Topic 13 · page 3 of 12

Heterogeneous systems

When a system has more than one phase, pure solids and pure liquids are left out of Kc and Kp, and Kp contains gases only. Check the state symbols first.

12 min read13.113.2

Read first: Writing Kc, Partial pressures and Kp

Solids are left out 13.1

Take the decomposition of calcium carbonate in a closed container:

\ce{CaCO3(s) <=> CaO(s) + CO2(g)}

The "concentration" of a pure solid is fixed by its density. Twice as much CaCO₃ is not twice as concentrated: it is the same material, packed the same way. Since the value never changes as the reaction goes, it is built into the constant and left out of the expression.

K_\text{c} = [\ce{CO2}]

MethodWriting an expression for a heterogeneous system

Look at every state symbol in the equation.
Cross out each (s), and each pure (l).
Write the expression from what is left, as on page 1.
For $K_\text{p}$ also cross out every (aq), so that only gases remain.

Worked exampleFour heterogeneous systems

\ce{3Fe(s) + 4H2O(g) <=> Fe3O4(s) + 4H2(g)} \qquad K_\text{c} = \frac{[\ce{H2}]^4}{[\ce{H2O}]^4}

\ce{C(s) + CO2(g) <=> 2CO(g)} \qquad K_\text{c} = \frac{[\ce{CO}]^2}{[\ce{CO2}]}

\ce{Fe^3+(aq) + Ag(s) <=> Fe^2+(aq) + Ag+(aq)} \qquad K_\text{c} = \frac{[\ce{Fe^2+}]\,[\ce{Ag+}]}{[\ce{Fe^3+}]}

\ce{CaCO3(s) <=> CaO(s) + CO2(g)} \qquad K_\text{c} = [\ce{CO2}]

Kp uses gases only 13.2

Partial pressures belong to gases, so $K_\text{p}$ contains only gases. The same equations as above give:

\ce{3Fe(s) + 4H2O(g) <=> Fe3O4(s) + 4H2(g)} \qquad K_\text{p} = \frac{p_{\ce{H2}}^{\,4}}{p_{\ce{H2O}}^{\,4}}

\ce{C(s) + CO2(g) <=> 2CO(g)} \qquad K_\text{p} = \frac{p_{\ce{CO}}^{\,2}}{p_{\ce{CO2}}}

\ce{CaCO3(s) <=> CaO(s) + CO2(g)} \qquad K_\text{p} = p_{\ce{CO2}}

The silver and iron(III) reaction has no gas in it, so there is no $K_\text{p}$ for it at all.

When you find a mole fraction in a heterogeneous system, count the gas only. Solids are not part of the total moles of gas. In the CaCO₃ system, CO₂ is the only gas, so $p_{\ce{CO2}}$ equals the total pressure. A pure liquid in equilibrium with its own vapour is the simplest case:

\ce{H2O(l) <=> H2O(g)} \qquad K_\text{p} = p_{\ce{H2O}}

This is the vapour pressure of water at that temperature.

Water as a solvent 13.1

Dissolved species stay in the expression. Water is different when it is the solvent, because there is so much of it that its concentration hardly changes. It is left out of $K_\text{c}$ .

\ce{Cr2O7^2-(aq) + H2O(l) <=> 2CrO4^2-(aq) + 2H+(aq)} \qquad K_\text{c} = \frac{[\ce{CrO4^2-}]^2\,[\ce{H+}]^2}{[\ce{Cr2O7^2-}]}

Water in the esterification on page 1 is not a solvent. It is one of four liquids in the mixture, and its amount changes noticeably as the reaction goes, so it stays in.

Check yourself

Write $K_\text{c}$ and $K_\text{p}$ for $\ce{C(s) + H2O(g) <=> CO(g) + H2(g)}$ .
Write $K_\text{c}$ for $\ce{Zn(s) + Cu^2+(aq) <=> Zn^2+(aq) + Cu(s)}$ . Explain why there is no $K_\text{p}$ .
Write $K_\text{p}$ for $\ce{2NaHCO3(s) <=> Na2CO3(s) + H2O(g) + CO2(g)}$ .
A student writes $K_\text{c} = \dfrac{[\ce{CaO}][\ce{CO2}]}{[\ce{CaCO3}]}$ for the decomposition of calcium carbonate. What is wrong, and why is it wrong?
In a closed container, solid CaCO₃ and CaO are in equilibrium with CO₂ at 1000 K, and the CO₂ pressure is 0.12 atm (illustrative). Give $K_\text{p}$ at 1000 K.
Write $K_\text{c}$ for the hydrolysis $\ce{CH3COOC2H5(l) + H2O(l) <=> CH3COOH(l) + C2H5OH(l)}$ , and say why water is included.

Answers cover these until you have tried the questions

$K_\text{c} = \dfrac{[\ce{CO}]\,[\ce{H2}]}{[\ce{H2O}]}$ and $K_\text{p} = \dfrac{p_{\ce{CO}}\;p_{\ce{H2}}}{p_{\ce{H2O}}}$ . Carbon is a solid, so it is left out of both.
$K_\text{c} = \dfrac{[\ce{Zn^2+}]}{[\ce{Cu^2+}]}$ . Both metals are solids, so they are left out. $K_\text{p}$ needs gases and there are none.
$K_\text{p} = p_{\ce{H2O}}\;p_{\ce{CO2}}$ . Both solids are left out, and the two gases are multiplied.
CaCO₃ and CaO are pure solids. Their concentrations are constant, so they are left out. The correct expression is $K_\text{c} = [\ce{CO2}]$ .
$K_\text{p} = p_{\ce{CO2}}$ = 0.12 atm. No other term is needed, because both solids are left out.
$K_\text{c} = \dfrac{[\ce{CH3COOH}]\,[\ce{C2H5OH}]}{[\ce{CH3COOC2H5}]\,[\ce{H2O}]}$ . Water is a reactant in a mixture of liquids, and its amount changes during the reaction, so it is not a solvent and stays in.