6.8: Describing a Reaction - Equilibria, Rates, and Energy Changes

Every chemical reaction can go in either a forward or reverse direction. Reactants can go forward to products, and products can revert to reactants. As you may remember from your general chemistry course, the position of the resulting chemical equilibrium is expressed by an equation in which K_eq, the equilibrium constant, is equal to the product concentrations multiplied together, divided by the reactant concentrations multiplied together, with each concentration raised to the power of its coefficient in the balanced equation. For the generalized reaction

$$a\text{A}+b\text{B}\rightleftharpoons c\text{C}+$$

we have

$$K_{\text{eq}}=\frac{{[\text{C}]}^c{[\text{D}]}^d}{{[\text{A}]}^a{[\text{B}]}^b}$$

The value of the equilibrium constant tells which side of the reaction arrow is energetically favored. If K_eq is much larger than 1, then the product concentration term [C]^c[D]^d is much larger than the reactant concentration term [A]^a[B]^b, and the reaction proceeds as written from left to right. If K_eq is near 1, appreciable amounts of both reactant and product are present at equilibrium. And if K_eq is much smaller than 1, the reaction does not take place as written but instead goes in the reverse direction, from right to left.

In the reaction of ethylene with HBr, for example, we can write the following equilibrium expression and determine experimentally that the equilibrium constant at room temperature is approximately 7.1 × 10⁷:

Because K_eq is relatively large, the reaction proceeds as written and more than 99.999 99% of the ethylene is converted into bromoethane. For practical purposes, an equilibrium constant greater than about 10³ means that the amount of reactant left over will be barely detectable (less than 0.1%).

What determines the magnitude of the equilibrium constant? For a reaction to have a favorable equilibrium constant and proceed as written, the energy of the products must be lower than the energy of the reactants. In other words, energy must be released. This situation is analogous to that of a rock poised precariously in a high-energy position near the top of a hill. When it rolls downhill, the rock releases energy until it reaches a more stable, low-energy position at the bottom.

The energy change that occurs during a chemical reaction is called the Gibbs free-energy change (ΔG), which is equal to the free energy of the products minus the free energy of the reactants: ΔG = G_products − G_reactants. For a favorable reaction, ΔG has a negative value, meaning that energy is lost by the chemical system and released to the surroundings, usually as heat. Such reactions are said to be exergonic. For an unfavorable reaction, ΔG has a positive value, meaning that energy is absorbed by the chemical system from the surroundings. Such reactions are said to be endergonic.

You might also recall from general chemistry that the standard free-energy change for a reaction is denoted as ΔG°, where the superscript ° means that the reaction is carried out under standard conditions, with pure substances in their most stable form at 1 atm pressure and a specified temperature, usually 298 K. For biological reactions, the standard free-energy change is denoted as ΔG°′ and refers to a reaction carried out at pH = 7.0 with solute concentrations of 1.0 M.

Because the equilibrium constant, K_eq, and the standard free-energy change, ΔG°, both measure whether a reaction is favorable, they are mathematically related by the equation

$$\text{Δ}{G}^{\circ }=-RT\ln K_{\text{eq}}\text{or}{K}_{\text{eq}}=e^{-\text{Δ}{G}^{\circ }/RT}$$

where

$$\begin{array}{l}R=8.314\text{J/(K}\cdot \text{mol)}=\text{1}\text{.987 cal/(K}\cdot \text{mol)}\\ T=\text{Kelvin temperature}\\ e=2.718\\ \text{ln}{K}_{\text{eq}}=\text{natural logarithm of}{K}_{\text{eq}}\end{array}$$

For example, the reaction of ethylene with HBr has K_eq = 7.1 × 10⁷, so ΔG° = −44.8 kJ/mol (−10.7 kcal/mol) at 298 K:

$$\begin{array}{l}{K}_{\text{eq}}=7.1\times {10}^7\text{and ln}{K}_{\text{eq}}=18.08\\ \Delta G^{\circ }=-RT\ln K_{\text{eq}}=-[8.314\text{J/(K}\cdot \text{mol)](298 K)(18}\text{.08)}\\ \text{=}-\text{44,800 J/mol=}-44.8\text{kJ/mol}\end{array}$$

