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# 20.3: Free Energy and Equilibrium

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

• Use the reaction quotient to determine which direction a reaction must run in order to reach equilibrium.
• Use the equilibrium constant of a reaction to calculate the change in Gibbs free energy.
• Use changes in Gibbs free energy to calculate equilibrium constants.
• Predict whether the reactants or products of a reaction are favored under a given set of conditions by looking at either the equilibrium constant or the change in Gibbs free energy.

## Vocabulary

• reaction quotient (Q): The ratio of the reactant and product concentrations in their non-equilibrium states, raised to their respective exponents.

• What data is needed to calculate the change in Gibbs free energy for a reaction?
• What information is needed to calculate an equilibrium constant?

## The Reaction Quotient

In the previous chapter, we learned how to set up an equilibrium constant expression. For the generic reaction shown below, the equilibrium constant expression is constructed as follows:

aA+bBcC+dD\begin{align*}aA+bB \rightleftarrows cC+dD\end{align*}Keq=[C]c[D]d[A]a[B]b\begin{align*}K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}\end{align*}

It should be emphasized that the equation above is only true when the reaction is at equilibrium. Given enough time, the indicated ratio of concentrations will eventually converge on the equilibrium constant, but for any reaction in which net changes are still occurring, this equation is not true. For a reaction that is still in progress, it is often useful to calculate the same ratio of concentrations, but in a non-equilibrium state. This value is called the reaction quotient (Q), and comparing the value of Q to the value of Keq tells us which direction the reaction needs to shift in order to reach equilibrium.

Keq=[C]c[D]d[A]a[B]b\begin{align*}K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}\end{align*} (Only if the reaction is at equilibrium)
Q=[C]c[D]d[A]a[B]b\begin{align*}Q=\frac{[C]^c[D]^d}{[A]^a[B]^b}\end{align*} (Always)

If Q > Keq, then the concentrations of the products (in the numerator) are too large compared to the concentrations of the reactants. Therefore, the reverse reaction must be favored (reducing the products and generating more reactants) in order for the reaction to achieve equilibrium. If Q < Keq, the situation is reversed. There are too many reactants and not enough products, so the reaction quotient is a small number compared to Keq, and the forward reaction will be favored until equilibrium is reached. When Q = Keq, the reaction is at equilibrium.

## Free Energy at Non-Standard Concentrations

In the previous section, we looked at values for ΔG°, which tells us the free energy change for a reaction being run at standard conditions. In addition to the requirements for temperature and pressure (25°C and 1 atm), standard conditions also specify the concentrations of each reaction component. Under standard conditions, the concentration of each reaction component in the aqueous or gaseous states is exactly 1 M.

Plugging a value of 1 M in for every concentration in the reaction quotient expression will give us a value of Q = 1. Unless Keq happens to be exactly 1, reaction mixtures are not at equilibrium under standard conditions. However, we can calculate the value of ΔG for any combination of concentrations if we know the values of ΔG° and Q. The following equation (which we will not derive) shows the relationships between these quantities:

ΔG = ΔG° + RT ln Q
ΔG = free energy change of the reaction under existing conditions
ΔG° = free energy change of the reaction under standard-state conditions
R = universal gas constant (8.314 J/K•mol)
T = absolute temperature (in Kelvin)
Q = reaction quotient

Although ΔG° is going to be a fixed value for a given temperature, ΔG will vary depending on the value of RT ln Q.

## Free Energy and the Equilibrium Constant

At equilibrium, the forward and reverse reactions proceed at equal rates. The driving force in each direction is equal, because the free energy of the reactants and products under equilibrium conditions is equivalent (ΔG = 0). We also know that, at equilibrium, Q = Keq. For a reaction that has reached equilibrium, the equation above becomes the following:

ΔG = ΔG° + RT ln Q
0 = ΔG° + RT ln K
ΔG° = - RT ln K

This equation provides us with a way to convert between ΔG° and the equilibrium constant for a given reaction. Knowing either ΔG° or Keq tells us whether the reactants or products are favored at equilibrium. This relationship is summarized in the following table:

< 1 negative positive reactants favored over products at equilibrium
1 0 0 equilibrium situation; products and reactants equally favored
>1 positive negative products favored over reactants at equilibrium

Example 20.4

The standard free energy of formation (∆G°f) for ammonia is -16.6 kJ/mol. Calculate the equilibrium constant for the following reaction at 25°C (298 K):

N2(g) + 3H2(g) \begin{align*}\mathrm{\rightleftarrows}\end{align*} 2NH3(g)

If we know the value of ∆G° for this reaction, we can calculate the equilibrium constant using the equation above. ∆G° can be calculated from ∆G°f values using the following equation:

ΔG°rxn = ΣnΔG°f(products) – ΣnΔG°f(reactants)

Because nitrogen and hydrogen are both in their standard elemental states, they have ΔG°f values of zero.

