School Chemistry — Grades 1 to 12 · Grades 1–12

43Chemical Equilibrium: Quotient and Constant

A sealed bottle of sparkling water keeps its fizz for months; opened, it goes flat in an afternoon. Nothing has gone wrong in either case. In the closed bottle, carbon dioxide leaves the water and returns to it at the same rate, and the composition stays put; open the cap, and the gas that leaves no longer comes back. Many reactions behave like this: they stop before any reactant is used up, at a balance between two opposite reactions. This chapter describes that balance with one number.

You already know

The extent $x$ of a reaction, its maximum $x_{\max}$ , and the total reactions that reach it (Chapter 27). Rates and kinetic factors (Chapter 41); a catalyst does not change the final state (Chapter 42).

Sealed, the water keeps its gas; opened, the gas escapes and does not come back.

43.1 Reactions that are not total

Definition 43.1 (Non-total reaction, final extent ratio)

A reaction is a non-total reaction if it stops at a final extent $x_f$ smaller than its maximum extent $x_{\max}$ , with all the reactants still present. Its final extent ratio is

\tau = \frac{x_f}{x_{\max}}, \qquad 0 < \tau < 1

( $\tau = 1$ for a total reaction).

Example 43.2 (An ester that stops halfway)

Heated with $1\,\mathrm{mol}$ of ethanol, $1\,\mathrm{mol}$ of ethanoic acid forms ethyl ethanoate and water, $\ce{CH3COOH + C2H5OH <=> CH3COOC2H5 + H2O}$ . The reaction stops when about two thirds of the acid have reacted: $x_f = 0.67\,\mathrm{mol}$ for $x_{\max} = 1\,\mathrm{mol}$ , so $\tau \approx 0.67$ . Starting from $1\,\mathrm{mol}$ of ester and $1\,\mathrm{mol}$ of water, the reverse reaction also stops, with about one third of the ester hydrolysed: the same final mixture.

History — Berthelot and Péan de Saint-Gilles, 1862

The chemists Marcellin Berthelot and Léon Péan de Saint-Gilles heated mixtures of ethanoic acid and ethanol, and of ethyl ethanoate and water, for long periods, and analysed them. From equal amounts of acid and alcohol, the reaction always stopped at about two thirds; from ester and water, at one third of hydrolysis; and they noticed that the amount of ester formed at each moment depended on the product of the amounts of the reactants. Their work was one of the first quantitative studies of a chemical equilibrium.

43.2 The equilibrium state

Definition 43.3 (Equilibrium state)

A system is in an equilibrium state when its composition no longer changes although both the forward and the reverse reactions are still going on, at the same rate. A reaction that can reach such a state is written with the double arrow $\rightleftharpoons$ .

The esterification (blue) and the hydrolysis of the ester (orange) reach the same equilibrium composition, from opposite sides. (The time scale is a model; the final value is the measured one.)

43.3 The reaction quotient

Notation 43.4 (Standard concentration)

$c^\circ = 1\,\mathrm{mol}/\mathrm{L}$ is the standard concentration. Dividing a concentration by $c^\circ$ gives a number without unit with the same value: $[\ce{A}]/c^\circ$ .

Definition 43.5 (Reaction quotient)

For a reaction in solution $a\,\ce{A} + b\,\ce{B} \rightleftharpoons c\,\ce{C} + d\,\ce{D}$ , the reaction quotient in a given state of the system is

Q = \frac{\left([\ce{C}]/c^\circ\right)^c \left([\ce{D}]/c^\circ\right)^d} {\left([\ce{A}]/c^\circ\right)^a \left([\ce{B}]/c^\circ\right)^b} .

Solids, and the solvent (water in an aqueous solution), do not appear in $Q$ .

Example 43.6 (Writing quotients)

For $\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$ in water (the ion $\ce{H3O+}$ is introduced in the next chapter), water is the solvent: $Q = \dfrac{[\ce{CH3COO-}]\,[\ce{H3O+}]}{[\ce{CH3COOH}]\,c^\circ}$ . For the dissolving of a solid, $\ce{AgCl(s) <=> Ag+ + Cl-}$ , the solid is left out: $Q = [\ce{Ag+}][\ce{Cl-}]/(c^\circ)^2$ . For the esterification, where all four species are in the same liquid mixture of volume $V$ , the volumes cancel and $Q = \dfrac{n_{\text{ester}}\, n_{\text{water}}}{n_{\text{acid}}\, n_{\text{alcohol}}}$ .

43.4 The equilibrium constant

Definition 43.7 (Equilibrium constant)

The value $Q_{eq}$ of the reaction quotient in the equilibrium state is the equilibrium constant $K$ of the reaction. It has no unit.

