Chemistry · General chemistry II · Worked example

Enthalpy of solution of NaCl from a cycle

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 $Δ H (soln)$ .

NaCl (s) \to {Na}^{+} (aq) + {Cl}^{-} (aq)

Break the lattice

The lattice energy describes forming the solid from gaseous ions. Going from the solid to gaseous ions is the reverse, so its sign flips and it costs $+ 787 kJ / mol$ .

NaCl (s) \to {Na}^{+} (g) + {Cl}^{-} (g) Δ H_{solute} = + 787 kJ/mol

\begin{array}{c} NaCl (s) \to {Na}^{+} (g) + {Cl}^{-} (g) \\ Δ H_{solute} = + 787 kJ/mol \end{array}

Hydrate the ions

Each gaseous ion releases energy as water surrounds it. $NaCl$ gives one of each, and the data give the pair together, $- 784 kJ / mol$ , because measurements fix the sum. A table that lists each ion on its own has to split that sum by a convention, so its values belong with its own lattice energy.

Δ H_{hydration} = - 784 kJ/mol

Close the cycle

Add the two paths. Hess’s law allows it, because the climb and the drop together lead from the solid to the dissolved ions. The lattice cost and the hydration payback almost cancel.

Δ H_{soln} = + 787 + (- 784) = + 3 kJ/mol

\begin{array}{c} Δ H_{soln} = + 787 + (- 784) \\ = + 3 kJ/mol \end{array}

Judge the result

$+ 3 kJ / mol$ is the difference of two numbers near $785$ , so a few $kJ / mol$ either way in a table would change it. The measured value is $+ 3.88 kJ / mol$ , slightly endothermic, with almost no temperature change. The salt still dissolves because entropy increases.

Result

$Δ H (soln) \approx + 3 kJ / mol$ from this data (measured $+ 3.88 kJ / mol$ ). Dissolving $NaCl$ is nearly thermoneutral, slightly endothermic.

Your turn

1. One like the example

For ${CaCl}_{2}$ , use a lattice energy of $- 2258 kJ / mol$ and hydration enthalpies of $- 1616 kJ / mol$ for ${Ca}^{2 +}$ and $- 364 kJ / mol$ for ${Cl}^{-}$ . Find $Δ H (soln)$ , and say whether the solution warms or cools.

Show the answer and explanation

$- 86 kJ / mol$ , so it warms.

$Δ H (solute)$ $= + 2258 kJ / mol$ . Hydration counts two chloride ions, $- 1616 + 2 (- 364)$ $= - 2344 kJ / mol$ . The sum is $2258 + (- 2344)$ $= - 86 kJ / mol$ . Dissolving releases heat, which is why ${CaCl}_{2}$ is used in hot packs and to melt ice.

\begin{array}{l} - 1616 kJ/mol + 2 (- 364 kJ/mol) = - 2344 kJ/mol \\ 2258 kJ/mol + (- 2344 kJ/mol) = - 86 kJ/mol \end{array}

\begin{array}{c} - 1616 kJ/mol \\ + 2 (- 364 kJ/mol) \\ = - 2344 kJ/mol \\ 2258 kJ/mol + (- 2344 kJ/mol) \\ = - 86 kJ/mol \end{array}

2. A variation

For $KBr$ , a table gives $Δ H (soln) = + 19.9 kJ / mol$ and a lattice energy of $- 671 kJ / mol$ . What is the combined hydration enthalpy of $K^{+}$ and ${Br}^{-}$ ?

Show the answer and explanation

$- 651 kJ / mol$ .

This time the hydration term is the unknown. Solve the cycle for it before putting in numbers, $Δ H (hydration)$ $= Δ H (soln) + Δ H (lattice)$ . Then $19.9 + (- 671)$ $= - 651 kJ / mol$ . It is slightly smaller in size than the lattice cost, which is why $KBr$ cools the water a little.

\begin{array}{l} Δ H_{soln} = - Δ H_{lattice} + Δ H_{hydration} \\ Δ H_{hydration} = Δ H_{soln} + Δ H_{lattice} \\ Δ H_{soln} = 19.9 kJ/mol \\ Δ H_{lattice} = - 671 kJ/mol \\ Δ H_{hydration} = 19.9 kJ/mol + (- 671 kJ/mol) = - 651 kJ/mol \end{array}

\begin{array}{c} Δ H_{soln} = - Δ H_{lattice} \\ + Δ H_{hydration} \\ Δ H_{hydration} = Δ H_{soln} \\ + Δ H_{lattice} \\ Δ H_{soln} = 19.9 kJ/mol \\ Δ H_{lattice} = - 671 kJ/mol \\ Δ H_{hydration} \\ = 19.9 kJ/mol + (- 671 kJ/mol) \\ = - 651 kJ/mol \end{array}

3. Somewhere new

$5.00 g$ of ${CaCl}_{2}$ ( $Δ H (soln) = - 86 kJ / mol$ , from the first item) dissolves in $100.0 g$ of water. About how much does the temperature rise? Take $c = 4.184 J / (g \cdot ° C)$ for the solution and assume no heat is lost.

Show the answer and explanation

About $8.8 ° C$ .

$5.00 g \div 110.98 g / mol$ $= 0.04505 mol$ . The dissolving releases $0.04505 \times 86 = 3.874 kJ$ , which is $3874 J$ and good to two figures. The $105.0 g$ of solution absorbs it, so $Δ T = 3874 \div (105.0 \times 4.184)$ $= 8.8 ° C$ .

\begin{array}{l} 5.00 g {CaCl}_{2} \times \frac{1 mol {CaCl}_{2}}{110.98 g {CaCl}_{2}} = 0.04505 mol {CaCl}_{2} \\ (0.04505 mol) (86 kJ/mol) = 3.874 kJ \\ \frac{3874 J}{(105.0 g) (4.184 J g^{- 1} \circ C^{- 1})} = 8.8^{\circ} C \end{array}

\begin{array}{c} 5.00 g {CaCl}_{2} \times \frac{1 mol {CaCl}_{2}}{110.98 g {CaCl}_{2}} \\ = 0.04505 mol {CaCl}_{2} \\ (0.04505 mol) (86 kJ/mol) \\ = 3.874 kJ \\ \frac{3874 J}{(105.0 g) (4.184 J g^{- 1} \circ C^{- 1})} \\ = 8.8^{\circ} C \end{array}

Keep exploring

Open the ${CaCl}_{2}$ cycle of the first practice item in Energy diagrams and watch $Δ H (soln)$ come out at $- 86 kJ / mol$ . Then set the chloride count to $1$ and see how far one missing ion moves it.

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Study it, then try it

Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

Study the 7 key terms Build the

{CaCl}_{2}

cycle in Energy diagrams Try the three-step model Check the cycle in a Chemistry box Open worked example on a board Chemistry formulas: energy and thermodynamics

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Break the lattice

Hydrate the ions

Close the cycle

Judge the result

Result

Your turn

1. One like the example

−86 kJ⁢/⁢mol−86 kJ/mol, so it warms.

2. A variation

−651 kJ⁢/⁢mol−651 kJ/mol.

3. Somewhere new

About 8.8 °⁢C8.8 ∘C.

Keep exploring

Study it, then try it

$- 86 kJ / mol$ , so it warms.

$- 651 kJ / mol$ .

About $8.8 ° C$ .