Chemistry Textbook

The status of a reversible reaction is conveniently assessed by evaluating its reaction quotient (Q). For a reversible reaction described by

the reaction quotient is derived directly from the stoichiometry of the balanced equation as

where the subscript c denotes the use of molar concentrations in the expression. If the reactants and products are gaseous, a reaction quotient may be similarly derived using partial pressures:

Note that the reaction quotient equations above are a simplification of more rigorous expressions that use relative values for concentrations and pressures rather than absolute values. These relative concentration and pressure values are dimensionless (they have no units); consequently, so are the reaction quotients. For purposes of this introductory text, it will suffice to use the simplified equations and to disregard units when computing Q. In most cases, this will introduce only modest errors in calculations involving reaction quotients.

 

The numerical value of Q varies as a reaction proceeds towards equilibrium; therefore, it can serve as a useful indicator of the reaction’s status. To illustrate this point, consider the oxidation of sulfur dioxide:

Two different experimental scenarios are depicted in Figure 11.5, one in which this reaction is initiated with a mixture of reactants only, SO₂ and O₂, and another that begins with only product, SO₃. For the reaction that begins with a mixture of reactants only, Q is initially equal to zero:

As the reaction proceeds toward equilibrium in the forward direction, reactant concentrations decrease (as does the denominator of Q_c), product concentration increases (as does the numerator of Q_c), and the reaction quotient consequently increases. When equilibrium is achieved, the concentrations of reactants and product remain constant, as does the value of Q_c.

If the reaction begins with only product present, the value of Q_c is initially undefined (immeasurably large, or infinite):

In this case, the reaction proceeds toward equilibrium in the reverse direction. The product concentration and the numerator of Q_c decrease with time, the reactant concentrations and the denominator of Q_c increase, and the reaction quotient consequently decreases until it becomes constant at equilibrium.

The constant value of Q exhibited by a system at equilibrium is called the equilibrium constant, K:

Comparison of the data plots in Figure 11.5 (a) and (b) shows that both experimental scenarios resulted in the same value for the equilibrium constant. This is a general observation for all equilibrium systems, known as the law of mass action: At a given temperature, the reaction quotient for a system at equilibrium is constant.

By its definition, the magnitude of an equilibrium constant explicitly reflects the composition of a reaction mixture at equilibrium, and it may be interpreted with regard to the extent of the forward reaction. A reaction exhibiting a large K will reach equilibrium when most of the reactant has been converted to product, whereas a small K indicates the reaction achieves equilibrium after very little reactant has been converted. It’s important to keep in mind that the magnitude of K does not indicate how rapidly or slowly equilibrium will be reached. Some equilibria are established so quickly as to be nearly instantaneous, and others so slowly that no perceptible change is observed over the course of days, years, or longer.

The equilibrium constant for a reaction can be used to predict the behavior of mixtures containing its reactants and/or products. As demonstrated by the sulfur dioxide oxidation process described above, a chemical reaction will proceed in whatever direction is necessary to achieve equilibrium. Comparing Q to K for an equilibrium system of interest allows prediction of what reaction (forward or reverse), if any, will occur.

To further illustrate this important point, consider the reversible reaction shown below:

The bar charts in Figure 11.6 represent changes in reactant and product concentrations for three different reaction mixtures. The reaction quotients for mixtures 1 and 3 are initially lesser than the reaction’s equilibrium constant, so each of these mixtures will experience a net forward reaction to achieve equilibrium. The reaction quotient for mixture 2 is initially greater than the equilibrium constant, so this mixture will proceed in the reverse direction until equilibrium is established.

A homogeneous equilibrium is one in which all reactants and products (and any catalysts, if applicable) are present in the same phase. By this definition, homogeneous equilibria take place in solutions. These solutions are most commonly either liquid or gaseous phases, as shown by the examples below:

$$\begin{array}{cccccccc}\hfill {\text{C}}_2{\text{H}}_2(aq)+2{\text{Br}}_2(aq)& \rightleftharpoons \hfill & {\text{C}}_2{\text{H}}_2{\text{Br}}_4(aq)\hfill & & & K_c\hfill & =\hfill & \frac{\left[{\text{C}}_2{\text{H}}_2{\text{Br}}_4\right]}{\left[{\text{C}}_2{\text{H}}_2\right]{\left[{\text{Br}}_2\right]}^2}\hfill \\ \hfill {\text{I}}_2(aq)+{\text{I}}^{\text{−}}(aq)& \rightleftharpoons \hfill & {\text{I}}_3{}^{\text{−}}(aq)\hfill & & & K_c\hfill & =\hfill & \frac{\left[{\text{I}}_3{}^{\text{−}}\right]}{[{\text{I}}_2][{\text{I}}^{\text{−}}]}\hfill \\ \hfill \text{HF}(aq)+{\text{H}}_2\text{O}(l)& \rightleftharpoons \hfill & {\text{H}}_3{\text{O}}^{\text{+}}(aq)+{\text{F}}^{\text{−}}(aq)\hfill & & & K_c\hfill & =\hfill & \frac{[{\text{H}}_3{\text{O}}^{\text{+}}]\left[{\text{F}}^{\text{−}}\right]}{\left[\text{HF}\right]}\hfill \\ \hfill {\text{NH}}_3(aq)+{\text{H}}_2\text{O}(l)& \rightleftharpoons \hfill & {\text{NH}}_4{}^{\text{+}}(aq)+{\text{OH}}^{\text{−}}(aq)\hfill & & & K_c\hfill & =\hfill & \frac{[{\text{NH}}_4{}^{\text{+}}]\left[{\text{OH}}^{\text{−}}\right]}{\left[{\text{NH}}_3\right]}\hfill \end{array}$$