Proposition 43.8 ( $K$ depends only on the temperature)

For a given reaction, $Q_{eq} = K$ in every equilibrium state at a given temperature, whatever the initial amounts.

Proof. Admitted; it is derived in the Year 2 volume. It is checked on the esterification: both runs of the figure, started from opposite sides, give the same $Q_{eq}$ . ∎

Example 43.9 ( $K$ of the esterification)

From $1\,\mathrm{mol}$ of acid and $1\,\mathrm{mol}$ of alcohol, the equilibrium holds $\frac{2}{3}$ mol of ester and of water and $\frac{1}{3}$ mol of acid and of alcohol:

K = \frac{(2/3)\times(2/3)}{(1/3)\times(1/3)} = 4 .

Method 43.10 (Final state of a non-total reaction)

Fill the progress table with the unknown final extent $x_f$ .
Write $Q_{eq}$ as a function of $x_f$ and set it equal to $K$ .
Solve for $x_f$ (often a quadratic equation), keeping the root between 0 and $x_{\max}$ .
Deduce the final amounts and $\tau = x_f/x_{\max}$ .

Example 43.11 (An excess of alcohol)

From $1\,\mathrm{mol}$ of acid and $2\,\mathrm{mol}$ of alcohol: $\dfrac{x^2}{(1-x)(2-x)} = 4$ , that is $3x^2 - 12x + 8 = 0$ , whose root between 0 and 1 is $x_f = 0.845\,\mathrm{mol}$ . Doubling the alcohol raises the yield, counted on the acid, from $67\,\%$ to $85\,\%$ .

43.5 Predicting and shifting the evolution

Proposition 43.12 (Direction of evolution)

If $Q < K$ , the system evolves in the forward direction (the products increase) until $Q = K$ ; if $Q > K$ , it evolves in the reverse direction; if $Q = K$ , it is in equilibrium and does not change.

Proof. Admitted: $Q$ increases when the forward reaction proceeds (products in the numerator grow, reactants in the denominator shrink), and the system moves towards the state where $Q = K$ . ∎

The quotient always moves towards the constant.

Remark 43.13 (Shifting an equilibrium)

To obtain more product from a non-total reaction, one can add an excess of one reactant (the cheaper one), or remove a product as it forms: each makes $Q < K$ again, and the system moves forward. Removing the water of an esterification (with a drying agent, or by distilling it off) can drive the reaction almost to completion. The open bottle of sparkling water does the same with the carbon dioxide that escapes.

Stalactites grow where water dripping in a cave loses carbon dioxide to the air: an equilibrium of dissolved limestone is shifted, and calcium carbonate is deposited.

43.6 Exercises

Exercise 43.1 ★

Write the reaction quotient of: (a) $\ce{N2 + 3H2 <=> 2NH3}$ (gases, use concentrations); (b) $\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$ in water; (c) $\ce{CaCO3(s) <=> Ca^{2+} + CO3^{2-}}$ ; (d) $\ce{Fe^{3+} + SCN- <=> FeSCN^{2+}}$ .

Solution

Solution of Exercise 43.1.

(a) $Q = \dfrac{[\ce{NH3}]^2 (c^\circ)^2}{[\ce{N2}][\ce{H2}]^3}$ ; (b) $Q = \dfrac{[\ce{CH3COO-}][\ce{H3O+}]}{[\ce{CH3COOH}]\,c^\circ}$ (water, the solvent, is left out); (c) $Q = [\ce{Ca^{2+}}][\ce{CO3^{2-}}]/(c^\circ)^2$ (the solid is left out); (d) $Q = \dfrac{[\ce{FeSCN^{2+}}]\,c^\circ}{[\ce{Fe^{3+}}][\ce{SCN-}]}$ .

Exercise 43.2 ★

A reaction has $x_{\max} = 0.050\,\mathrm{mol}$ and reaches $x_f = 0.020\,\mathrm{mol}$ . Compute $\tau$ . Is the reaction total?

Solution

Solution of Exercise 43.2.

$\tau = 0.020/0.050 = 0.40$ : not total.

Exercise 43.3 ★

For a reaction with $K = 10$ , in which direction does a mixture evolve if $Q = 2$ ? If $Q = 50$ ? If $Q = 10$ ?

Solution

Solution of Exercise 43.3.

$Q = 2 < K$ : forward. $Q = 50 > K$ : reverse. $Q = K$ : no change.

Exercise 43.4 ★

Explain why an equilibrium is said to be “dynamic”.

Solution

Solution of Exercise 43.4.

The composition is constant, but the forward and reverse reactions go on at the same rate: molecules keep reacting both ways.

Exercise 43.5 ★

Does a catalyst change $K$ ? The time needed to reach equilibrium? Explain.

