4.3: Day 29- Gibbs Free Energy, Chemical Equilibrium

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Day 29: Gibbs Free Energy, Chemical Equilibrium

D29.1 Temperature Dependence of Gibbs Free Energy

Whether a reaction is product-favored, that is, whether the reactants are converted to products under standard-state conditions, is reflected in the arithmetic sign of its Δ_rG°. This equation

Δ_rG° = Δ_rH° − TΔ_rS°

shows that the sign of Δ_rG° depends on the signs of Δ_rH° and Δ_rS°, and, in some cases, the absolute temperature (which can only have positive values). Four possibilities exist:

1. Both Δ_rH° and Δ_rS° are positive—an endothermic process with an increase in system entropy. Δ_rG° is negative if TΔ_rS° > Δ_rH°, and positive if TΔ_rS° < Δ_rH°. Such a process is product-favored at high temperatures and reactant-favored at low temperatures.
2. Both Δ_rH° and Δ_rS° are negative—an exothermic process with a decrease in system entropy. Δ_rG° is negative if |TΔ_rS°| < |Δ_rH°| and positive if |TΔ_rS°| > |Δ_rH°|. Such a process is product-favored at low temperatures and reactant-favored at high temperatures. (Remember that |TΔ_rS°| represents the magnitude of TΔ_rS°, ignoring mathematical sign.)
3. Δ_rH° is positive and Δ_rS° is negative—an endothermic process that with a decrease in system entropy. Δ_rG° is positive regardless of the temperature. Such a process is reactant-favored at all temperatures.
4. Δ_rH° is negative and Δ_rS° is positive—an exothermic process with an increase in system entropy. Δ_rG° is negative regardless of the temperature. Such a process is product-favored at all temperatures.

These four scenarios are summarized in Figure 1.

Figure 1. There are four possible combinations of signs of Δ_rH° and Δ_rS°. For two combinations, whether a process is reactant-favored or product-favored depends on temperature.

Activity 1: Temperature and Product-favored or Reactant-favored Reactions

Query \(\PageIndex{1}\)

Figure 2 illustrates the four scenarios graphically, where Δ_rG° is plotted versus temperature:

ΔrG°	=	− ΔrS°(T)	+	ΔrH°
y	=	m(x)	+	b

Query \(\PageIndex{2}\)

Figure 2. These plots show the variation in Δ_rG^o with temperature for the four possible combinations of arithmetic sign for Δ_rG^o. Click on each “+” for more information.

For most reactions, neither Δ_rH° nor Δ_rS° change significantly as temperature changes. Thus, in Figure 2, the lines representing Δ_rG° are linear because the slope of each line (−Δ_rS°) is the same at all temperatures. The orange and green plots (representing examples of scenario 1 and 2, respectively) cross from product-favored to reactant-favored (as reflected by the sign of Δ_rG°) at a temperature that is characteristic of the specific process. This temperature is represented by the x-intercept, the value of T for which Δ_rG° is zero:

Δ_rG° = 0 = Δ_rH°− TΔ_rS°

\[T_{\Delta_{r}G^{\circ}=0} = \dfrac{\Delta_{r}H^{\circ}}{\Delta_{r}S^{\circ}} \nonumber \]

Hence, saying a process is product-favored at “high” or “low” temperatures is simply indicating whether the temperature is above or below T_ΔrG°=0. These relative terms are reaction-specific, that is, what is a “high” temperature for one reaction may very well be a “low” temperature for another reaction.

Exercise 1: Estimating Boiling Point

D29.2 Chemical Equilibrium

A chemical reaction is usually written with a single arrow, which suggests it proceeds in one direction, the direction of the arrow. But all chemical reactions are reversible, and both the forward and reverse reaction occur simultaneously. When reactions involve gases or solutions, where concentrations change as the reaction proceeds, the reaction eventually reaches a dynamic chemical equilibrium.

In a chemical equilibrium, the forward and reverse reactions occur at the same rates, and the concentrations of products and reactants remain constant over time. This implies that, if a reaction occurs in a closed system so that the products cannot escape, the reaction often does not yield 100% products. Instead, some reactants remain after the concentrations stop changing. At this point, when there is no further change in concentrations of reactants and products, we say the reaction is at equilibrium.

For example, when we place a sample of dinitrogen tetraoxide (N₂O₄, a colorless gas) in a glass vessel, the color becomes darker as N₂O₄ is converted to nitrogen dioxide (NO₂, a red-brown gas) by the reaction:

\[\text{N}_2\text{O}_4(g)\;\xrightleftharpoons[k_r]{k_f}\;2\text{NO}_2(g) \nonumber \]

Activity 2: Reaction Energy Diagram

Query \(\PageIndex{4}\)

At the beginning of this reaction, there is pure N₂O₄. As soon as the forward reaction produces some NO₂, at rate = k_f[N₂O₄]_t, the reverse reaction begins to occur at rate = k_r[NO₂]_t², and NO₂ starts to react to form N₂O₄. (The subscripts, t, indicate a time before equilibrium is reached.) As the reaction proceeds, the rate of the forward reaction decreases as [N₂O₄]_t decreases and the rate of the reverse reaction increases as [NO₂]_t increases. When the system reaches equilibrium, both N₂O₄ and NO₂ are present.

