Chemical Equilibrium

Lessons

Senior 6 · Chemistry

Chemical Equilibrium

5 lessons

Lesson 1 of 5

Reversible Reactions and Dynamic Equilibrium

25 min

Why does a lime kiln need a way out for its gas?

Quicklime, calcium oxide, is made by heating limestone in a kiln:

$\mathrm{CaCO_3(s) \longrightarrow CaO(s) + CO_2(g)} \qquad \Delta H = +178\,\mathrm{kJ\,mol^{-1}}$

Heat limestone to about 900 °C in a sealed steel box and something surprising happens: the decomposition runs only until the carbon dioxide in the box reaches a pressure of 1 atm, and then it stops. Part of the limestone stays unchanged however long you wait. At 1000 °C the limit is higher, 3.8 atm, but there is still a limit. In a kiln the carbon dioxide is carried away with the hot furnace gases, its pressure never builds up, and every lump of limestone is converted.

So the same reaction is "complete" in one container and "incomplete" in another. The reason is that carbon dioxide and calcium oxide also react with each other, re-forming calcium carbonate. This lesson explains what happens when a reaction can run both ways, and why, in a closed container, it settles into a state called dynamic equilibrium.

Definition

Reversible reaction: a reaction in which the products can react to re-form the reactants, so it can go in either direction. It is written with the sign $\rightleftharpoons$ . Example: $\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}$ .

Irreversible reaction: a reaction that, in practice, goes only forwards, either because a product escapes or because so little of the reactants can re-form that none can be detected. Example: $\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)}$ .

Closed system: a system that no matter can enter or leave, although energy (heat) can. Example: a stoppered flask or a sealed steel vessel.

Dynamic equilibrium: the state of a reversible reaction in a closed system in which the forward and backward reactions go on at equal rates, so the concentrations of all the reactants and products stay constant.

Position of equilibrium: the proportions of reactants and products present at equilibrium. "The equilibrium lies to the right" means the mixture is mainly products.

Homogeneous equilibrium: one in which every species is in the same phase, e.g. $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$ . Heterogeneous equilibrium: one involving more than one phase, e.g. $\mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)}$ .

Reversible and irreversible reactions in the laboratory

Table 1.1 lists reactions you have met before. Notice what decides whether a reaction "comes back".

Reaction	What you observe	Type
$\mathrm{CuSO_4{\cdot}5H_2O(s) \rightleftharpoons CuSO_4(s) + 5H_2O(g)}$	Blue crystals turn white on heating; adding water to the white powder turns it blue again, and it gets hot.	reversible
$\mathrm{NH_4Cl(s) \rightleftharpoons NH_3(g) + HCl(g)}$	Heated in a test tube, the white solid disappears from the hot bottom and re-forms on the cool upper wall.	reversible
$\mathrm{Fe^{3+}(aq) + SCN^{-}(aq) \rightleftharpoons [Fe(SCN)]^{2+}(aq)}$	A blood-red colour forms; adding more of either ion makes it deeper, adding a reagent that removes $\mathrm{Fe^{3+}}$ makes it fade.	reversible, homogeneous
$\mathrm{Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)}$	In an open beaker the hydrogen bubbles away; the magnesium is used up completely.	irreversible (open system)
$\mathrm{2Mg(s) + O_2(g) \rightarrow 2MgO(s)}$	A white flame and white ash; the oxide does not decompose back on cooling.	irreversible (equilibrium lies very far to the right)

Every reaction is reversible in principle. We call a reaction irreversible when a product leaves the system, or when the equilibrium lies so far to one side that the reverse reaction cannot be detected. The limestone example shows that one reaction can be either: reversible in a sealed box, irreversible in an open kiln.

Approaching equilibrium from both sides

Hydrogen and iodine vapour react at 430 °C to form hydrogen iodide, a colourless gas: $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$ . Two sealed flasks are kept at 430 °C.

Flask A starts with $0.0100\,\mathrm{mol\,dm^{-3}}$ each of hydrogen and iodine. The violet colour of the iodine vapour fades quickly at first, then more slowly, and then stops fading.
Flask B starts with $0.0200\,\mathrm{mol\,dm^{-3}}$ of hydrogen iodide alone. A violet colour appears and deepens, then stops changing.

Analysis of the two flasks gives the same final mixture: $[\mathrm{H_2}] = [\mathrm{I_2}] = 0.0021$ and $[\mathrm{HI}] = 0.0157\,\mathrm{mol\,dm^{-3}}$ . Both flasks contain the same numbers of hydrogen and iodine atoms per cubic decimetre, so whichever side you start from, the system ends in the same equilibrium state (Figure 1.1).

Figure 1.1: Concentration against time for $\mathrm{H_2 + I_2 \rightleftharpoons 2HI}$ at 430 °C, started from each side. The final concentrations are the same.

Why does the change stop? At the start of flask A only the forward reaction can happen, and it is fast because $[\mathrm{H_2}]$ and $[\mathrm{I_2}]$ are high. As they fall, the forward rate falls. Meanwhile HI builds up, so the backward reaction, $\mathrm{2HI \rightarrow H_2 + I_2}$ , speeds up. When the two rates become equal, HI is made exactly as fast as it is used up, and no concentration changes any more (Figure 1.2). Both reactions are still running: that is why the equilibrium is called dynamic.

