Chemical Equilibrium

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Senior 6 · Chemistry

Chemical Equilibrium

5 lessons

Lesson 2 of 5

The Equilibrium Constant Kc and the Law of Mass Action

25 min

Why does the ester stop at two-thirds?

In 1862, in Paris, Marcellin Berthelot and Léon Péan de Saint-Gilles heated ethanoic acid with ethanol in sealed glass tubes to make the ester ethyl ethanoate:

$\mathrm{CH_3COOH(l) + C_2H_5OH(l) \rightleftharpoons CH_3COOC_2H_5(l) + H_2O(l)}$

Starting from one mole of each, they never obtained more than about two-thirds of a mole of ester, however long they heated. Starting from the ester and water, they reached the same mixture from the other side. Changing the starting amounts changed the final amounts, but not at random. Table 2.1 shows three such experiments at one temperature.

No.	Start: acid, ethanol, water / mol	Equilibrium: acid, ethanol, ester, water / mol	$\dfrac{n(\text{ester}) \times n(\text{water})}{n(\text{acid}) \times n(\text{ethanol})}$
1	1.00, 1.00, 0	0.333, 0.333, 0.667, 0.667	4.0
2	1.00, 2.00, 0	0.155, 1.155, 0.845, 0.845	4.0
3	1.00, 1.00, 1.00	0.457, 0.457, 0.543, 1.543	4.0

Table 2.1: Illustrative results for the esterification, calculated by Ganzaa to match the value found experimentally.

The equilibrium amounts differ from one experiment to the next, yet one particular ratio is the same every time. In 1864 two Norwegian chemists, Cato Guldberg and Peter Waage, stated the general rule behind it: the law of mass action.

Definition

Law of mass action (equilibrium law): for a reversible reaction at equilibrium at a given temperature, the product of the equilibrium concentrations of the products, each raised to the power of its coefficient in the balanced equation, divided by the same product for the reactants, is a constant.

Equilibrium constant, $K_{c}$ : the value of that ratio, using concentrations in $\mathrm{mol\,dm^{-3}}$ . It depends only on temperature. Example: $K_{c} = 54.3$ at 430 °C for $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$ .

Equilibrium concentration, $[\mathrm{X}]$ : the concentration of X, in $\mathrm{mol\,dm^{-3}}$ , in the equilibrium mixture. Square brackets always mean equilibrium values in a $K_{c}$ expression.

ICE table: a table of the Initial amounts, the Change and the Equilibrium amounts of every species, used to work out equilibrium compositions.

Writing the expression for $K_{c}$

For a general reaction $a\mathrm{A} + b\mathrm{B} \rightleftharpoons c\mathrm{C} + d\mathrm{D}$ at equilibrium:

$K_c = \frac{[\mathrm{C}]^{c}\,[\mathrm{D}]^{d}}{[\mathrm{A}]^{a}\,[\mathrm{B}]^{b}}$

Products go on top, reactants underneath, and each concentration is raised to the power of its coefficient.
Pure solids and pure liquids are left out, because their "concentration" (amount per unit volume of the substance itself) is fixed. So is the solvent in a dilute solution.
The units of $K_{c}$ are $(\mathrm{mol\,dm^{-3}})^{\Delta n}$ , where $\Delta n$ = (sum of the powers on top) − (sum of the powers underneath). If $\Delta n = 0$ , $K_{c}$ has no units.
$K_{c}$ changes only when the temperature changes. Concentrations, pressure and catalysts do not alter it.

