What happens to the value of K when ΔG becomes more positive?

What happens to the value of K when ΔG becomes more positive?

• A. It gets larger
• B. It stays the same
• C. It gets smaller
• D. It depends on the reaction

Answer: (C)

The value of the equilibrium constant, K, is related to the change in Gibbs free energy, ΔG, through the equation: \[ ΔG = ΔG^\circ + RT \ln Q \] At equilibrium, ΔG equals zero, and this leads to the relationship: \[ 0 = ΔG^\circ + RT \ln K \] From this equation, we can derive that: \[ ΔG^\circ = -RT \ln K \] This indicates that if ΔG becomes more positive, ΔG^\circ must also become more positive. A more positive ΔG^\circ indicates that the reaction is less favorable in the forward direction. Since ΔG^\circ is inversely related to K through the equation, a more positive ΔG^\circ will result in a smaller value of K. This means the equilibrium position favors the reactants over the products, reflecting a decrease in K. Thus, as ΔG becomes more positive, the value of K indeed gets smaller, confirming the validity of the answer.

The value of the equilibrium constant, K, is related to the change in Gibbs free energy, ΔG, through the equation:

[

ΔG = ΔG^\circ + RT \ln Q

]

At equilibrium, ΔG equals zero, and this leads to the relationship:

[

0 = ΔG^\circ + RT \ln K

]

From this equation, we can derive that:

[

ΔG^\circ = -RT \ln K

]

This indicates that if ΔG becomes more positive, ΔG^\circ must also become more positive. A more positive ΔG^\circ indicates that the reaction is less favorable in the forward direction. Since ΔG^\circ is inversely related to K through the equation, a more positive ΔG^\circ will result in a smaller value of K. This means the equilibrium position favors the reactants over the products, reflecting a decrease in K.

Thus, as ΔG becomes more positive, the value of K indeed gets smaller, confirming the validity of the answer.