Figure 1.2: Rates of the forward and backward reactions against time. They become equal at equilibrium; neither falls to zero.

Characteristics of dynamic equilibrium

It is reached only in a closed system.
The forward and backward reactions go on at equal rates; neither rate is zero.
The concentrations of all species are constant, but they are not usually equal to each other.
Observable (macroscopic) properties such as colour, pressure and density stay constant.
The same equilibrium can be reached from either side, as in flasks A and B.
A catalyst speeds up the forward and backward reactions equally. It shortens the time taken to reach equilibrium but does not change the equilibrium mixture.

Evidence that the equilibrium is dynamic

Constant concentrations alone could mean that the reaction had simply stopped. Radioactive tracers show that it has not. Add a trace of iodine labelled with radioactive iodine-131 to an equilibrium mixture of $\mathrm{H_2}$ , $\mathrm{I_2}$ and HI: the radioactivity soon appears in the hydrogen iodide as well, although the concentrations stay practically the same. Iodine atoms are still passing into HI and back out again.

Physical equilibria

Equilibria are not only chemical. Water in a sealed bottle reaches $\mathrm{H_2O(l) \rightleftharpoons H_2O(g)}$ , with evaporation and condensation at equal rates. A saturated salt solution in contact with undissolved crystals reaches $\mathrm{NaCl(s) \rightleftharpoons Na^{+}(aq) + Cl^{-}(aq)}$ : a crystal with a chipped corner slowly changes shape while its mass stays the same. The partition of a solute between two immiscible solvents (Unit 6) is another physical equilibrium.

Worked example

Problem. Dinitrogen tetroxide, a colourless gas, dissociates into brown nitrogen dioxide: $\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}$ . A sealed flask at 60 °C initially contains $\mathrm{N_2O_4}$ only. These are illustrative readings:

time / s	0	20	40	60	80	100
$[\mathrm{N_2O_4}]$ / $\mathrm{mol\,dm^{-3}}$	0.0400	0.0300	0.0240	0.0220	0.0220	0.0220

(a) When is equilibrium reached? (b) Calculate $[\mathrm{NO_2}]$ at equilibrium. (c) What is happening to the molecules after 60 s?

Solution.

(a) The concentration stops changing at 60 s, so equilibrium is reached by 60 s (it may have been reached a little earlier; the readings cannot show exactly when).

(b) $\mathrm{N_2O_4}$ used up $= 0.0400 - 0.0220 = 0.0180\,\mathrm{mol\,dm^{-3}}$ . Each $\mathrm{N_2O_4}$ gives two $\mathrm{NO_2}$ :

$[\mathrm{NO_2}]_{\text{eq}} = 2 \times 0.0180 = 0.0360\,\mathrm{mol\,dm^{-3}}$

(c) $\mathrm{N_2O_4}$ molecules are still dissociating and $\mathrm{NO_2}$ molecules are still combining, at equal rates, so the brown colour stays constant.

Answer: equilibrium by 60 s; $[\mathrm{NO_2}] = 0.0360\,\mathrm{mol\,dm^{-3}}$ . Note that $[\mathrm{NO_2}] \ne [\mathrm{N_2O_4}]$ at equilibrium: constant does not mean equal.

Worked example

Problem. 50.0 g of calcium carbonate is sealed in an evacuated 10.0 dm³ steel vessel and heated to 1000 °C, where the equilibrium pressure of carbon dioxide over $\mathrm{CaCO_3}$ and CaO is 3.8 atm. (a) What mass of $\mathrm{CaCO_3}$ decomposes? (b) What would happen with 20.0 g instead? Ignore the volume of the solids; $R = 8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}$ , 1 atm = 101 325 Pa, $M(\mathrm{CaCO_3}) = 100.09\,\mathrm{g\,mol^{-1}}$ .

Solution.

(a) Amount of $\mathrm{CO_2}$ needed to reach 3.8 atm, from $p V = n R T$ :

$n(\mathrm{CO_2}) = \frac{pV}{RT} = \frac{(3.8 \times 101\,325\,\mathrm{Pa}) \times (1.00 \times 10^{-2}\,\mathrm{m^3})}{8.314\,\mathrm{J\,K^{-1}\,mol^{-1}} \times 1273\,\mathrm{K}} = 0.364\,\mathrm{mol}$

One mole of $\mathrm{CaCO_3}$ gives one mole of $\mathrm{CO_2}$ , so mass decomposed $= 0.364 \times 100.09 = 36\,\mathrm{g}$ (2 s.f., because 3.8 atm has 2 s.f.). About 14 g of limestone remains, in equilibrium with CaO and the gas.

(b) 20.0 g is only $\frac{20.0}{100.09} = 0.200\,\mathrm{mol}$ . Even if all of it decomposed, the pressure would be

$p = \frac{nRT}{V} = \frac{0.200 \times 8.314 \times 1273}{1.00 \times 10^{-2}} = 2.1 \times 10^{5}\,\mathrm{Pa} = 2.1\,\mathrm{atm}$

This is below 3.8 atm, so the equilibrium pressure is never reached.

