Energy on Pressure The entropy of 1 mole of gas in a 20.0 L container.

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Presentation on theme: "Text pages 205 3.3Pressure, Equilibrium and Gibbs Free Energy Dependence of Free Energy on Pressure The entropy of 1 mole of gas in a 20.0 L container."— Presentation transcript:

1 Text pages 205 3.3Pressure, Equilibrium and Gibbs Free Energy Dependence of Free Energy on Pressure The entropy of 1 mole of gas in a 20.0 L container is larger than the entropy of 1 mole of the same gas in a 10.0 L container (more microstates or ways to arrange the gas molecules in a larger volume). Entropy depends on the volume and pressure; therefore, so will the free energy. Experimentally, it can be seen that: How does the free energy change for a reaction depend upon pressure? Methane combusts in oxygen to form carbon dioxide and water: A derivation is found in the work text and we end up with..The expression in brackets is the reaction quotient, Q

2 Text pages 205 3.3Pressure, Equilibrium and Gibbs Free Energy Dependence of Free Energy on Pressure The derivation is provided to show how the reaction quotient, Q, appears in the equation when a change in the Gibbs free energy is considered in a reaction. For any reaction with species present at non-standard pressures: You can calculate ∆G°for a reaction from reference values of free energies of formation of reactants and products. Then you can use this as well as Q to determine ∆G. Refer to sample problem 3.3.1 to review how to calculate and use ∆G.

3 Text pages 207 3.3Pressure, Equilibrium and Gibbs Free Energy Free Energy and the Equilibrium Constant The relationship between the reaction quotient, Q, and free energy is: At equilibrium, ∆G = 0. The reaction mixture has reached a minimum of free energy. Also, at equilibrium, Q = K eq so the reaction quotient is equal to the equilibrium constant. Then, at equilibrium, becomes

4 Text pages 207-208 3.3Pressure, Equilibrium and Gibbs Free Energy Free Energy and the Equilibrium Constant Therefore, With this expression, we can use the reference values of ∆G f ° to calculate ∆G° Then you could calculate K, the equilibrium constant. (This could actually be Kp if we are dealing with gases at equilibrium or even Ksp for low solubility salts or Ka for weak acids.) Note: Some references prefer to deal with logs to the base 10, rather than natural logs to the base e. So, since 2.303log x = ln x, the expression becomes:

5 Text pages 208 3.3Pressure, Equilibrium and Gibbs Free Energy Predicting Reaction Direction Using K and ∆G° There are three possible situations when calculating the change in Gibbs free energy at equilibrium:

6 Text pages 208 3.3Pressure, Equilibrium and Gibbs Free Energy Predicting Reaction Direction Using K and ∆G° Here is a summary of reaction direction, K, and ∆G° Refer to sample problems 3.3.2(a) and (b) to carefully review examples of this content...

7 Text pages 208 3.3Pressure, Equilibrium and Gibbs Free Energy Predicting Reaction Direction Using K and ∆G°

8 Text pages 209 3.3Pressure, Equilibrium and Gibbs Free Energy Predicting Reaction Direction Using K and ∆G°

9 Text pages 210 3.3Pressure, Equilibrium and Gibbs Free Energy Variations in Total Free Energy for Reversible Reactions In an exergonic reaction, ∆G°is... negative. Here is a graph of free energy changes during the course of a reaction at constant T: The products (have less free energy than the reactants) are more stable than the reactants. The difference between point 1 (in the forward direction) or point 2 (in the reverse direction) and any point on the curve is ∆G. The mixture of reactants and products present at the minimum of the curve (point 3) has the least free energy and is the most stable of all. This represents the equilibrium mixture.

10 Text pages 210 3.3Pressure, Equilibrium and Gibbs Free Energy Variations in Total Free Energy for Reversible Reactions In an exergonic reaction, ∆G°is... negative. Here is a graph of free energy changes during the course of a reaction at constant T: At any point on the curve, a comparison of the reaction quotient, Q, to the equilibrium constant K indicates the direction in which the reaction must proceed to reach equilibrium...it will always the direction that is downhill in the free energy diagram! For an exergonic reaction, K must be greater than one so the equilibrium mixture will always contain more products than reactants.

11 Text pages 210 3.3Pressure, Equilibrium and Gibbs Free Energy Variations in Total Free Energy for Reversible Reactions In an endergonic reaction, ∆G°is... positive. Here is a graph of free energy changes during the course of a reaction at constant T: The products (point 2) will have more free energy than the reactants (point 1). In this case, K is less than one and the standard reaction is reactant favored, resulting in an equilibrium mixture containing more reactants than products.


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