Chemistry · General chemistry II · Concept

Enthalpy of solution and lattice energy

Dissolving a solid can warm or cool the water, and the enthalpy of solution, $Δ H (soln)$ , says which and by how much. Build it as a cycle: break the solid into separate particles, then let the water surround them. For an ionic solid, $Δ H (soln)$ is minus the lattice energy plus the hydration enthalpies of the ions.

A concrete case

A cold pack, a hot pack and table salt

Stir $10.0 g$ of ammonium nitrate, ${NH}_{4} {NO}_{3}$ , into $100.0 g$ of water at $22.0 ° C$ . The salt dissolves, and the water cools to about $15.0 ° C$ . Squeeze an instant cold pack and the same thing happens inside it.

Stir $5.00 g$ of calcium chloride, ${CaCl}_{2}$ , into $100.0 g$ of water instead. This time the water warms, by almost $9 ° C$ . Calcium chloride is what spreads on icy roads and fills some hot packs.

Stir in $5.00 g$ of table salt, $NaCl$ , and the thermometer falls by less than a degree.

Three white solids, three different answers. Every one of them dissolved, so dissolving itself does not decide the temperature. The answer comes from two energy changes that pull in opposite directions. Pulling the solid apart costs energy, and letting water molecules crowd around the freed particles releases energy. Whichever is larger wins.

Each solid stirred into $100.0 g$ of water
Solid	Mass	What the water does
Ammonium nitrate, ${NH}_{4} {NO}_{3}$	$10.0 g$	Cools by about $7 ° C$
Calcium chloride, ${CaCl}_{2}$	$5.00 g$	Warms by about $9 ° C$
Sodium chloride, $NaCl$	$5.00 g$	Cools by less than $1 ° C$

The rule

The energy cycle for dissolving

The enthalpy of solution, $Δ H (soln)$ , is the enthalpy change when one mole of solute dissolves. A negative value means heat is released and the solution warms. A positive value means heat is absorbed from the water and the solution cools.

Any dissolving can be split into three steps. Separating the solute particles costs energy, ΔH(solute), which is positive. Pushing solvent molecules apart to make room also costs energy, ΔH(solvent). Letting solute and solvent attract each other releases energy, ΔH(mix), usually negative.

For an ionic solid in water, two numbers do the job. The lattice energy, ΔH(lattice), is the enthalpy change for forming the solid from its gaseous ions, so it is negative. Breaking the lattice reverses it, $Δ H (solute)$ $= - Δ H (lattice)$ . The hydration enthalpy of an ion is the enthalpy change when the gaseous ion enters water. It combines the last two steps and is always negative.

Δ H_{soln} = Δ H_{solute} + Δ H_{solvent} + Δ H_{mix}, Δ H_{soln} = - Δ H_{lattice} + Δ H_{hydration}

\begin{array}{c} Δ H_{soln} = Δ H_{solute} + Δ H_{solvent} \\ + Δ H_{mix} \\ Δ H_{soln} = - Δ H_{lattice} + Δ H_{hydration} \end{array}

Why it works

Why a cycle gives the answer

Enthalpy is a state function, so $Δ H$ depends only on where a process starts and ends. The solid and the water at the start, and the dissolved ions at the end, are fixed. Any path between them gives the same total, as Hess’s law says. The cycle simply picks a path whose steps have known values.

The path for an ionic solid climbs first. Pulling the crystal apart into gaseous ions costs hundreds of kilojoules per mole, because every ion is held by its neighbors of opposite charge. Then the path drops. Each gaseous ion attracts the polar water molecules, which turn their oppositely charged ends toward it. That ion–dipole attraction releases about as much energy as the lattice cost.

$Δ H (soln)$ is the small difference between the climb and the drop. That is why three similar-looking salts can cool, warm or barely change the water.

Signs of the steps
Step	What happens	Sign
$Δ H (solute)$ $= - Δ H (lattice)$	Solute particles pulled apart	$+ (always)$
$Δ H (solvent)$	Solvent molecules pulled apart	$+ (always)$
$Δ H (mix)$	Solute–solvent attractions form	$- (usually)$
$Δ H (hydration)$	$Δ H (solvent) + Δ H (mix)$ for an ion in water	$- (always)$
$Δ H (soln)$	The sum	either sign

The method

Building the cycle from a table

1. Write the dissolving with states, such as ${CaCl}_{2} (s)$ $\to {Ca}^{2 +} (aq) + 2 {Cl}^{-} (aq)$ . The coefficients say how many of each ion form.