Answer: (a) 36 g decomposes. (b) All 20.0 g decomposes: a heterogeneous equilibrium exists only while some of every phase, here solid $\mathrm{CaCO_3}$ , is present.

Warning

Three common mistakes

"At equilibrium the reaction stops." No: both reactions continue at equal rates.
"At equilibrium the concentrations of reactants and products are equal." No: they are constant. In Figure 1.1, $[\mathrm{HI}]$ is more than seven times $[\mathrm{H_2}]$ .
"A catalyst gives more product at equilibrium." No: it speeds up both directions equally, so the mixture is reached sooner but is the same mixture.

Tip

When a question asks you to describe dynamic equilibrium, an examiner looks for three points: closed system; rate of forward reaction = rate of backward reaction; concentrations constant. Learn them as one sentence: "In a closed system, the forward and backward rates are equal, so the concentrations stay constant."

University of Surrey: Dynamic Equilibrium. The animation shows water molecules evaporating and condensing in a closed container. Watch how the two rates become equal, so that the amount of vapour stays constant while molecules keep moving both ways.

✦

Practice

State three characteristics of a system in dynamic equilibrium.
Classify each equilibrium as homogeneous or heterogeneous: (a) $\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$ ; (b) $\mathrm{H_2O(l) \rightleftharpoons H_2O(g)}$ ; (c) $\mathrm{Fe^{3+}(aq) + SCN^{-}(aq) \rightleftharpoons [Fe(SCN)]^{2+}(aq)}$ ; (d) $\mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)}$ .
Explain why magnesium reacts completely with excess dilute hydrochloric acid in an open beaker.
Sketch, and describe in words, the concentration–time graph for $\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}$ started from pure $\mathrm{NO_2}$ .
A little solid sodium chloride containing radioactive sodium-24 is added to a saturated solution of ordinary sodium chloride. An hour later the solution is radioactive, but the mass of undissolved solid is unchanged. Explain.
2.00 g of $\mathrm{CaCO_3}$ is heated to 900 °C in a sealed, evacuated 5.00 dm³ vessel, where the equilibrium pressure of $\mathrm{CO_2}$ is 1.0 atm. Show that all of it decomposes. What mass decomposes if 10.0 g is used instead?
A student writes: "The colour of the flask stopped changing, so the reaction has finished." Evaluate this statement.

Show answersHide answers

Any three: closed system; forward and backward rates equal and not zero; concentrations constant (not necessarily equal); macroscopic properties constant; can be reached from either side.
(a) homogeneous; (b) heterogeneous (liquid and gas); (c) homogeneous (all aqueous); (d) heterogeneous.
The hydrogen escapes from the open beaker, so it cannot react back; the system is open and the reaction goes to completion.
$[\mathrm{NO_2}]$ starts high and falls, quickly at first, then levels off; $[\mathrm{N_2O_4}]$ starts at zero and rises, levelling off at the same moment. The fall in $[\mathrm{NO_2}]$ is twice the rise in $[\mathrm{N_2O_4}]$ .
Dissolving and crystallising go on at equal rates. Radioactive ions dissolve from the solid while ordinary ions crystallise onto it, so the label spreads into the solution but the mass of solid stays constant: the equilibrium is dynamic.
$n = \frac{2.00}{100.09} = 0.0200\,\mathrm{mol}$ ; if all decomposes, $p = \frac{0.0200 \times 8.314 \times 1173}{5.00 \times 10^{-3}} = 3.9 \times 10^{4}\,\mathrm{Pa} = 0.38\,\mathrm{atm}$ , below 1.0 atm, so all decomposes. With 10.0 g, complete decomposition would give 1.9 atm, so equilibrium is reached: $n = \frac{101\,325 \times 5.00 \times 10^{-3}}{8.314 \times 1173} = 0.0519\,\mathrm{mol}$ , so 5.2 g decomposes and 4.8 g remains.
Not correct. A constant colour shows that the concentrations are constant, which fits equilibrium as well as a finished reaction. If the reaction is reversible and the flask is closed, both reactions are still running at equal rates; only a tracer experiment could show this directly.

Sources and credits

Lime-burning temperatures and pressures: Lime kiln, Wikipedia contributors, Wikipedia, licensed under CC BY-SA 4.0, adapted by Ganzaa.
Equilibrium constant for $\mathrm{H_2 + I_2 \rightleftharpoons 2HI}$ used for Figure 1.1: $K_{c} = 54.3$ at 430 °C, the value quoted in Raymond Chang, Chemistry (McGraw-Hill). The readings in the first worked example are illustrative.
Image: 2017 Wapiennik w Mielniku 2.jpg by Jacek Halicki, via Wikimedia Commons, licensed under CC BY-SA 4.0.
Video: Dynamic Equilibrium, University of Surrey, embedded from YouTube.
The text, graphs, table, 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 a lime kiln need a way out for its gas?

2Reversible and irreversible reactions in the laboratory

3Approaching equilibrium from both sides

4Characteristics of dynamic equilibrium