Query \(\PageIndex{5}\)

Figure 3. A container of N₂O₄(g) reacts toward equilibrium at 75°C. Colorless N₂O₄ converts to brown NO₂. As the reaction proceeds toward equilibrium, the color of the mixture darkens due to the increasing concentration of NO₂. Once equilibrium has been reached, there is no further darkening of color.

At equilibrium, [N₂O₄] and [NO₂] no longer change over time because the rate of NO₂ formation is exactly equal to the rate of NO₂ consumption, and the rate of N₂O₄ formation is exactly equal to the rate of N₂O₄ consumption. Chemical equilibrium is a dynamic process: the numbers of reactant and product molecules remain constant, but the forward and reverse reactions do not stop.

We use the ⇌ arrow when writing an equation for a reversible reaction. Such a reaction may or may not be at equilibrium. When we wish to speak about one particular aspect of a reversible reaction, we use a single arrow. For example, when the reaction in Figure 3 is at equilibrium, the rate of the forward reaction:

N₂O₄(g) → 2 NO₂(g)

is equal to the rate of the reverse reaction:

2 NO₂(g) → N₂O₄(g)

An equilibrium can be established for a physical change as well as for a chemical reaction. For example:

Br₂(l) ⇌ Br₂(g)

Figure 4 shows a sample of liquid Br₂ at equilibrium with Br₂ vapor in a closed container. When we pour liquid Br₂ into an empty bottle in which there is no bromine vapor, some liquid evaporates: the amount of liquid decreases and the amount of vapor increases. If we seal the container so no vapor escapes, the amount of liquid and vapor will eventually stop changing; at that point an equilibrium between the liquid and the vapor has been established. If the container were not sealed, the bromine vapor would escape and no equilibrium would be reached.

Figure 4. Sample of bromine in a sealed glass tube. Liquid bromine is in the bottom of the tube with gaseous bromine above it.

D29.3 Concentration Equilibrium Constants

A concentration equilibrium constant (K_c ) is a ratio of equilibrium concentrations of products and reactants that is constant for a given reaction at a given temperature. For example, consider this generic reversible reaction:

m A + n B ⇌ x C + y D

For this reaction, the concentration equilibrium constant, K_c, is:

\[K_c = \dfrac{[\text{C}]_e^{\;x}\;[\text{D}]_e^{\;y}}{[\text{A}]_e^{\;m}\;[\text{B}]_e^{\;n}} \nonumber \]

This mathematical expression is called the equilibrium constant expression. The “[…]_e” expression indicates explicitly equilibrium concentration of a reactant or product.

Query \(\PageIndex{6}\)

As a reaction approaches equilibrium, concentrations of reactants and products need to change until the rates of forward and reverse reactions are equal. Therefore, only substances whose concentrations can change as a reaction occurs are included in an equilibrium constant expression. For example, consider the following reaction at equilibrium:

2 HgO(s) ⇌ 2 Hg(l) + O₂(g) K_c = [O₂]_e

Because HgO is a pure solid, the number of HgO formula units in a given volume of HgO is the same throughout the reaction; it depends only on the density of HgO at the temperature of the reaction. Similarly, the number of Hg atoms per unit volume of pure Hg(l) is constant throughout the reaction. Thus, these concentrations are not included in the K_c expression. It is necessary for some HgO(s) and some Hg(l) to be present for the equilibrium to be maintained, but the quantity of each does not matter.

In general, K_c expressions do not contain terms for pure solids or pure liquids. In addition, for dilute solutions, the concentration of solvent remains constant throughout an equilibrium reaction and is also not included in the K_c expression, even though the solvent may appear in the reaction equation.

