Exercises: Free Energy of Dissolution
When ammonium nitrate dissolves in water to make an instant cold pack, $\Delta H^\circ_{\text{soln}} = +25.7\ \text{kJ/mol}$ and…
When ammonium nitrate dissolves in water to make an instant cold pack, $\Delta H^\circ_{\text{soln}} = +25.7\ \text{kJ/mol}$ and $\Delta S^\circ_{\text{soln}} = +108\ \text{J/(mol·K)}$. Calculate $\Delta G^\circ_{\text{soln}}$ at 298 K and state whether dissolving is favorable. Explain why the cold pack gets cold.
For each case, state whether dissolving is favorable at all temperatures, at no temperature, at high temperature only, or at low temperature only: (i)…
For each case, state whether dissolving is favorable at all temperatures, at no temperature, at high temperature only, or at low temperature only:
(i) $\Delta H_{\text{soln}} < 0$ and $\Delta S_{\text{soln}} > 0$
(ii) $\Delta H_{\text{soln}} > 0$ and $\Delta S_{\text{soln}} > 0$
(iii) $\Delta H_{\text{soln}} < 0$ and $\Delta S_{\text{soln}} < 0$
(iv) $\Delta H_{\text{soln}} > 0$ and $\Delta S_{\text{soln}} < 0$
For $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$, the standard free energy change is about $+56\ \text{kJ/mol}$ at 298 K. Is dissolving…
For $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$, the standard free energy change is about $+56\ \text{kJ/mol}$ at 298 K. Is dissolving favorable under standard conditions? Is $K_{sp}$ greater than 1 or less than 1? What does this say about the solubility of AgCl?
For the dissolution of AgCl at 298 K, $\Delta G^\circ = +55.6\ \text{kJ/mol}$. Use $\Delta G^\circ = -RT\ln K$ with $R = 8.314\ \text{J/(mol·K)}$ to calculate $K_{sp}$.…
For the dissolution of AgCl at 298 K, $\Delta G^\circ = +55.6\ \text{kJ/mol}$. Use $\Delta G^\circ = -RT\ln K$ with $R = 8.314\ \text{J/(mol·K)}$ to calculate $K_{sp}$. Then calculate the molar solubility $s$ of AgCl in pure water.
The table gives approximate data for dissolving two salts at 298 K. | Salt | $\Delta H^\circ_{\text{soln}}$ (kJ/mol) | $\Delta S^\circ_{\text{soln}}$ (J/(mol·K)) |…
The table gives approximate data for dissolving two salts at 298 K.
| Salt | $\Delta H^\circ_{\text{soln}}$ (kJ/mol) | $\Delta S^\circ_{\text{soln}}$ (J/(mol·K)) |
|---|---|---|
| NaCl | $+3.9$ | $+43$ |
| LiF | $+4.9$ | $-35.6$ |
(a) Calculate $\Delta G^\circ_{\text{soln}}$ for each salt. (b) Which salt is more soluble? (c) Explain how a negative $\Delta S^\circ_{\text{soln}}$ can occur when a solid dissolves, using the ordering of water molecules around small, highly charged ions.
For potassium nitrate, $\Delta H^\circ_{\text{soln}} = +34.9\ \text{kJ/mol}$ and $\Delta S^\circ_{\text{soln}} = +117\ \text{J/(mol·K)}$. Assume both are constant with…
For potassium nitrate, $\Delta H^\circ_{\text{soln}} = +34.9\ \text{kJ/mol}$ and $\Delta S^\circ_{\text{soln}} = +117\ \text{J/(mol·K)}$. Assume both are constant with temperature.
(a) Calculate $\Delta G^\circ_{\text{soln}}$ at 298 K and at 323 K. (b) Find the temperature at which $\Delta G^\circ_{\text{soln}} = 0$. (c) Explain how your results are consistent with the fact that KNO$_3$ is much more soluble in hot water than in cold water.
Two hypothetical salts, P and Q, have these dissolution data (assume constant $\Delta H^\circ$ and $\Delta S^\circ$): | Salt | $\Delta H^\circ_{\text{soln}}$ (kJ/mol) |…
Two hypothetical salts, P and Q, have these dissolution data (assume constant $\Delta H^\circ$ and $\Delta S^\circ$):
| Salt | $\Delta H^\circ_{\text{soln}}$ (kJ/mol) | $\Delta S^\circ_{\text{soln}}$ (J/(mol·K)) |
|---|---|---|
| P | $-20.0$ | $-120$ |
| Q | $+15.0$ | $+60$ |
(a) Calculate $\Delta G^\circ$ and $K$ for each at 298 K. (b) Calculate $\Delta G^\circ$ for each at 348 K. (c) Predict how the solubility of each salt changes as temperature rises, and justify it with $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$. (d) Identify which factor, enthalpy or entropy, drives dissolution of Q and which opposes dissolution of P.
The enthalpy of dissolution can be estimated as $\Delta H_{\text{soln}} = \Delta H_{\text{lattice}} + \Delta H_{\text{hydration}}$, where $\Delta H_{\text{lattice}}$ is…
The enthalpy of dissolution can be estimated as $\Delta H_{\text{soln}} = \Delta H_{\text{lattice}} + \Delta H_{\text{hydration}}$, where $\Delta H_{\text{lattice}}$ is the energy needed to separate the solid into gaseous ions. Approximate values:
| Salt | $\Delta H_{\text{lattice}}$ (kJ/mol) | $\Delta H_{\text{hydration}}$ (kJ/mol) |
|---|---|---|
| NaCl | $+787$ | $-784$ |
| AgCl | $+916$ | $-850$ |
(a) Calculate $\Delta H_{\text{soln}}$ for each salt. (b) For AgCl, $\Delta G^\circ_{\text{soln}} = +55.6\ \text{kJ/mol}$ at 298 K. Using your $\Delta H_{\text{soln}}$, calculate $\Delta S^\circ_{\text{soln}}$. (c) Explain why AgCl is only sparingly soluble even though $\Delta S^\circ_{\text{soln}}$ is positive. (d) Predict whether AgCl becomes more or less soluble as temperature increases, and justify your answer.