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Adding or removing a substance moves the reaction quotient away from a fixed constant, and the net reaction that follows partly undoes the change.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to predict which way an equilibrium shifts when a reactant or product is added or removed, explain the prediction by comparing the reaction quotient with the equilibrium constant, and calculate the new equilibrium mixture for a simple case. You will also be able to say why the constant itself does not change and why the shift only partly undoes the disturbance.
You can write $K_c$ for a reaction, calculate it from an ICE table, and compare a reaction quotient $Q_c$ with it: if $Q_c$ is less than $K_c$ the forward reaction runs net, and if it is greater the reverse reaction does. Everything in this lesson is that comparison, applied to a mixture that was at equilibrium until somebody interfered with it.
The position of equilibrium is the actual mixture — how much of each substance there is. It is different from the equilibrium constant, which is the fixed value that position has to satisfy at a given temperature. A disturbance is anything done to the mixture: adding or removing a substance, changing the pressure, changing the temperature. A shift to the right means net forward reaction until equilibrium is restored; a shift to the left means net reverse reaction. Le Chatelier's principle summarises the result: a system at equilibrium responds to a disturbance in the direction that partly undoes it.
Take $\mathrm{A \rightleftharpoons B}$ with $K_c = 2$: at equilibrium there is always twice as much B as A. Suppose the flask holds 10 mmol of A and 20 mmol of B, and 15 mmol more A is poured in.
Immediately afterwards there are 25 mmol of A and 20 mmol of B. The quotient is $20 \div 25 = 0.8$, well below 2. The constant has not changed — nothing about the temperature did — so the mixture is simply not at equilibrium any more. The forward reaction now outruns the reverse, turning A into B, until the ratio is 2 again.
How far does it go? If $x$ mmol of A becomes B, then $(20 + x) \div (25 - x) = 2$, so $x = 10$. The new equilibrium has 15 mmol of A and 30 mmol of B.
Look at what happened to the A. It went up by 15 when it was added and then down by 10 as the system responded: the response partly undid the disturbance, and the new mixture still has more A than the old one. That is exactly what Le Chatelier's principle says, and now it has a reason attached.
The rules that follow:
Adding a substance, removing one or squeezing the gas moves the position of an equilibrium and leaves the equilibrium constant exactly where it was. Only a change of temperature changes the constant itself.
Another way: picture
Picture a see-saw with a spring that always pulls it back to the same angle. Put a weight on one end and it tips; the spring then pulls it most of the way back, but not all the way, because the weight is still there. The angle the spring returns it towards is the constant; the tilt you can see is the position.
Another way: steps
To predict a shift after a change of concentration:
Iron(III) ions, which are pale yellow, and thiocyanate ions, which are colourless, form a deep blood-red complex:
$$\mathrm{Fe^{3+}(aq) + SCN^-(aq) \rightleftharpoons FeSCN^{2+}(aq)}$$
Because only the complex is strongly coloured, the colour is a direct reading of the position.
| Disturbance | Effect on $Q_c$ | Net reaction | Colour |
|---|---|---|---|
| iron(III) chloride added | falls | forward | deeper red |
| potassium thiocyanate added | falls | forward | deeper red |
| hydroxide added, removing $\mathrm{Fe^{3+}}$ as a precipitate | rises | reverse | paler |
| silver ions added, removing $\mathrm{SCN^-}$ as a precipitate | rises | reverse | paler |
The last two rows are worth noticing. Nothing was added to the equilibrium itself; something was added that took a reactant out, and removing a reactant pushes the position left just as surely as adding a product does.
In the Haber process ammonia is cooled out of the gas mixture as a liquid as soon as it leaves the converter, and the unreacted nitrogen and hydrogen go back round. Removing the product keeps the quotient below the constant, so the forward reaction never gets the chance to balance the reverse, and over many passes nearly all the nitrogen ends up as ammonia.
The same idea explains a fizzy drink going flat: the carbon dioxide dissolved in a sealed bottle is at equilibrium with the gas above it, and opening the cap lets the gas escape. With a product of the dissolving removed, more gas comes out of solution — the fizz. Neither case changes any constant; both change a concentration and let the quotient do the rest.
"Equilibrium moves towards the products." It has no favourite side. It moves towards the products after a reactant is added or a product removed, and towards the reactants after the opposite.
"Adding a reactant changes $K_c$." The constant is fixed by the temperature. What changes is $Q_c$, and it is the gap between the two that drives the shift.
"The shift cancels the disturbance completely." It only partly does. After extra A is added, the new equilibrium still has more A than before — otherwise the ratio could not be back at the constant with more B present.
"At the new equilibrium the reactions stop." They carry on at equal rates, exactly as before the disturbance.
"A solid added to the mixture shifts it." A pure solid does not appear in $K_c$, so adding more of one leaves the quotient, and the position, where they were.
