Topic 12 · page 16 of 17

Will it dissolve?

Dissolving is one more change that happens when the total entropy change is positive. The enthalpy of solution fixes the surroundings term, and the system term is ions dispersing against water becoming ordered.

12 min read12.19

Read first: Charge and radius

The same test as before 12.19

Page 6 gave the criterion for any change: it can happen if $\Delta S_\text{total}$ is positive. Dissolving an ionic solid is no exception.

\Delta S_\text{total} = \Delta S_\text{system} - \frac{\Delta_\text{sol}H}{T}

$\Delta_\text{sol}H$ (in J mol⁻¹, at temperature $T$ in K) fixes $\Delta S_\text{surroundings} = -\Delta_\text{sol}H / T$ .
$\Delta S_\text{system}$ has two competing parts: the ions disperse through the solution (entropy rises), but water molecules become ordered round the ions (entropy falls).

Worked exampleAmmonium nitrate at 298 K

$\Delta_\text{sol}H$ = +25.7 kJ mol⁻¹ and $\Delta S_\text{system}$ = +108.7 J K⁻¹ mol⁻¹.

Convert to joules: +25.7 kJ = +25 700 J mol⁻¹.
$\Delta S_\text{surroundings}$ = −25 700 ÷ 298 = −86.2 J K⁻¹ mol⁻¹
$\Delta S_\text{total}$ = +108.7 + (−86.2) = +22.5 J K⁻¹ mol⁻¹

$\Delta S_\text{total}$ is positive, so ammonium nitrate dissolves, even though the process is endothermic and the solution gets cold. This is the chemistry of an instant cold pack.

Endothermic dissolving takes energy from the surroundings, so the temperature of the solution falls. Entropy increases enough in the system to pay for the loss.

Temperature and the four cases 12.19

The same four cases as page 7 apply, with $\Delta_\text{sol}H$ in place of $\Delta H$ :

$\Delta_\text{sol}H$	$\Delta S_\text{system}$	Dissolves?
negative	positive	at all temperatures
negative	negative	only at lower temperatures
positive	positive	only at higher temperatures
positive	negative	never

The ammonium nitrate example is in the third row: the surroundings term becomes less negative as $T$ rises, so it dissolves better when warmer. At 308 K, −25 700 ÷ 308 = −83.4 and $\Delta S_\text{total}$ = +25.3 J K⁻¹ mol⁻¹, a little larger than at 298 K.

MethodPredicting solubility from data

Convert $\Delta_\text{sol}H$ to J mol⁻¹ and the temperature to K.
Calculate $\Delta S_\text{surroundings} = -\Delta_\text{sol}H / T$ .
Add $\Delta S_\text{system}$ to get $\Delta S_\text{total}$ .
State the conclusion with the sign: positive means it dissolves; negative means it does not.

Check yourself

A salt has $\Delta_\text{sol}H$ = −20 kJ mol⁻¹ and $\Delta S_\text{system}$ = −30 J K⁻¹ mol⁻¹ at 298 K (illustrative values). Does it dissolve?
Calculate $\Delta S_\text{total}$ for ammonium nitrate at 308 K, using the data in the worked example.
A salt has $\Delta_\text{sol}H$ = +30.0 kJ mol⁻¹ and $\Delta S_\text{system}$ = +70 J K⁻¹ mol⁻¹ (illustrative values). Is it soluble at 298 K? Above what temperature does $\Delta S_\text{total}$ become positive?
Explain why the entropy of the system can fall when a salt dissolves, even though the ions spread out.
Why does the solution get cold when ammonium nitrate dissolves, although it dissolves spontaneously?
Explain why a very exothermic enthalpy change of hydration does not by itself guarantee that a salt is soluble.

Answers cover these until you have tried the questions

$\Delta S_\text{surroundings}$ = +20 000 ÷ 298 = +67.1 J K⁻¹ mol⁻¹. $\Delta S_\text{total}$ = −30 + 67.1 = +37.1 J K⁻¹ mol⁻¹. It is positive, so the salt dissolves.
$\Delta S_\text{surroundings}$ = −25 700 ÷ 308 = −83.4 J K⁻¹ mol⁻¹. $\Delta S_\text{total}$ = 108.7 − 83.4 = +25.3 J K⁻¹ mol⁻¹.
$\Delta S_\text{surroundings}$ = −30 000 ÷ 298 = −100.7. $\Delta S_\text{total}$ = +70 − 100.7 = −30.7 J K⁻¹ mol⁻¹: negative, so it is not soluble at 298 K. $\Delta S_\text{total}$ = 0 when $T$ = 30 000 ÷ 70 = 429 K, so positive above 429 K (156 °C), beyond the boiling point of water at normal pressure, so in practice it would not dissolve in water.
Water molecules become ordered round the ions (they are held in hydration shells), and this loss of freedom is a fall in entropy. It can outweigh the increase from the ions dispersing, especially for small, highly charged ions.
Dissolving is endothermic, so energy is taken from the surroundings and their temperature falls. It is still spontaneous because $\Delta S_\text{system}$ is positive enough that $\Delta S_\text{total}$ is positive.
Strong hydration releases energy (making $\Delta S_\text{surroundings}$ positive) but it also orders the water strongly, making $\Delta S_\text{system}$ more negative. Solubility depends on the total, which combines both terms, and on the lattice energy that has to be overcome first.