Equilibrium	$K_{c}$	Units
$\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$	$\dfrac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]}$	none
$\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$	$\dfrac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3}$	$\mathrm{mol^{-2}\,dm^{6}}$
$\mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)}$	$[\mathrm{CO_2}]$	$\mathrm{mol\,dm^{-3}}$
$\mathrm{Fe^{3+}(aq) + SCN^{-}(aq) \rightleftharpoons [Fe(SCN)]^{2+}(aq)}$	$\dfrac{[\mathrm{Fe(SCN)^{2+}}]}{[\mathrm{Fe^{3+}}][\mathrm{SCN^{-}}]}$	$\mathrm{mol^{-1}\,dm^{3}}$
$\mathrm{CH_3COOH(l) + C_2H_5OH(l) \rightleftharpoons CH_3COOC_2H_5(l) + H_2O(l)}$	$\dfrac{[\mathrm{CH_3COOC_2H_5}][\mathrm{H_2O}]}{[\mathrm{CH_3COOH}][\mathrm{C_2H_5OH}]}$	none

In the esterification, water is included: it is a product mixed with three other liquids, not a solvent present in huge excess. Since $\Delta n = 0$ , the volume cancels, which is why Table 2.1 could use moles directly.

What the size of $K_{c}$ tells you

A large $K$ means the top of the fraction, the products, dominates the equilibrium mixture; a small $K$ means the reactants dominate. As a rough guide, when $K$ lies between about $10^{-3}$ and $10^{3}$ , both are present in significant amounts (Figure 2.1). $K$ says nothing about how fast equilibrium is reached: hydrogen and chlorine can stay unreacted in the dark for a long time, although $K$ is enormous.

Figure 2.1: Equilibrium constants on a logarithmic scale. The values for the NO and HCl reactions were calculated from standard Gibbs energies.

Changing the equation changes $K$

Reversing the equation gives $\frac{1}{K}$ . For $\mathrm{2HI \rightleftharpoons H_2 + I_2}$ at 430 °C, $K_c = \frac{1}{54.3} = 0.0184$ .
Multiplying every coefficient by $n$ raises $K$ to the power $n$ . For $\mathrm{\tfrac{1}{2}H_2 + \tfrac{1}{2}I_2 \rightleftharpoons HI}$ , $K_c = \sqrt{54.3} = 7.37$ .
Adding two equations multiplies their constants: $K_3 = K_1 \times K_2$ .

So a value of $K$ means nothing unless the equation it belongs to is written beside it.

Solving equilibrium problems with an ICE table

Write the balanced equation and the $K_{c}$ expression.
I: write the initial amounts (or concentrations).
C: write the changes. They must be in the ratio of the coefficients; use the one measured value, or an unknown $x$ .
E: add to get the equilibrium amounts, then divide by the volume to get concentrations.
Substitute into $K_{c}$ , solve, and give units and sensible significant figures.

Worked example

Problem (finding $K_{c}$ ). 0.400 mol of sulfur dioxide and 0.200 mol of oxygen are sealed in a 2.00 dm³ vessel at 1000 K. At equilibrium the vessel contains 0.293 mol of sulfur trioxide. Calculate $K_{c}$ for $\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}$ .

Solution. The ratio of changes is 2 : 1 : 2.

	$\mathrm{2SO_2}$	$\mathrm{O_2}$	$\mathrm{2SO_3}$
Initial / mol	0.400	0.200	0
Change / mol	−0.293	−0.1465	+0.293
Equilibrium / mol	0.107	0.0535	0.293
Equilibrium / $\mathrm{mol\,dm^{-3}}$	0.0535	0.02675	0.1465

$K_c = \frac{[\mathrm{SO_3}]^2}{[\mathrm{SO_2}]^2[\mathrm{O_2}]} = \frac{(0.1465)^2}{(0.0535)^2 \times 0.02675} = 2.80 \times 10^{2}$

Units: $\frac{(\mathrm{mol\,dm^{-3}})^2}{(\mathrm{mol\,dm^{-3}})^3} = \mathrm{mol^{-1}\,dm^{3}}$ .

Answer: $K_c = 2.80 \times 10^{2}\,\mathrm{mol^{-1}\,dm^{3}}$ at 1000 K. Because $\Delta n \ne 0$ , the moles had to be converted to concentrations first.

Worked example

Problem (finding equilibrium amounts). 1.00 mol of ethanoic acid is mixed with 2.00 mol of ethanol and left to reach equilibrium, where $K_{c} = 4.0$ . Calculate the amount of ester formed and its percentage yield.