$\mathrm{A \rightleftharpoons B}$ with $K_c = 3$ holds 10 mmol of A and 30 mmol of B. Then 12 mmol of B is removed.
Start from the mixture just after the change: 10 of A, 18 of B.
$Q_c = 18 \div 10 = 1.8$, below 3, so the forward reaction runs net.
Removing a product lowers the quotient.
If $x$ of A becomes B: $(18 + x) \div (10 - x) = 3$, so $4x = 12$ and $x = 3$. New mixture: 7 mmol of A, 21 mmol of B.
Part of the removed B is replaced — 3 of the 12 — not all of it.
$\mathrm{H_2 + I_2 \rightleftharpoons 2HI}$ has $[\mathrm{H_2}] = [\mathrm{I_2}] = 0.10$ and $[\mathrm{HI}] = 0.70$ mol/L, so $K_c = 49$.
Check the constant from the equilibrium values first.
Iodine is added until $[\mathrm{I_2}] = 0.20$ mol/L. Now $Q_c = 0.70^2 \div (0.10 \times 0.20) = 24.5$.
Same expression, new iodine concentration.
$Q_c < K_c$, so net forward reaction: some hydrogen and iodine combine and the hydrogen iodide rises until the quotient is 49 again.
The comparison decides the direction.
Just after: 20 of A and 30 of B, so $Q_c = 1.5$, above 1.
Adding a product raises the quotient.
Net reverse: $x$ of B becomes A, and $(30 - x) \div (20 + x) = 1$, so $x = \ldots$
…$5$, which leaves 25 mmol of A and 25 mmol of B at the new equilibrium.
An isomerisation $\mathrm{A \rightleftharpoons B}$ sits in a flask of fixed volume, where $K_c = 3$, so at equilibrium there is $3$ times as much B as A. It was at equilibrium with $20$ mmol of A and $60$ mmol of B, and then $80$ mmol more A was added. Use the net-reaction steps to bring the mixture back to equilibrium.
This task has no paper form; do it on a device.
Iron(III) ions and thiocyanate ions are at equilibrium with a blood-red complex: $\mathrm{Fe^{3+}(aq) + SCN^-(aq) \rightleftharpoons FeSCN^{2+}(aq)}$. Then more iron(III) thiocyanate complex is added from a stock bottle. What happens to the position of equilibrium?
The iron(III) thiocyanate equilibrium, $\mathrm{Fe^{3+} + SCN^- \rightleftharpoons FeSCN^{2+}}$ (blood red), is set up in three tubes, and each tube gets a different disturbance. For each, say which way the position moves and what happens to the colour.
| which way does the position move? | does the red get deeper or paler? | |
|---|---|---|
| a few crystals of potassium thiocyanate are dissolved in it | ||
| more iron(III) thiocyanate complex is added from a stock bottle | ||
| a few drops of concentrated iron(III) chloride solution are added |
$\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$ is at equilibrium with $[\mathrm{H_2}] = [\mathrm{I_2}] = 0.2$ mol/L and $[\mathrm{HI}] = 0.8$ mol/L, so $K_c = 16$. Hydrogen is injected until $[\mathrm{H_2}] = 1$ mol/L, with nothing else changed yet. What is the reaction quotient $Q_c$ at that moment?
Answer:
For the equilibrium $\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$ at a fixed temperature, match each disturbance with the net reaction that follows it.
| net forward reaction: more ammonia forms | net reverse reaction: some ammonia decomposes | |
|---|---|---|
| More nitrogen is pumped in | ||
| Ammonia is condensed out and removed | ||
| Extra ammonia is injected | ||
| Some hydrogen is drawn off | ||
| More hydrogen is pumped in |
A learner explains what happens when extra ethanol is added to an ester equilibrium, $\mathrm{CH_3COOH + C_2H_5OH \rightleftharpoons CH_3COOC_2H_5 + H_2O}$. Select every sentence that is correct.
This task has no paper form; do it on a device.
In the blood, haemoglobin carries oxygen by the equilibrium $\mathrm{Hb + 4O_2 \rightleftharpoons Hb(O_2)_4}$. A climber arrives at $5\,000$ m, where the air holds much less oxygen per litre than at sea level. What happens to the equilibrium in the climber's lungs and blood?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
The isomerisation $\mathrm{A \rightleftharpoons B}$ has $K_c = 2$ in a flask of fixed volume. At equilibrium there are $20$ mmol of A and $40$ mmol of B. Then $15$ mmol of A is added. How much B is there once equilibrium is restored?
Answer: unit: mol / mmol
You can turn Le Chatelier's principle into a comparison of $Q_c$ with $K_c$. Say out loud why adding iron(III) ions deepens the red of the thiocyanate complex while $K_c$ stays exactly the same. Next: the one disturbance that does change the constant — temperature.
10. Your turn: $\mathrm{A \rightleftharpoons B}$ with $K_c = 1$ holds 20 mmol of each. Then 10 mmol of B is added. What is the new equilibrium?, step 3