KEY2CHEM

Le Chatelier’s Principle and Coupled Reactions

Coupling reactions can allow a thermodynamically favorable reaction to provide sufficient free energy to “drive” a thermodynamically unfavorable reaction. Coupled reactions have common intermediates; changing the concentration of these intermediates via coupling allows the equilibrium position to shift toward the desired product.

Example 1.

In the formation of sucrose from glucose and fructose, the reaction is coupled to $$\require{mhchem}\ce{ATP}$$ hydrolysis. Which is common intermediate for these coupled reactions?

\begin{align}\require{mhchem}\ce{glucose\; + ATP \;\;&<=> glucose-P + ADP \\ glucose-P + fructose\;\; &<=> sucrose + P}_i\end{align}

A. $$\require{mhchem}\ce{ glucose}$$

B. $$\require{mhchem}\ce{ ATP}$$

C. $$\require{mhchem}\ce{ glucose-P}$$

Solution

C. $$\require{mhchem}\ce{ glucose-P}$$

The common intermediate “cancels out” when the reactions are added together since it is a product in one step and a reactant in another step. Consuming the $$\require{mhchem}\ce{glucose-P }$$ shifts the equilibrium position toward products ($$\require{mhchem}\ce{sucrose}$$ and inorganic phosphate, $$P_i$$).

Example 2.

In the coupled reactions below, which statement is correct?

$$\require{mhchem}\ce{A + B <=> C } \;\;\;\;\; \Delta G = -10 \text{ kJ}$$

$$\require{mhchem}\ce{C <=>E } \;\;\;\; \;\;\;\; \Delta G = 5 \text{ kJ}$$

A. As $$C$$ is consumed, the equilibrium position shifts toward forming $$E$$.

B. The overall reaction $$\Delta G = 15 \text{ kJ}$$.

C. As $$C$$ is consumed, the equilibrium position shifts toward forming $$A$$ and $$B$$.

Solution

A. As $$C$$ is consumed, the equilibrium position shifts toward forming $$E$$.

$$C$$ is a common intermediate. As $$C$$ is consumed, the equilibrium position shifts to replace the lost amount of $$C$$ to a greater extent, which ultimately generates more $$E$$.

Example 3.

Consuming a common intermediate in a coupled reaction _______ the amount of product formed.

A. does not change

B. increases

C. decreases

Solution

B. increases

Consuming the common intermediate shifts the equilibrium position toward formation of the product; this can aid “driving” a thermodynamically unfavorable reaction.