Question

The K_sp of ZnF₂ is$\mathbf{3}\mathbf{.0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}$at${\mathbf{25}}^{\mathbf{°}}\mathbf{\text{C}}$. What is$\mathit{\Delta }{\mathit{G}}^{\mathbf{°}}$? Is it possible to prepare a solution that contains Zn²⁺(aq) and F^-(aq) at their standard-state concentrations?

Short answer

For this reaction, the $\Delta G_{rxn}^{°}$ is 8.688kJ/mol.

The forward reaction is non-spontaneous because $Q>>K_{sp}$. As a result, preparing a solution of conventional 1M concentrations after dissolving ZnF₂ salt is difficult.

Step-by-step solution

Definition of standard-state concentrations

When the concentration of a solute is 1M, it is considered to be under standard state conditions.

Calculate ΔGrxn°

Since the solubility-product constant K_sp is given and the temperature is specified, Gibb's free energy of the reaction can be calculated as:

$\Delta G^{°}=-RT\ln K_p$

Then, at $298\text{K}\left({25}^{°}\text{C}\right)$

$\begin{array}{l}\Delta G^{°}=-8.314\frac{\text{J}}{\text{mol}\cdot \text{K}}\cdot 298\text{K}\cdot \ln (3.0\cdot {10}^{-2})\\ \Delta G^{°}=8.68775\cdot {10}^3\frac{\text{J}}{\text{mol}}\backslash \\ =8.688\frac{\text{kJ}}{\text{mol}}.\end{array}$

Therefore, the$\Delta G_{rxn}^{°}$ for this reaction is 8.688kJ/mol.

Whether the reaction is spontaneous or not

At standard conditions, the reaction is non-spontaneous with a delta $G^{°}>0$. As a result, preparing a solution of standard concentration - 1M would be impossible.

In ordinary conditions, the response quotient Q is

$\begin{array}{l}Q=\left[Z{n}^{2+}\right]\cdot {\left[F^{-}\right]}^2\\ Q=\left[1\text{M}\right]\text{Invalid <msup> element}\cdot \\ Q=1.\end{array}$

Therefore, the forward reaction is non-spontaneous because $Q>>K\_\left\{sp\right\}$. As a result, preparing a solution of standard 1M concentrations after dissolvingZnF₂ salt is impossible.