2. Reverse the lattice energy. $Δ H (solute)$ $= - Δ H (lattice)$ , a positive number.

3. Add the hydration enthalpies of the ions, each times its coefficient.

4. Add the two results to get $Δ H (soln)$ , in $kJ$ per mole of solute.

5. Read the sign. Negative warms the water, positive cools it, and a value near zero barely changes it.

Comparing the two terms
If	Then $Δ H (soln)$	The solution
$∣ Δ H (solute) ∣$ $< ∣ Δ H (hydration) ∣$	negative	warms
$∣ Δ H (solute) ∣$ $> ∣ Δ H (hydration) ∣$	positive	cools
$∣ Δ H (solute) ∣$ $\approx ∣ Δ H (hydration) ∣$	near zero	barely changes temperature

From $Δ H (soln)$ to a temperature

1. Convert the mass of solute to moles.

2. Multiply by $Δ H (soln)$ . That is the heat the dissolving absorbs, $q = n Δ H (soln)$ .

3. The water supplies that heat, so the solution’s heat is the opposite, $q (soln) = - q$ .

4. Divide by the mass of the whole solution, solute included, and its specific heat, $Δ T = q (soln) / (m c)$ . Take $c$ as water’s $4.184 J / (g \cdot ° C)$ for a dilute solution.

5. Add $Δ T$ to the starting temperature. This assumes no heat escapes to the cup or the air.

Where it breaks

$Δ H (soln)$ is a small difference of two large numbers. Published lattice and hydration enthalpies differ from table to table by several $kJ / mol$ , enough to change the sign of a near-zero answer. One consistent set gives $NaCl$ $+ 3 kJ / mol$ , close to its measured $+ 3.88 kJ / mol$ . The same lattice energy paired with hydration enthalpies from another table, $- 406$ and $- 364 kJ / mol$ , gives $+ 17 kJ / mol$ . Take every value from one table, and trust the sign only when the answer is well away from zero.

A positive $Δ H (soln)$ does not mean a solid will not dissolve. Ammonium nitrate dissolves readily while absorbing heat. Dissolving spreads the ions through the water, and that gain in entropy can outweigh the cost in enthalpy.

$Δ H (soln)$ also depends a little on how much solute is already present, because more water molecules surround each ion in a dilute solution. Tables quote it at a stated concentration or for infinite dilution.

The cycle describes physical dissolving. A solid that reacts with water, as sodium metal does, needs the enthalpy of that reaction instead.

Check your answer

Compare the size. $Δ H (soln)$ should be much smaller than either the lattice term or the hydration term. An answer near $- 1571 kJ / mol$ for $NaCl$ means the lattice energy was not reversed.

Compare the sign with what you would see. A salt sold for cold packs must come out positive, and a de-icing salt that warms its meltwater negative.

Count the ions. ${CaCl}_{2}$ gives two chloride ions, so its hydration sum has two ${Cl}^{-}$ terms.

In a Chemistry box, write the cycle as a relation with each value on its own row. When a table lists the ions separately, add their hydration enthalpies on a row of their own first. Under $Δ H (soln)$ $= - Δ H (lattice) + Δ H (hydration)$ , the row $Δ H (soln)$ $= - (- 787 kJ / mol)$ $+ (- 784 kJ / mol)$ $= 3 kJ / mol$ reads ✓. The Checker says “Substitution matches row 1 with the values defined above.”

The $3 kJ / mol$ is good to the ones place, because both values are. It is not trusted for that reason alone. Tables differ by several $kJ / mol$ , as Where it breaks says, and that spread is larger than the answer.

Charge, size and the second law

Both terms of the cycle grow with ionic charge and shrink with ionic size. A small, highly charged ion holds water dipoles tightly, so its hydration enthalpy is large. The same ion also binds its neighbors in the crystal tightly, so its lattice energy is large too. The two trends nearly cancel, which is why $Δ H (soln)$ values are so much smaller than either term.