Solution

Solution of Exercise 43.5.

No, $K$ is unchanged (the final state is the same); the time to reach equilibrium is shorter.

Exercise 43.6 ★★

$2.0\,\mathrm{mol}$ of ethanoic acid and $2.0\,\mathrm{mol}$ of ethanol are mixed ( $K = 4$ ). Compute the final amounts of the four species.

Solution

Solution of Exercise 43.6.

$\dfrac{x^2}{(2-x)^2} = 4$ , so $\dfrac{x}{2-x} = 2$ and $x = 1.33\,\mathrm{mol}$ : $0.67\,\mathrm{mol}$ of acid and of alcohol, $1.33\,\mathrm{mol}$ of ester and of water (again two thirds).

Exercise 43.7 ★★

A mixture contains $0.5\,\mathrm{mol}$ each of ethanoic acid, ethanol, ethyl ethanoate and water. Compute $Q$ . In which direction does it evolve?

Solution

Solution of Exercise 43.7.

$Q = (0.5 \times 0.5)/(0.5 \times 0.5) = 1 < 4$ : forward, more ester forms.

Exercise 43.8 ★★

Using the figure of the two runs, read the amount of ester after $20\,\mathrm{h}$ in each run. When can the system be considered at equilibrium?

Solution

Solution of Exercise 43.8.

About $0.52\,\mathrm{mol}$ from the acid side and $0.79\,\mathrm{mol}$ from the ester side. After about $100\,\mathrm{h}$ both curves are on the dashed line: equilibrium.

Exercise 43.9 ★★

Starting from $1\,\mathrm{mol}$ of ester and $1\,\mathrm{mol}$ of water ( $K = 4$ for the esterification, so $1/4$ for the hydrolysis), compute the final amounts by the method of the chapter, and compare with the esterification run.

Solution

Solution of Exercise 43.9.

With $y$ mol hydrolysed: $\dfrac{y^2}{(1-y)^2} = \dfrac{1}{4}$ , so $\dfrac{y}{1-y} = \dfrac{1}{2}$ , $y = \frac{1}{3}$ : $\frac{2}{3}$ mol of ester and of water, $\frac{1}{3}$ mol of acid and of alcohol. The same final mixture as the esterification.

Exercise 43.10 ★★

Why does the solid not appear in the quotient of $\ce{AgCl(s) <=> Ag+ + Cl-}$ ? Does adding more solid shift the equilibrium?

Solution

Solution of Exercise 43.10.

A solid is a pure phase, whose “concentration” does not change: it is left out of $Q$ . Adding more solid does not change $Q$ , so it does not shift the equilibrium (as long as some solid is present).

Exercise 43.11 ★★

In a closed bottle of sparkling water, carbon dioxide leaves and enters the water at the same rate. Write the equilibrium $\ce{CO2(aq) <=> CO2(g)}$ and explain, with $Q$ and $K$ , why the water goes flat when the bottle is opened.

Solution

Solution of Exercise 43.11.

$K$ fixes the ratio of carbon dioxide in the gas to that in the water. Opened, the gas above the water escapes into the air: the carbon dioxide in the gas falls, $Q < K$ , and more gas leaves the water, until almost none is left.

Exercise 43.12 ★★★

An esterification of $1\,\mathrm{mol}$ of acid and $1\,\mathrm{mol}$ of alcohol reaches equilibrium. Half of the water present is then removed. Compute the new $Q$ , say in which direction the system evolves, and find the new final amount of ester.

Solution

Solution of Exercise 43.12.

At equilibrium: acid and alcohol $\frac{1}{3}$ , ester and water $\frac{2}{3}$ . Removing half of the water leaves $\frac{1}{3}$ : $Q = \dfrac{(2/3)(1/3)}{(1/3)(1/3)} = 2 < 4$ , forward. With $y$ more reacting: $(2/3 + y)(1/3 + y) = 4(1/3 - y)^2$ , that is $27y^2 - 33y + 2 = 0$ , $y = 0.064$ : the ester rises to $0.73\,\mathrm{mol}$ .

Exercise 43.13 ★★★

Show that the constant of the reverse reaction is $1/K$ . What is the constant of the hydrolysis of ethyl ethanoate?

Solution

Solution of Exercise 43.13.

The quotient of the reverse reaction has numerator and denominator exchanged: $Q' = 1/Q$ , so at equilibrium $K' = 1/K$ . Hydrolysis: $1/4 = 0.25$ .

Exercise 43.14 ★★★

A chemist mixes $1\,\mathrm{mol}$ of acid, $1\,\mathrm{mol}$ of alcohol, $2\,\mathrm{mol}$ of ester and $0.5\,\mathrm{mol}$ of water. Will anything happen? Explain.

