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Extended Topics in Thermodynamics and Reaction Mechanisms · Thermodynamic Control and Coupling

Free Energy of Dissolution

Squeeze an instant cold pack and the bag turns icy in seconds. Ammonium nitrate dissolves in water and absorbs heat, yet it still dissolves. Energy is absorbed, so why is dissolving favorable? By the end of this page you will be able to answer that with a single equation.

Dissolving as a thermodynamic process

Dissolving a salt is a process like any other, so it has an enthalpy change, an entropy change, and a free energy change:

$$\Delta G^\circ_{\text{soln}} = \Delta H^\circ_{\text{soln}} - T\,\Delta S^\circ_{\text{soln}}$$

Each term has a physical meaning.

  • $\Delta H^\circ_{\text{soln}}$: heat absorbed or released. Breaking the ionic lattice costs energy. Surrounding the ions with water molecules (hydration) releases energy. $\Delta H_{\text{soln}}$ is the net result.

  • $\Delta S^\circ_{\text{soln}}$: change in dispersal of matter and energy. Ions leaving a crystal usually mean more disorder, so $\Delta S$ is often positive.

  • $\Delta G^\circ_{\text{soln}}$: if it is negative, dissolving is thermodynamically favorable. If it is positive, it is unfavorable under standard conditions.

Practice★☆☆Not started
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

Signs of ΔH and ΔS

Because $T$ in kelvins is always positive, the signs of $\Delta H$ and $\Delta S$ tell you how temperature affects favorability.

$\Delta H_{\text{soln}}$ $\Delta S_{\text{soln}}$ Favorable?
− + At all temperatures
+ + At high temperature only
− − At low temperature only
+ − At no temperature

The pattern: a negative $\Delta H$ helps, a positive $\Delta S$ helps, and the $T\Delta S$ term grows with temperature. Whichever term is larger decides the sign of $\Delta G$.

Practice★☆☆Not started
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

From ΔG° to K_sp and solubility

Dissolving a slightly soluble salt is an equilibrium, so its equilibrium constant is $K_{sp}$. The link between free energy and $K$ is

$$\Delta G^\circ = -RT\ln K \qquad\Longleftrightarrow\qquad K = e^{-\Delta G^\circ/RT}$$

with $R = 8.314\ \text{J/(mol·K)}$, so $\Delta G^\circ$ must be in J/mol here.

$\Delta G^\circ$ $K$ Meaning
negative $K > 1$ products favored, very soluble
zero $K = 1$
positive $K < 1$ reactants favored, sparingly soluble
Practice★☆☆Not started
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★★☆Not started
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

Temperature and solubility

If you assume $\Delta H^\circ$ and $\Delta S^\circ$ do not change much with temperature, you can predict how solubility changes as the temperature changes.

Compare with a salt whose $\Delta H$ and $\Delta S$ are both negative: its $\Delta G^\circ$ becomes less negative on heating, so it is less soluble in hot water. This is uncommon for ionic solids (a few, like Li$_2$CO$_3$, do it), and it is typical of gases dissolving in water.

Practice★★☆Not started
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

Why ΔS° can be negative

Does a crystal turning into free ions always raise the entropy? Not always. Ions with a small radius and a large charge, like $\text{Li}^+$, $\text{F}^-$, or $\text{Mg}^{2+}$, grab water molecules strongly. The water around them becomes locked into ordered shells. That lost freedom of the water can outweigh the freedom gained by the ions, so $\Delta S^\circ_{\text{soln}}$ is negative.

You can also estimate $\Delta H_{\text{soln}}$ by splitting the process into two steps:

$$\Delta H_{\text{soln}} = \Delta H_{\text{lattice}} + \Delta H_{\text{hydration}}$$

The lattice energy term is positive (energy is needed to pull the ions apart). The hydration term is negative (energy is released as water surrounds the ions). The two are large and nearly cancel, so $\Delta H_{\text{soln}}$ is usually small, and $T\Delta S$ often decides the outcome.

Energy steps of dissolving: lattice energy raises the energy, hydration lowers it, and the net change is small Solid salt Gaseous ions Aqueous ions + lattice − hydration
Practice★★☆Not started
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★★★Not started
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★★★Not started
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

Back to the cold pack: ammonium nitrate absorbs heat ($\Delta H>0$), but the entropy gain makes $T\Delta S$ larger, so $\Delta G<0$. The bag gets cold because heat flows from the water into the dissolving solid.

Summary

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