Solution. Let $x$ mol of ester form.

	acid	ethanol	ester	water
I / mol	1.00	2.00	0	0
C / mol	$- x$	$- x$	$+ x$	$+ x$
E / mol	$1.00 - x$	$2.00 - x$	$x$	$x$

The volume $V$ cancels because $\Delta n = 0$ :

$4.0 = \frac{x^2}{(1.00 - x)(2.00 - x)}$

$x^2 = 4.0\,(2.00 - 3.00x + x^2) \;\Rightarrow\; 3x^2 - 12x + 8 = 0$

$x = \frac{12 \pm \sqrt{144 - 96}}{6} = 0.845 \text{ or } 3.15$

The root 3.15 is impossible: it would use more acid than the 1.00 mol present. Check: $\frac{0.845^2}{0.155 \times 1.155} = 3.99 \approx 4.0$ .

Answer: 0.845 mol of ester, a yield of 84.5 % based on the acid, compared with 66.7 % when equal amounts are used. An excess of the cheaper reactant raises the yield of the dearer one.

Warning

Common mistakes in $K_{c}$ calculations

Using moles instead of concentrations when $\Delta n \ne 0$ . The volume cancels only when the powers on top and underneath add up to the same number.
Putting initial amounts into $K_{c}$ . Only equilibrium values go in.
Taking changes in the wrong ratio: in the $\mathrm{SO_3}$ example, oxygen falls by half as much as $\mathrm{SO_2}$ .
Including solids such as $\mathrm{CaCO_3}$ , or the solvent water in a dilute solution, in the expression.
Keeping both roots of a quadratic. Reject any root that makes an amount negative.

Tip

After any $K_{c}$ calculation, substitute your equilibrium values back into the expression. If you do not get $K_{c}$ back (to rounding), a change has the wrong sign or ratio. Then check the units: they come from $\Delta n$ alone.

Allery Chemistry: Calculating the number of moles at equilibrium from Kc. Watch how the initial moles, the changes in the ratio of the equation and the equilibrium moles are set out before substituting into Kc.

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Practice

Write $K_{c}$ , with its units, for: (a) $\mathrm{2NO_2(g) \rightleftharpoons N_2O_4(g)}$ ; (b) $\mathrm{PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)}$ ; (c) $\mathrm{C(s) + H_2O(g) \rightleftharpoons CO(g) + H_2(g)}$ ; (d) $\mathrm{Ag^{+}(aq) + 2NH_3(aq) \rightleftharpoons [Ag(NH_3)_2]^{+}(aq)}$ .
$K_{c} = 54.3$ at 430 °C for $\mathrm{H_2 + I_2 \rightleftharpoons 2HI}$ . Find $K_{c}$ at 430 °C for (a) $\mathrm{2HI \rightleftharpoons H_2 + I_2}$ ; (b) $\mathrm{HI \rightleftharpoons \tfrac{1}{2}H_2 + \tfrac{1}{2}I_2}$ .
Use the readings in Lesson 1's first worked example to calculate $K_{c}$ for $\mathrm{N_2O_4 \rightleftharpoons 2NO_2}$ at 60 °C.
0.100 mol of $\mathrm{PCl_5}$ is heated in a sealed 1.00 dm³ vessel. At equilibrium 0.042 mol of $\mathrm{Cl_2}$ is present. Calculate $K_{c}$ .
30.0 g of ethanoic acid ( $M = 60.05$ ) and 46.0 g of ethanol ( $M = 46.07$ ) reach equilibrium with $K_{c} = 4.0$ . Calculate the mass of ethyl ethanoate ( $M = 88.11$ ) formed.
A flask at 430 °C starts with $[\mathrm{H_2}] = 0.0300$ and $[\mathrm{I_2}] = 0.0100\,\mathrm{mol\,dm^{-3}}$ . Calculate all three equilibrium concentrations ( $K_{c} = 54.3$ ).
A student calculates $K_{c}$ for $\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}$ by putting equilibrium moles from a 2.00 dm³ vessel into the expression. Analyse the error: by what factor is the student's answer wrong, and is it too large or too small?

