KEY2CHEM

Equilibrium in Coupled Reactions

Coupled reactions are those in which a reaction that is thermodynamically favored “drives” a reaction that is not thermodynamically favored by providing sufficient free energy. The set of reactions contains a common intermediate; consumption of the intermediate drives the reaction toward formation of product. This means that the product of the equilibrium constants will yield an overall equilibrium constant that favors the products ($$K > 1$$).

Example 1.

A thermodynamically favored redox reaction is $$\require{mhchem}\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}$$. The half reactions can be thought of as an equilibrium of the metal with its ions:

$$\require{mhchem}\ce{Zn(s) <=> Zn^2+(aq) + 2 e-} \;\; K_1$$

$$\require{mhchem}\ce{Cu(s) <=> Cu^2+(aq) + 2 e-} \;\; K_2$$

Based on these equilibrium constants, which statement is true?

A. $$K_1 > K_2$$

B. $$K_1 < K_2$$

C. $$K_1 = K_2$$

Solution

A. $$K_1 > K_2$$

Since the overall reaction is thermodynamically favorable, the $$\require{mhchem}\ce{Zn}$$ reaction will proceed to generate electrons, while the $$\require{mhchem}\ce{Cu}$$ reaction will shift, based on Le Chatelier’s Principle, to consume electrons and generate its reactant.

Example 2.

What is the missing value of $$K$$?

$$\require{mhchem}\ce{N2 + O2 <=> 2 NO}\;\;\;\;\;\;\;\;\;\;K_1 = 3.2 \times 10^ {-24}$$

$$\require{mhchem}\ce{2NO + O2 <=> 2NO2}\;\;\;\;\;K_2 = 1.6 \times 10^ {9}$$

$$\require{mhchem}\ce{N2 + 2O2<=> 2NO2}\;\;\;\;\;\;K_{missing} = ?$$

A. $$5.1 \times 10 ^{-15}$$

B. $$1.6 \times 10^{9}$$

C. $$1.6 \times 10^{-15}$$

Solution

A. $$5.1 \times 10 ^{-15}$$

Adding chemical equations together requires multiplication of their equilibrium constants. The small value of the overall equilibrium constant demonstrates that this reaction will not be very thermodynamically favorable.

$$K_{missing} = K_1 \times K_2 = (3.2 \times 10^ {-24}) \times (1.6 \times 10^ {9}) = 5.1 \times 10 ^{-15}$$

Example 3.

An overall reaction that is thermodynamically favorable will have a _____ value of equilibrium constant $$K$$.

A. negative

B. small

C. large

Solution

C. large

A reaction that is thermodynamically favorable has a negative value of $$\Delta G^{\circ}$$ and a relatively large value of $$K$$, as described quantitatively by $$\Delta G^{\circ} = -RT ln K$$.