Whether a solid dissolves is decided by the Gibbs energy, $Δ G = Δ H - T Δ S$ , not by $Δ H$ alone. For salts such as $NaCl$ and ${NH}_{4} {NO}_{3}$ , ΔS(soln) is positive, so at room temperature a modest positive $Δ H (soln)$ can still give a negative $Δ G$ . The Gibbs free energy lesson treats this balance in full.

IUPAC writes the enthalpy of solution with the subscript sol, $Δ_{sol} H$ , and marks an infinitely dilute water solution as aq, $\infty$ . The same cycle also explains a familiar rule. A polar solvent dissolves ionic and polar solutes because only it can pay back the energy needed to pull them apart.

Common mistakes

Taking ΔH(solute) as the lattice energy itself. For $NaCl$ that gives $- 787 + (- 784)$ $= - 1571 kJ / mol$ . Breaking the lattice costs energy, so $Δ H (solute) = + 787 kJ / mol$ and $Δ H (soln)$ is about $+ 3 kJ / mol$ .
Mixing values from different tables. The $- 787 kJ / mol$ lattice energy of $NaCl$ with hydration enthalpies of $- 406$ and $- 364 kJ / mol$ from another table gives $+ 17 kJ / mol$ , more than four times the measured $+ 3.88 kJ / mol$ . Take every value from one table, or the difference is meaningless.
Counting one chloride for ${CaCl}_{2}$ . $- 1616 + (- 364)$ gives a hydration sum of $- 1980 kJ / mol$ and $Δ H (soln) = + 278 kJ / mol$ . ${CaCl}_{2}$ releases two ${Cl}^{-}$ ions, so the sum is $- 2344 kJ / mol$ and $Δ H (soln) = - 86 kJ / mol$ .
Concluding that a positive $Δ H (soln)$ means a solid will not dissolve. Ammonium nitrate dissolves readily with $Δ H (soln) = + 25.7 kJ / mol$ , because the entropy gained by spreading the ions out pays for the heat absorbed.
Reading a positive $Δ H (soln)$ as a warmer solution. Positive means the dissolving absorbs heat, and it takes that heat from the water, so the cold pack falls from $22.0 ° C$ to $15.0 ° C$ .
Using only the water’s mass in $q = m c Δ T$ . $10.0 g$ of salt in $100.0 g$ of water makes $110.0 g$ of solution, and all of it changes temperature. Using $100.0 g$ gives a drop of $7.67 ° C$ instead of $6.97 ° C$ .

Key terms

Enthalpy of solution: The heat absorbed or released when a solute dissolves in a solvent. For an ionic solid in water, $Δ H_{soln}$ $= - Δ H_{lattice} + Δ H_{hydration}$ : the energy to pull the ions apart plus the energy released as water surrounds them.
Lattice energy: The energy change between an ionic solid and its separate gaseous ions. Forming the solid releases this energy and separating it absorbs the same amount, so its sign depends on which direction a book uses.
Enthalpy of hydration: The enthalpy change when one mole of gaseous ions is surrounded by water molecules and dissolves, as in ${Na}^{+} (g) \to {Na}^{+} (aq)$ . It is negative, because ion–dipole attractions to water release energy.
Hess’s law: If a reaction is the sum of other reactions, its $Δ H$ is the sum of their $Δ H$ values, because enthalpy is a state function. Reversing a reaction flips the sign of $Δ H$ . Multiplying a reaction scales $Δ H$ by the same factor.
Enthalpy: $H = U + P V$ , where $U$ is the system’s internal energy, $P$ its pressure and $V$ its volume. Enthalpy is a state function. At constant pressure its change, $Δ H$ , equals the heat the system absorbs (positive) or releases (negative).
Exothermic reaction: A reaction that releases heat to its surroundings, so its enthalpy change is negative $(Δ H < 0)$ and the surroundings warm.
Endothermic reaction: A reaction that absorbs heat from its surroundings, so its enthalpy change is positive $(Δ H > 0)$ and the surroundings cool.

Work through an example

One consistent set of data gives a lattice energy of $- 787 kJ / mol$ for $NaCl$ and a hydration enthalpy of $- 784 kJ / mol$ for one ${Na}^{+}$ ion and one ${Cl}^{-}$ ion together. Build the energy cycle and find $Δ H (soln)$ .