Solution

Solution of Exercise 43.14.

$Q = (2 \times 2)/(1 \times 1) = 4 = K$ : the mixture is already in equilibrium; its composition does not change.

Exercise 43.15 ★★★

What excess of alcohol (amount per mole of acid) is needed to esterify $95\,\%$ of the acid? Solve $\dfrac{0.95^2}{0.05(n - 0.95)} = 4$ .

Solution

Solution of Exercise 43.15.

$n - 0.95 = 0.9025 / (4 \times 0.05) = 4.51$ , so $n = 5.5\,\mathrm{mol}$ of alcohol per mole of acid.

43.7 Problem: Berthelot’s Ester

Problem 43.1

Weekend problem — how much ester when the alcohol is put in threefold excess?

Ethyl ethanoate, a solvent and a flavouring, is made from ethanoic acid and ethanol. In 1862 Berthelot and Péan de Saint-Gilles found that from equal amounts of acid and alcohol the reaction stops when about two thirds of the acid have reacted.

Part I — The reaction.

Write the equation of the esterification, with semi-structural formulas.
Why is it written with a double arrow?
Write its reaction quotient in amounts.
From $1\,\mathrm{mol}$ of acid and $1\,\mathrm{mol}$ of alcohol, what is $x_{\max}$ ? What is $x_f$ , from Berthelot’s result?
Compute $\tau$ .

Part II — The constant.

Give the final amounts of the four species.
Compute $K$ .
Starting from $1\,\mathrm{mol}$ of ester and $1\,\mathrm{mol}$ of water, Berthelot found one third hydrolysed. Show that this agrees with the same $K$ .
Why must $K$ be the same in both cases?

Part III — A threefold excess of alcohol.

From $1\,\mathrm{mol}$ of acid and $3\,\mathrm{mol}$ of alcohol, fill the progress table.
Write the equation $Q_{eq} = K$ and show that it becomes $3x^2 - 16x + 12 = 0$ .
Solve it, and keep the meaningful root.
What fraction of the acid is esterified?
Compute the mass of ester obtained.

Part IV — Removing water.

From the equimolar equilibrium, the water is removed as it forms. Explain with $Q$ why the reaction goes on.
If all the water could be removed, what would the yield become?
Why is the alcohol, not the acid, usually put in excess? (Think of the cost and of separating the products.)
Does heating the mixture change the final state, or only the time to reach it? (The esterification gives out almost no heat.)
State the final answer: what fraction of the acid is esterified with a threefold excess of alcohol?

Solution

Solution of Problem 43.1.

1. $\ce{CH3-COOH + CH3-CH2-OH <=> CH3-COO-CH2-CH3 + H2O}$ .

2. It is not total: it stops in an equilibrium where both directions go on.

3. $Q = \dfrac{n_{\text{ester}}\, n_{\text{water}}}{n_{\text{acid}}\, n_{\text{alcohol}}}$ .

4. $x_{\max} = 1\,\mathrm{mol}$ ; $x_f = \frac{2}{3}$ mol.

5. $\tau = 0.67$ .

6. Acid and alcohol $\frac{1}{3}$ mol each, ester and water $\frac{2}{3}$ mol each.

7. $K = \dfrac{(2/3)^2}{(1/3)^2} = 4$ .

8. One third hydrolysed: ester and water $\frac{2}{3}$ , acid and alcohol $\frac{1}{3}$ : the same mixture, $Q_{eq} = 4$ .

9. It is the same reaction at the same temperature, and $K$ depends only on the temperature.

10. Acid $1 - x$ ; alcohol $3 - x$ ; ester $x$ ; water $x$ .

11. $x^2 = 4(1 - x)(3 - x) = 4(3 - 4x + x^2)$ , so $3x^2 - 16x + 12 = 0$ .

12. $x = (16 \pm \sqrt{256 - 144})/6 = (16 \pm 10.58)/6$ : 0.903 or 4.43; only $x_f = 0.903\,\mathrm{mol}$ lies between 0 and 1.

13. $90\,\%$ .

14. $M = 88.0\,\mathrm{g}/\mathrm{mol}$ : $0.903 \times 88.0 = 79.5\,\mathrm{g}$ .

15. Each removal of water lowers the numerator of $Q$ , which falls below $K$ : the system moves forward again.

16. The reaction would go on until the acid is used up: $100\,\%$ .

17. Ethanol is cheaper, and its excess is easily separated from the ester (and recycled).

18. Only the time: a reaction that gives out almost no heat has a constant that hardly changes with temperature, and heating only makes it reach equilibrium sooner.

19. About $90\,\%$ of the acid.

Terms defined in this chapter

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