Problem 6 - Chemistry An Atoms First Approach

Question

You remember that \(\Delta G^{\circ}\) is related to \(R T \ln (K)\) but cannot remember if it's \(R T \ln (K)\) or \(-R T \ln (K) .\) Realizing what \(\Delta G^{\circ}\) and \(K\) mean, how can you figure out the correct sign?

Step-by-Step Solution

Answer

Understanding the meanings of ΔG° and K, we can determine that the correct relationship between Gibbs free energy and the equilibrium constant is ΔG° = -RT ln(K). This is because, for a spontaneous reaction (products favored, K > 1), ΔG° must be negative, and for a non-spontaneous reaction (reactants favored, K < 1), ΔG° must be positive. This relationship is consistent with these conditions.

Step 1: Understand Gibbs free energy ΔG°

Gibbs free energy (ΔG°) is a thermodynamic potential that measures the maximum or reversible work that may be done by a system at constant temperature T and pressure P. A negative ΔG° value implies that the reaction is spontaneous, while a positive ΔG° value indicates that the reaction is non-spontaneous. When ΔG° = 0, the reaction is at equilibrium.

Step 2: Understand equilibrium constant K

The equilibrium constant (K) is a measure of the concentration of reaction products and reactants in the equilibrium state. If K > 1, the reaction favors the formation of products, while if K < 1, the reaction favors the formation of reactants. At equilibrium, both reactants and products are present in a ratio governed by K.

Step 3: Relate ΔG° and K

Since we know that a negative ΔG° corresponds to a spontaneous reaction, which means that products are favored (K > 1), and a positive ΔG° corresponds to a non-spontaneous reaction, which means reactants are favored (K < 1), we can deduce the correct relationship between ΔG° and K. If ΔG° = RT ln(K): - For K > 1 (products favored), ln(K) > 0, and RT > 0, so ΔG° would be positive, which contradicts our understanding of spontaneity (ΔG° should be negative). - For K < 1 (reactants favored), ln(K) < 0, and RT > 0, so ΔG° would be negative, which also contradicts our understanding of spontaneity (ΔG° should be positive). If ΔG° = -RT ln(K): - For K > 1 (products favored), ln(K) > 0, and RT > 0, so ΔG° would be negative, which is consistent with our understanding of spontaneity. - For K < 1 (reactants favored), ln(K) < 0, and RT > 0, so ΔG° would be positive, which is also consistent with our understanding of spontaneity. Based on this analysis, the correct relationship between ΔG° and K must be:

Step 4: Conclusion

ΔG° = -RT ln(K)