**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 \(\Delta G^{\circ} = -RT ln K\).