A homogeneous equilibrium is one in which all of the reactants and products are present in a single phase. Examples of homogeneous equilibria are reactions in the gas phase and reactions in liquid solutions. For example:

C₂H₂(g) + 2 Br₂(g) ⇌ C₂H₂Br₄(g)

\[K_c = \dfrac{[\text{C}_2\text{H}_2\text{Br}_4]_e}{[\text{C}_2\text{H}_2]_e[\text{Br}_2]_e^{\;2}} \nonumber \]

I₂(aq) + I‾(aq) ⇌ I₃‾(aq)

\[K_c = \dfrac{[\text{I}_3^{-}]_e}{[\text{I}_2]_e[\text{I}^{-}]_e} \nonumber \]

Hg₂²⁺(aq) + NO₃‾(aq) + 3 H₃O⁺(aq) ⇌ 2 Hg²⁺(aq) + HNO₂(aq) + 4H₂O(l)

\[K_c = \dfrac{[\text{Hg}^{2+}]_e^{\;2}[\text{HNO}_2]_e}{[\text{Hg}_2^{2+}]_e[\text{NO}_3^{-}]_e[\text{H}_3\text{O}^{+}]_e^{\;3}} \nonumber \]

HF(aq) + H₂O(l) ⇌ H₃O⁺(aq) + F‾(aq)

\[K_c = \dfrac{[\text{H}_3\text{O}^{+}]_e[\text{F}^{-}]_e}{[\text{HF}]_e} \nonumber \]

NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH‾(aq)

\[K_c = \dfrac{[\text{NH}_4^{+}]_e[\text{OH}^{-}]_e}{[\text{NH}_3]_e} \nonumber \]

In the aqueous equilibrium systems, H₂O(l) is the solvent. Its concentration does not appear in the K_c expression.

A heterogeneous equilibrium is a system in which reactants and products are found in two or more phases. Some heterogeneous equilibria involve chemical changes, for example:

PbCl₂(s) ⇌ Pb²⁺(aq) + 2Cl‾(aq) K_c = [Pb²⁺]_e[Cl‾]_e²

CaO(s) + CO₂(g) ⇌ CaCO₃(s)

\[K_c = \dfrac{1}{[\text{CO}_2]_e} \nonumber \]

C(s) + 2S(g) ⇌ CS₂(g)

\[K_c = \dfrac{[\text{CS}_2]_e}{[\text{S}]_e^{\;2}} \nonumber \]

Other heterogeneous equilibria involve phase changes, for example:

Br₂(l) ⇌ Br₂(g) K_c = [Br₂(g)]_e

Working with K_c

When all the coefficients in a balanced chemical equation are multiplied by some factor n, then the new K_c is the original K_c raised to the n^th power. For example:

2 NO₂(g) ⇌ N₂O₄(g)

\[K_{c,1} = \dfrac{[\text{N}_2\text{O}_4]_e}{[\text{NO}_2]_e^{\;2}} \nonumber \]

NO₂(g) ⇌ ½ N₂O₄(g)

\[K_{c,2} = \dfrac{[\text{N}_2\text{O}_4]_e^{\;1/2}}{[\text{NO}_2]_e} = (K_{c,1})^{1/2} \nonumber \]

When a reaction’s direction is reversed, the K_c for the new reaction is the reciprocal (inverse) of the original reaction K_c. For example:

A + 2B ⇌ AB₂

\[K_{c,1} = \dfrac{[\text{AB}_2]_e}{[\text{A}]_e[\text{B}]_e^{\;2}} \nonumber \]

AB₂ ⇌ A + 2B

\[K_{c,2} = \dfrac{[\text{A}]_e[\text{B}]_e^{\;2}}{[\text{AB}_2]_e} = \dfrac{1}{K_{c,1}} \nonumber \]

When two reactions occur sequentially to yield a new overall reaction, the K_c for the overall reaction is the product of the K_c values for the individual steps. For example:

A + C ⇌ AC

\[K_{c,1} = \dfrac{[\text{AC}]_e}{[\text{A}]_e[\text{C}]_e} \nonumber \]

AC + C ⇌ AC₂

\[K_{c,2} = \dfrac{[\text{AC}_2]_e}{[\text{AC}]_e[\text{C}]_e} \nonumber \]

A + 2C ⇌ AC₂

\[K_{c,3} = \dfrac{[\text{AC}_2]_e}{[\text{A}]_e[\text{C}]_e^{\;2}} = K_{c,1}\times K_{c,2} \nonumber \]

Exercise 2: Properties of Chemical Equilibrium

D29.4 Equilibrium Constant and Partial Pressure

Reactions in which all reactants and products are in the gas phase are another class of homogeneous equilibria. In these cases, the partial pressure of each gas can be used instead of its concentration in the equilibrium constant equation. At constant temperature, the partial pressure of a gas is directly proportional to its molar concentration (c), which is the amount of a substance (moles) per unit volume (liters): c = n/V. This proportionality of pressure and concentration can be derived from the ideal gas equation:

\[\begin{array}{rcl} PV & = & nRT \\[0.5em] P & = & (\dfrac{n}{V})\;RT \\[1em] P & = & cRT \end{array} \nonumber \]

For an example, consider the following reaction:

C₂H₆(g) ⇌ C₂H₄(g) + H₂(g)

