Question

If \(\Delta H^{\circ}=T \Delta S^{\circ}\), what is the value of \(\Delta G^{\circ}\) ? Predict the value of \(K\) in this case.

Short answer

The value of \(\Delta G^{\circ}\) is 0, and the equilibrium constant K is 1.

Step-by-step solution

Solve for \(\Delta G^{\circ}\)

Given the condition that the enthalpy change \(\Delta H^{\circ}\) is equal to the product of temperature and entropy change \(T \Delta S^{\circ}\). Substitute this condition into the Gibbs free energy equation \( \Delta G^{\circ}= \Delta H^{\circ}- T\Delta S^{\circ}\). The \(\Delta H^{\circ}\) and \(T\Delta S^{\circ}\) cancel out, which leaves us with \(\Delta G^{\circ} = 0\).

Predict the value of K

The relationship between Gibbs free energy and the equilibrium constant K can be represented as \(\Delta G^{\circ} = -RT \ln K\). Since we now know that \(\Delta G^{\circ} = 0\), we can substitute this value into the equation. This results in \(0 = -RT \ln K\), or rearranged as \(\ln K = 0\). The value of K that makes this equation hold true is 1, since the natural logarithm of 1 is 0.