Back to homeAP® Chemistry · AP Chemistry · Change subject
Start for free
Notes

Exercises: Free Energy of Dissolution

Practice★☆☆
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.

Solve
Practice★☆☆
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 $\Delta S_{\text{soln}} < 0$

Solve
Practice★☆☆
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?

Solve
Practice★★☆
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.

Solve
Practice★★☆
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.

Solve
Practice★★☆
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.

Solve
Practice★★★
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.

Solve
Practice★★★
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.

Solve