Show answersHide answers

(a) $\frac{[\mathrm{N_2O_4}]}{[\mathrm{NO_2}]^2}$ , $\mathrm{mol^{-1}\,dm^{3}}$ ; (b) $\frac{[\mathrm{PCl_3}][\mathrm{Cl_2}]}{[\mathrm{PCl_5}]}$ , $\mathrm{mol\,dm^{-3}}$ ; (c) $\frac{[\mathrm{CO}][\mathrm{H_2}]}{[\mathrm{H_2O}]}$ (carbon is a solid), $\mathrm{mol\,dm^{-3}}$ ; (d) $\frac{[\mathrm{Ag(NH_3)_2^{+}}]}{[\mathrm{Ag^{+}}][\mathrm{NH_3}]^2}$ , $\mathrm{mol^{-2}\,dm^{6}}$ .
(a) $\frac{1}{54.3} = 0.0184$ ; (b) $\frac{1}{\sqrt{54.3}} = 0.136$ .
$K_c = \frac{0.0360^2}{0.0220} = 0.0589\,\mathrm{mol\,dm^{-3}}$ .
At equilibrium: $\mathrm{PCl_5}$ 0.058, $\mathrm{PCl_3}$ 0.042, $\mathrm{Cl_2}$ 0.042 mol dm⁻³; $K_c = \frac{0.042^2}{0.058} = 0.030\,\mathrm{mol\,dm^{-3}}$ .
$n(\text{acid}) = 0.4996$ , $n(\text{ethanol}) = 0.9985$ ; $4.0 = \frac{x^2}{(0.4996 - x)(0.9985 - x)}$ gives $3 x^{2} - 5.992 x + 1.995 = 0$ , $x = 0.422\,\mathrm{mol}$ ; mass $= 0.422 \times 88.11 = 37.2\,\mathrm{g}$ .
$54.3 = \frac{(2x)^2}{(0.0300 - x)(0.0100 - x)}$ gives $50.3 x^{2} - 2.172 x + 0.01629 = 0$ , $x = 0.00966$ . $[\mathrm{H_2}] = 0.0203$ , $[\mathrm{I_2}] = 3.4 \times 10^{-4}$ , $[\mathrm{HI}] = 0.0193\,\mathrm{mol\,dm^{-3}}$ . The excess hydrogen uses up almost all the iodine.
Each concentration is $\frac{n}{V}$ , so $K_c = \frac{n(\mathrm{NH_3})^2}{n(\mathrm{N_2})\,n(\mathrm{H_2})^3} \times V^{2}$ . The student left out $V^{2} = 4.00$ , so the answer is 4 times too small.

Sources and credits

History: the esterification studies of M. Berthelot and L. Péan de Saint-Gilles (1862) and the law of mass action of C. M. Guldberg and P. Waage (1864), as described in Léon Péan de Saint-Gilles and Law of mass action, Wikipedia contributors, Wikipedia, licensed under CC BY-SA 4.0, adapted by Ganzaa.
Equilibrium constants: $K_{c} = 54.3$ (HI, 430 °C) and $2.8 \times 10^{2}$ ( $\mathrm{SO_3}$ , 1000 K) as quoted in Raymond Chang, Chemistry (McGraw-Hill); the NO and HCl values in Figure 2.1 were calculated by Ganzaa from standard Gibbs energies of formation in the CRC Handbook of Chemistry and Physics. Table 2.1 is illustrative.
Image: Guldberg und Waage 01.jpg, photogravure by H. Riffarth, Berlin, via Wikimedia Commons, public domain.
Video: Calculating the number of moles at equilibrium from Kc, Allery Chemistry, embedded from YouTube.
The text, scale diagram, tables, worked examples and Practice items were written and drawn by Ganzaa.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1Why does the ester stop at two-thirds?

2Writing the expression for Kc K_c Kc​

3What the size of Kc K_c Kc​ tells you

Changing the equation changes K K K

4Solving equilibrium problems with an ICE table

Why does the ester stop at two-thirds?

Writing the expression for $K_{c}$

What the size of $K_{c}$ tells you

Changing the equation changes $K$

Solving equilibrium problems with an ICE table