These examples all involve aqueous solutions, those in which water functions as the solvent. In the last two examples, water also functions as a reactant, but its concentration is not included in the reaction quotient. The reason for this omission is related to the more rigorous form of the Q (or K) expression mentioned previously in this chapter, in which relative concentrations for liquids and solids are equal to 1 and needn’t be included. Consequently, reaction quotients include concentration or pressure terms only for gaseous and solute species.

The equilibria below all involve gas-phase solutions:

$$\begin{array}{cccccccc}\hfill {\text{C}}_2{\text{H}}_6(g)& \rightleftharpoons \hfill & {\text{C}}_2{\text{H}}_4(g)+{\text{H}}_2(g)\hfill & & & K_c\hfill & =\hfill & \frac{\left[{\text{C}}_2{\text{H}}_4\right]\left[{\text{H}}_2\right]}{\left[{\text{C}}_2{\text{H}}_6\right]}\hfill \\ \hfill 3{\text{O}}_2(g)& \rightleftharpoons \hfill & 2{\text{O}}_3(g)\hfill & & & K_c\hfill & =\hfill & \frac{{\left[{\text{O}}_3\right]}^2}{{\left[{\text{O}}_2\right]}^3}\hfill \\ \hfill {\text{N}}_2(g)+3{\text{H}}_2(g)& \rightleftharpoons \hfill & 2{\text{NH}}_3(g)\hfill & & & K_c\hfill & =\hfill & \frac{{\left[{\text{NH}}_3\right]}^2}{\left[{\text{N}}_2\right]{\left[{\text{H}}_2\right]}^3}\hfill \\ \hfill {\text{C}}_3{\text{H}}_8(g)+5{\text{O}}_2(g)& \rightleftharpoons \hfill & 3{\text{CO}}_2(g)+4{\text{H}}_2\text{O}(g)\hfill & & & K_c\hfill & =\hfill & \frac{{\left[{\text{CO}}_2\right]}^3{\left[{\text{H}}_2\text{O}\right]}^4}{\left[{\text{C}}_3{\text{H}}_8\right]{\left[{\text{O}}_2\right]}^5}\hfill \end{array}$$

For gas-phase solutions, the equilibrium constant may be expressed in terms of either the molar concentrations (K_c) or partial pressures (K_p) of the reactants and products. A relation between these two K values may be simply derived from the ideal gas equation and the definition of molarity:

$$PV=nRT$$

$$P=\left(\frac{n}{V}\right)RT$$

$$=MRT$$

where P is partial pressure, V is volume, n is molar amount, R is the gas constant, T is temperature, and M is molar concentration.

For the gas-phase reaction $m\text{A}+n\text{B}\rightleftharpoons x\text{C}+y\text{D:}$

$$K_P=\frac{{\left(P_C\right)}^x{\left(P_D\right)}^y}{{\left(P_A\right)}^m{\left(P_B\right)}^n}$$

$$=\frac{{\left([\text{C]}\times RT\right)}^x([\text{D]}\times RT{)}^y}{([\text{A]}\times RT{)}^m{\left([\text{B]}\times RT\right)}^n}$$

$$=\frac{[\text{C}{]}^x{[\text{D}]}^y}{[\text{A}{]}^m{[\text{B}]}^n}\times \frac{{\left(RT\right)}^{x\text{+}y}}{{\left(RT\right)}^{m\text{+}n}}$$

$$=K_c{(RT)}^{\left(x\text{+}y\right)-\left(m\text{+}n\right)}$$

$$=K_c{(RT)}^{\text{Δ}n}$$

And so, the relationship between K_c and K_P is

$$K_P=K_c{\left(RT\right)}^{\text{Δ}n}$$

where Δn is the difference in the molar amounts of product and reactant gases, in this case:

$$\text{Δ}n=(x\text{+}y)-(m\text{+}n)$$

EXAMPLE 11.4

Calculation of K_P Write the equations relating K_c to K_P for each of the following reactions:

(a) ${\text{C}}_2{\text{H}}_6(g)\rightleftharpoons {\text{C}}_2{\text{H}}_4(g)+{\text{H}}_2(g)$

(b) $\text{CO}(g)+{\text{H}}_2\text{O}(g)\rightleftharpoons {\text{CO}}_2(g)+{\text{H}}_2(g)$

(c) ${\text{N}}_2(g)+3{\text{H}}_2(g)\rightleftharpoons 2{\text{NH}}_3(g)$

(d) K_c is equal to 0.28 for the following reaction at 900 °C:

$${\text{CS}}_2(g)+4{\text{H}}_2(g)\rightleftharpoons {\text{CH}}_4(g)+2{\text{H}}_2\text{S}(g)$$

What is K_P at this temperature?