We can write the equilibrium constant, K_p, using the equilibrium partial pressures of the gases, by following the same guidelines as for K_c expressions:

\[K_{\text{p}} = \dfrac{P_{\text{C}_2\text{H}_4}P_{\text{H}_2}}{P_{\text{C}_2\text{H}_6}} \nonumber \]

The two equilibrium constants, K_c and K_p, are directly related to each other. For the generic gas-phase reaction:

m A + n B ⇌ x C + y D

\[\begin{array}{rcl} K_{\text{p}} & = & \dfrac{(P_C)^x(P_D)^y}{(P_A)^m(P_B)^n} \\[1em] & = & \dfrac{([\text{C}]\;\times\;RT)^x([\text{D}]\;\times\;RT)^y}{([\text{A}]\;\times\;RT)^m([\text{B}]\;\times\;RT)^n} \\[1em] & = & \dfrac{[\text{C}]^x[\text{D}]^y}{[\text{A}]^m[\text{B}]^n}\;\times\;\dfrac{(RT)^{x+y}}{(RT)^{m+n}} \\[1em] & = & K_{\text{c}}(RT)^{(x+y)\;-\;(m+n)} \\[0.5em] K_{\text{p}} & = & K_{\text{c}}(RT)^{{\Delta}n} \end{array} \nonumber \]

where Δn is the difference between the sum of the coefficients of the gaseous products and the sum of the coefficients of the gaseous reactants in the reaction (that is, the change in amount of gas between the reactants and the products).

Note that the gas constant, R, can be expressed in different units. Use the R value and associated units that match the partial pressure units used in the K_p expression.

For heterogeneous equilibria that involve gases, equilibrium constants can also be expressed using partial pressures instead of concentrations. Two examples are:

CaO(s) + CO₂(g) ⇌ CaCO₃(s)

\[K_p = \dfrac{1}{P_{CO_2}} \nonumber \]

C(s) + 2 S(g) ⇌ CS₂(g)

\[K_p = \dfrac{P_{CS_2}}{(P_S)^2} \nonumber \]

Exercise 3: Pressure and Concentration Equilibrium Constant Expressions

Exercise 4: Converting Pressure and Concentration Equilibrium Constants

D29.5 Calculations Involving Equilibrium Constants

One way to determine the value for an equilibrium constant is to measure the concentrations (or partial pressures) of all reactants and all products at equilibrium.

Exercise 5: Calculating an Equilibrium Constant

Calculate the equilibrium constant K_c for the decomposition of PCl₅ at 250 °C.

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)

At equilibrium, [PCl₅]_e = 4.2 × 10^-5 M, [PCl₃]_e = 1.3 × 10^-2 M, [Cl₂]_e = 3.9 × 10^-3 M

Query \(\PageIndex{10}\)

A known equilibrium constant can be used to calculate an unknown equilibrium concentration, provided the equilibrium concentrations of all other reactants and products are known.

Exercise 6: Equilibrium Constant and Air Pollution

D29.6 Equilibrium Constants and Product-favored Reactions

The magnitude of an equilibrium constant is a measure of the yield of a reaction when it reaches equilibrium. A very large value for K_c (K_c >> 1) indicates that product concentrations are much larger than reactant concentrations when equilibrium has been achieved: nearly all reactants have been converted into products. If K_c is large enough, the reaction has gone essentially to completion when it reaches equilibrium.

Earlier we defined a product-favored reaction as one that proceeds spontaneously in the forward direction when all concentrations (or partial pressures) have the standard-state value of 1 M. If K_c > 1,so that the concentrations of products are greater than the concentrations of reactants, then when all concentrations are 1 M the reaction needs to produce greater concentrations of products to reach equilibrium. That is, the reaction needs to proceed in the forward direction. Thus, K_c > 1 (or K_p > 1) means that a reaction is product-favored.

A very small value of K_c, (K_c << 1) indicates that equilibrium is achieved when only a small fraction of the reactants has been converted into products. Such a reaction is reactant-favored. If K_c is small enough, essentially no reaction has occurred when equilibrium is reached. When K_c ≈ 1, both reactant and product concentrations are significant and it is necessary to use the equilibrium constant to calculate equilibrium concentrations.

By an argument similar to the one in the next to last paragraph above, K_c < 1 (or K_p < 1) means that a reaction is reactant-favored.

Exercise 7: Identifying Reactant-Favored and Product-Favored Processes

Podia Question

One way to remove silver oxide tarnish from silver is to heat the silver to a high temperature.

https://mediaspace.wisc.edu/id/0_p9myzb11

The reaction is:

2 Ag₂O → 4 Ag(s) + O₂(g)

At 298 K, the reaction is reactant-favored. Obtain data from the Appendix and determine at what temperature the reaction becomes product-favored. What assumptions need to be made to solve this problem?

Query \(\PageIndex{13}\)

Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.

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