0.23 Free energy and thermodynamic equilibrium  (Page 6/6)

Thermodynamics can even provide a quantitative prediction of the equilibrium constant. Recall that the condition for equilibrium is that ΔG = 0 and that ΔG depends on the pressures of the gases in the reaction mixture, because ΔS depends on these pressures. Above, we showed that the entropy of a gas changes with the pressure by

S(P) - S° = -Rln(P/(1 atm))

For the reaction of N ₂ and H ₂ to form NH ₃ , let’s use the above equation to calculate ∆S for whatever partial pressures at which the reactants and products happen to be.

∆S = 2S _NH3 - 3S _H2 - S _N2

∆S = ∆S° - 2Rln(P _NH3 ) + 3Rln(P _H2 ) + Rln(P _N2 )

∆S = ∆S° - Rln(P _NH3 ² /P _H2 ³ P _N2 )

In the last step, we used the properties of logarithms to combine all the terms and to use the coefficients as exponents. The term in parentheses looks just like the equilibrium constant expression. However, it is not equal to the equilibrium constant because these partial pressures are not necessarily at equilibrium – they are whatever we choose them to be. As such, we call this term in parentheses the “reaction quotient,” designated Q :

Q = (P _NH3 ² /P _H2 ³ P _N2 )

So,

∆S = ∆S° - RlnQ

It turns out that this is a general expression for any reaction involving gases, provided that we form Q from the balanced equation with appropriate exponents. It also turns out that this expression works for any reaction involving solutes or ions in solution provided that we take the concentrations of the solutes or ions in solution with the appropriate exponents.

We can use this expression to calculate ∆G, if we recall that ∆H does not depend on pressure:

∆G = ∆H - T∆S = ∆H° - T(∆S° - RlnQ)

∆G = ∆H° - T∆S° + RTlnQ

Or, finally,

∆G = ∆G° + RTlnQ

This last equation is an exceptionally powerful relationship for predicting in what direction a reaction will proceed spontaneously and when the reaction will come to equilibrium.

Let’s first consider the reaction at equilibrium. At equilibrium, ∆G = 0. Also Q = K _p by definition of Q and K _p . So

∆G° = -RTlnK _p

This is a remarkable equation as it relates a thermodynamic quantity involving enthalpy changes and entropy changes to the experimentally observed reaction equilibrium constant. In fact, this proves why the equilibrium constant we experimentally observed is indeed a constant. Because this result turns out to be general, we can drop the subscript “p” referring to partial pressures. K is then any equilibrium constant.

If ∆Gº is a large negative number, K will be a large positive number and the reaction is thermodynamically favorable. If ∆Gº is a large positive number, K will be a number much less than one, and the reaction is thermodynamically unfavorable.

Let’s plug the last equation for ∆Gº into the immediately preceding equation for ∆G :

∆G = -RTlnK + RT lnQ = RTln(Q/K)

We can use this equation to predict when a reaction will be spontaneous from reactants to products, when it will be spontaneous from products to reactants, and when it will be at equilibrium. ∆G will be negative when Q<K , or in other words, when the numerator in Q is too small and the denominator is too large to equal K . This means we need more products and less reactants, so the reaction proceeds spontaneously forward, as we expect when ∆G is negative. Similarly, ∆G will be positive when Q>K , and the reaction proceeds spontaneously in reverse. Only when Q = K will we have ∆G = 0, so equilibrium requires that Q = K . This is what we observed experimentally. Therefore, thermodynamics both predicts and explains the position of equilibrium for chemical reactions. This is an amazing result!

Note that the thermodynamic description of equilibrium and the dynamic description of equilibrium are complementary. Both predict the same equilibrium. In general, the thermodynamic arguments give us an understanding of the conditions under which equilibrium occurs, and the dynamic arguments help us understand how the equilibrium conditions are achieved.

Review and discussion questions

1. Why does the entropy of a gas increase as the volume of the gas increases? Why does the entropy decrease as the pressure increases?
2. For each of the following reactions, calculate the values of ΔS˚, ΔH˚, and ΔG˚ at T = 298 K and use these to predict whether equilibrium will favor products or reactants at T = 298 K. Also calculate K _p .
   1. 2CO(g) + O ₂ (g) → 2CO ₂ (g)
   2. O ₃ (g) + NO(g) → NO ₂ (g) + O ₂ (g)
   3. 2O ₃ (g) → 3O ₂ (g)
3. For each of the reactions in Question 3, predict whether increases in temperature will shift the reaction equilibrium more towards products or more towards reactants.
4. Show that for a given set of initial partial pressures where Q is larger than K _p , the reaction will spontaneously create more reactants. Also show that if Q is smaller than K _p , the reaction will spontaneously create more products.
5. In our study of phase equilibrium we found that a substance with weaker intermolecular forces has a greater vapor pressure than a substance with strong intermolecular forces. This was explained using dynamic equilibrium arguments. Use thermodynamic equilibrium arguments involving ∆H, ∆S, and ∆G to explain why a substance with weaker intermolecular forces has a greater vapor pressure than one with stronger intermolecular forces.