The free-energy change ΔG is made up of two terms, an enthalpy term, ΔH, and a temperature-dependent entropy term, TΔS. Of the two terms, the enthalpy term is often larger and more dominant.

$$\text{Δ}{G}^{\circ }=\text{Δ}{H}^{\circ }-T\text{Δ}{S}^{\circ }$$

For the reaction of ethylene with HBr at room temperature (298 K), the approximate values are

The enthalpy change (ΔH), also called the heat of reaction, is a measure of the change in total bonding energy during a reaction. If ΔH is negative, as in the reaction of HBr with ethylene, the products have less energy than the reactants. Thus, the products are more stable and have stronger bonds than the reactants, heat is released, and the reaction is said to be exothermic. If ΔH is positive, the products are less stable and have weaker bonds than the reactants, heat is absorbed, and the reaction is said to be endothermic. For example, if a reaction breaks reactant bonds with a total strength of 380 kJ/mol and forms product bonds with a total strength of 400 kJ/mol, then ΔH for the reaction is 400 kJ/mol – 380 kJ/mol = −20 kJ/mol and the reaction is exothermic.

The entropy change (ΔS) is a measure of the change in the amount of molecular randomness, or freedom of motion, that accompanies a reaction. For example, in an elimination reaction of the type

$$\mathbf{A}⟶\mathbf{B}+\mathbf{C}$$

there is more freedom of movement and molecular randomness in the products than in the reactant because one molecule has split into two. Thus, there is a net increase in entropy during the reaction and ΔS has a positive value.

On the other hand, for an addition reaction of the type

$$\mathbf{A}+\mathbf{B}⟶\mathbf{C}$$

the opposite is true. Because such reactions restrict the freedom of movement of two molecules by joining them together, the product has less randomness than the reactants and ΔS has a negative value. The reaction of ethylene and HBr to yield bromoethane, which has ΔS° = −0.132 kJ/(K · mol), is an example. Table 6.2 describes the thermodynamic terms more fully.

Table 6.2 Explanation of Thermodynamic Quantities: ΔG° = ΔH° − TΔS°

Term	Name	Explanation
ΔG°	Gibbs free-energy change	The energy difference between reactants and products. When ΔG° is negative, the reaction is exergonic, has a favorable equilibrium constant, and can occur spontaneously. When ΔG° is positive, the reaction is endergonic, has an unfavorable equilibrium constant, and cannot occur spontaneously.
ΔH°	Enthalpy change	The heat of reaction, or difference in strength between the bonds broken in a reaction and the bonds formed. When ΔH° is negative, the reaction releases heat and is exothermic. When ΔH° is positive, the reaction absorbs heat and is endothermic.
ΔS°	Entropy change	The change in molecular randomness during a reaction. When ΔS° is negative, randomness decreases. When ΔS° is positive, randomness increases.

Knowing the value of K_eq for a reaction is useful, but it’s important to realize its limitations. An equilibrium constant tells only the position of the equilibrium, or how much product is theoretically possible. It doesn’t tell the rate of reaction, or how fast the equilibrium is established. Some reactions are extremely slow even though they have favorable equilibrium constants. Gasoline is stable at room temperature, for instance, because the rate of its reaction with oxygen is slow at 298 K. Only at higher temperatures, such as contact with a lighted match, does gasoline react rapidly with oxygen and undergo complete conversion to the equilibrium products water and carbon dioxide. Rates (how fast a reaction occurs) and equilibria (how much a reaction occurs) are entirely different.

$$\text{Rate}\to \text{Is the reaction fast or slow?}$$

$$\text{Equilibrium}\to \text{In what direction does the reaction proceed?}$$

Which reaction is more energetically favored, one with ΔG° = −44 kJ/mol or one with ΔG° = +44 kJ/mol?