ΔG°rxn = ΣnΔG°f(products) – ΣnΔG°f(reactants)
ΔG°rxn = 2(-16.6 kJ/mol) - [(0 kJ/mol) + 3(0 kJ/mol)]
ΔG°rxn = -33.2 kJ/mol

Then solve ΔG° = - RT ln Keq for Keq:

ΔG=RT ln Keq\begin{align*}\mathrm{\Delta G ^\circ=-RT \ ln \ K_{eq}}\end{align*}
ln Keq=ΔGRT\begin{align*}\mathrm{ln \ K_{eq}= - \frac{\Delta G ^\circ}{RT}}\end{align*}
Keq=eΔGRT\begin{align*}\mathrm{K_{eq} = e^{- \frac{\Delta G ^\circ}{RT}}}\end{align*}

Before plugging values into this equation, note that we are using a value of 8.314 J/K•mol for R, but our value for ΔG° is -33.2 kJ/mol. In order for our units to cancel correctly, we need to convert -33.2 kJ/mol to -33,200 J/mol. Then, Keq can be calculated as follows:

Keq=eΔGRT\begin{align*}\mathrm{K_{eq} = e^{- \frac{\Delta G ^\circ}{RT}}}\end{align*}
Keq=e33,200 J/mol(8.314 J/Kmol)(298 K)\begin{align*}\mathrm{K_{eq} = e^{- \frac{-33,200 \ J/mol}{(8.314 \ J/K\cdot mol)(298 \ K)}}}\end{align*}
Keq=e13.4\begin{align*}\mathrm{K_{eq} = e^{13.4}}\end{align*}
Keq=6.6×105\begin{align*}\mathrm{K_{eq} = 6.6 \times 10^5}\end{align*}

Keq is very large, indicating that the products are heavily favored at 25°C.

Looking at the equation above, we see that increasing the temperature for any reaction makes the fraction in the exponent smaller, thus moving the equilibrium constant closer to 1. If we were to run the ammonia formation reaction above, known as the Haber process, at a higher temperature, the equilibrium constant would be smaller, and the desired products would be less favored. However, the industrial manufacture of ammonia uses this reaction at temperatures of 400-450°C. Why are elevated temperatures used? Knowing whether reactants or products are energetically favored is one important factor for optimizing production of a desired product, but it is not the only relevant variable. As it turns out, this reaction is quite slow, so even if the products are heavily favored at equilibrium, it takes a long time to reach that equilibrium. Increasing the temperature speeds up the reaction, and this advantage makes up for the slightly less favorable equilibrium constant.

Example 20.5

The equilibrium constant for the following reaction is 1.6 × 10-10 at 25°C:

AgCl(s) \begin{align*}\mathrm{\rightleftarrows}\end{align*} Ag+(aq) + Cl-(aq)

Calculate ΔG°.

Simply plug the relevant values into the following equation:

ΔG=RT ln Keq\begin{align*}\mathrm{\Delta G ^\circ=-RT \ ln \ K_{eq}}\end{align*}
ΔG=(8.314 J/Kmol)(298 K) ln (1.6×1010)\begin{align*}\mathrm{\Delta G ^\circ=-(8.314 \ J/K\cdot mol)(298 \ K) \ ln \ (1.6 \times 10^{-10})}\end{align*}
ΔG=55,884 J/mol\begin{align*}\mathrm{\Delta G ^\circ=-55,884 \ J/mol}\end{align*}

Note that the answer will be in J/mol (not kJ/mol) if we cancel the units correctly. Rounding the correct number of significant figures and converting to the usual units for free energy values, ΔG° = 56 kJ/mol for this reaction. This is a very large number, indicating that the reactants are highly favored, as would be expected for a relatively insoluble ionic compound.

One type of equilibrium that we will be studying extensively in the following chapter is the ionization of an acid in water:

HA(aq)+H2O(l)A(aq)+H3O+(aq)\begin{align*}\mathrm{HA(aq)+H_2O(l) \rightleftarrows A^-(aq)+H_3O^+(aq)}\end{align*}

where HA is a generic acid, and A- is its conjugate base.

## Lesson Summary

• The reaction quotient can be compared to the equilibrium constant to determine which direction a reaction will shift in order to reach equilibrium.
• Gibbs free energy values can be used to determine equilibrium constants.
• Equilibrium constants can be used to calculate changes in Gibbs free energy.
• Equilibrium constants and Gibbs free energy values allow us to predict the extent to which reactants or products will be favored once a reaction reaches equilibrium.

## Lesson Review Questions

1. What is the difference between the equilibrium constant and the reaction quotient?
2. In which direction will the reaction proceed based on the three following condtions?
1. Q>Keq
2. Q<Keq
3. Q=Keq
3. What conditions does the notation ΔG° imply? Explain the difference between ΔG° and ΔG.
4. Define each value in the equation that relates ΔG° and ΔG.
5. How can ΔG° or Keqbe used to tell whether reactants or products are favored at equilibrium?
6. Calculate ΔG° for the process 2H2O(l) \begin{align*}\mathrm{\rightleftarrows}\end{align*} H3O+(aq) + OH-(aq) if the equilibrium constant at 25°C is 1×10-14.
7. Calculate ΔG° and Keq for the following reaction at 25°C: 2H2O(g) \begin{align*}\mathrm{\rightleftarrows}\end{align*} 2H2(g) + O2(g)
8. Calculate the equilibrium constant for the following reaction at 25°C: 2O3(g) \begin{align*}\mathrm{\rightleftarrows}\end{align*} 3O2(g)
9. Calculate ΔG° for the ionization of acetic acid (CH3CO2H). The equilibrium constant for the following reaction is 1.8 × 10-5:
CH3CO2H(aq)+H2O(l)CH3CO2(aq)+H3O+(aq)\begin{align*}\mathrm{CH_3CO_2H(aq)+H_2O(l) \rightleftarrows CH_3CO_2^-(aq)+H_3O^+(aq)}\end{align*}

## Points to Consider

• How do equilibrium processes influence the behavior of acids and bases?

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