Solution (a) Δn = (2) − (1) = 1
K_P = K_c (RT)^Δn = K_c (RT)¹ = K_c (RT)

(b) Δn = (2) − (2) = 0
K_P = K_c (RT)^Δn = K_c (RT)⁰ = K_c

(c) Δn = (2) − (1 + 3) = −2
K_P = K_c (RT)^Δn = K_c (RT)⁻² = $\frac{K_c}{{(RT)}^2}$

(d) K_P = K_c (RT)^Δn = (0.28)[(0.0821)(1173)]⁻² = 3.0 $\times$ 10⁻⁵

Check Your Learning Write the equations relating K_c to K_P for each of the following reactions:

(a) $2{\text{SO}}_2(g)+{\text{O}}_2(g)\rightleftharpoons 2{\text{SO}}_3(g)$

(b) ${\text{N}}_2{\text{O}}_4(g)\rightleftharpoons 2{\text{NO}}_2(g)$

(c) ${\text{C}}_3{\text{H}}_8(g)+5{\text{O}}_2(g)\rightleftharpoons 3{\text{CO}}_2(g)+4{\text{H}}_2\text{O}(g)$

(d) At 227 °C, the following reaction has K_c = 0.0952:

$${\text{CH}}_3\text{OH}(g)\rightleftharpoons \text{CO}(g)+2{\text{H}}_2(g)$$

What would be the value of K_P at this temperature?

Answer:

(a) K_P = K_c (RT)⁻¹; (b) K_P = K_c (RT); (c) K_P = K_c (RT); (d) 160 or 1.6 $\times$ 10²

A heterogeneous equilibrium involves reactants and products in two or more different phases, as illustrated by the following examples:

$$\begin{array}{cccccccc}\hfill {\text{PbCl}}_2(s)& \rightleftharpoons \hfill & {\text{Pb}}^{\mathrm{2+}}(aq)+2{\text{Cl}}^{\text{−}}(aq)\hfill & & & K_c\hfill & =\hfill & [{\text{Pb}}^{\mathrm{2+}}]{\left[{\text{Cl}}^{\text{−}}\right]}^2\hfill \\ \hfill \text{CaO}(s)+{\text{CO}}_2(g)& \rightleftharpoons \hfill & {\text{CaCO}}_3(s)\hfill & & & K_c\hfill & =\hfill & \frac{1}{{[\text{CO}}_2]}\hfill \\ \hfill \text{C}(s)+2\text{S}(g)& \rightleftharpoons \hfill & {\text{CS}}_2(g)\hfill & & & K_c\hfill & =\hfill & \frac{{[\text{CS}}_2]}{{\left[\text{S}\right]}^2}\hfill \\ \hfill {\text{Br}}_2(l)& \rightleftharpoons \hfill & {\text{Br}}_2(g)\hfill & & & K_c\hfill & =\hfill & [{\text{Br}}_2(g)]\hfill \end{array}$$

Again, note that concentration terms are only included for gaseous and solute species, as discussed previously.

Two of the above examples include terms for gaseous species only in their equilibrium constants, and so K_p expressions may also be written:

$$\begin{array}{cccccccc}\hfill \text{CaO}(s)+{\text{CO}}_2(g)& \rightleftharpoons \hfill & {\text{CaCO}}_3(s)\hfill & & & K_P\hfill & =\hfill & \frac{1}{P_{{\text{CO}}_2}}\hfill \\ \hfill \text{C}(s)+2\text{S}(g)& \rightleftharpoons \hfill & {\text{CS}}_2(g)\hfill & & & K_P\hfill & =\hfill & \frac{P_{{\text{CS}}_2}}{{\left(P_{\text{S}}\right)}^2}\hfill \end{array}$$

The composition of a reaction mixture may be represented by a mathematical function known as the reaction quotient, Q. For a reaction at equilibrium, the composition is constant, and Q is called the equilibrium constant, K.

A homogeneous equilibrium is an equilibrium in which all components are in the same phase. A heterogeneous equilibrium is an equilibrium in which components are in two or more phases.

• $Q_c=\frac{[\text{C}{]}^x{[\text{D}]}^y}{[\text{A}{]}^m{[\text{B}]}^n}\text{for the reaction}m\text{A}+n\text{B}\rightleftharpoons x\text{C}+y\text{D}$
• $Q_P=\frac{{\left(P_C\right)}^x{\left(P_D\right)}^y}{{\left(P_A\right)}^m{\left(P_B\right)}^n}\text{for the reaction}m\text{A}+n\text{B}\rightleftharpoons x\text{C}+y\text{D}$
• P = MRT
• K_c = Q_c at equilibrium
• K_p = Q_p at equilibrium
• K_P = K_c (RT)^Δn