Chapter 55: Problem 9

For a chemical reaction with \(K>1\), is \(\Delta G^{\circ}\) positive or negative?

Short Answer

Expert verified

For a chemical reaction with \(K>1\), the standard Gibbs free energy change (\(\Delta G^{\circ}\)) is negative.

Step by step solution

Interpret the given condition

In the given exercise, it is mentioned that the equilibrium constant (K) for a chemical reaction is greater than 1. An equilibrium constant greater than 1 indicates that the reaction proceeds predominantly in the forward direction, that is, the concentration of the products is higher than the reactants at equilibrium.

Understand the Gibbs free energy

The Gibbs Free Energy (\(\Delta G^{\circ}\)) is a thermodynamic potential that measures the maximum reversible work that a system can perform at constant temperature and pressure. It combines the concepts of entropy (\(S\)) and enthalpy (\(H\)). The relationship is given as \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\). Here, \(\Delta H^{\circ}\) is the change in enthalpy, \(T\) is the absolute temperature, and \(\Delta S^{\circ}\) is the change in entropy. Negative \(\Delta G^{\circ}\) indicates that the reaction is spontaneous, while a positive value means it’s non-spontaneous.

Relate \(K\) and \(\Delta G^{\circ}\)

Considering the given value of \(K>1\) (implying more products than reactants at equilibrium), it corresponds to the reaction being spontaneous in the forward direction. The standard Gibbs free energy change, \(\Delta G^{\circ}\), is related to the equilibrium constant \(K\) by the equation \(\Delta G^{\circ} = -RT \ln K\), where \(R\) is the universal gas constant, \(T\) is the absolute temperature, and \(K\) is the equilibrium constant. Since \(K>1\), \(\ln K\) is positive. The negative sign in front of the right-hand side of the equation means \(\Delta G^{\circ}\) will be negative for \(K>1\).

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For a chemical reaction with \(K>1\), is \(\Delta G^{\circ}\) positive or negative?

Short Answer

Step by step solution

Interpret the given condition

Understand the Gibbs free energy

Relate \(K\) and \(\Delta G^{\circ}\)

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