Le Chatelier's Principle – Chemical Equilibrium (Senior 6 Chemistry) | Ganzaa

Chemical Equilibrium

Lessons

Senior 6 · Chemistry

Chemical Equilibrium

5 lessons

Lesson 4 of 5

Le Chatelier's Principle

25 min

One sealed ampoule, five colours

The photograph below shows the same sealed glass ampoule of nitrogen dioxide photographed at five temperatures. Nothing can get in or out, yet the gas is almost colourless when cold and deep red-brown when warm. Brown $\mathrm{NO_2}$ and colourless $\mathrm{N_2O_4}$ are in equilibrium:

$\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)} \qquad \Delta H = +57\,\mathrm{kJ\,mol^{-1}}$

So warming the ampoule must change the equilibrium mixture in favour of $\mathrm{NO_2}$ . Squeezing the gas into a smaller volume changes it too, and so does adding more of one substance. In 1884 the French chemist Henry Le Chatelier summed up all such changes in one rule. In this lesson you will use the rule to predict, and $Q$ and $K$ to explain, what happens.

Definition

Le Chatelier's principle: if a system at equilibrium is disturbed by a change in concentration, pressure or temperature, the position of equilibrium shifts in the direction that tends to oppose the change.

Shift of equilibrium: a net reaction in one direction after a disturbance, until equilibrium is re-established. "Shifts to the right" means more products form. Example: heating the ampoule shifts $\mathrm{N_2O_4 \rightleftharpoons 2NO_2}$ to the right.

Endothermic direction: the direction of a reversible reaction that absorbs heat ( $\Delta H > 0$ ); the reverse direction is exothermic by the same amount. Example: $\mathrm{N_2O_4 \rightarrow 2NO_2}$ is endothermic, so $\mathrm{2NO_2 \rightarrow N_2O_4}$ is exothermic, $\Delta H = -57\,\mathrm{kJ\,mol^{-1}}$ .

1. Changing a concentration

Mix very dilute iron(III) and thiocyanate solutions and you get a pale orange-red solution of $\mathrm{Fe^{3+}(aq) + SCN^{-}(aq) \rightleftharpoons [Fe(SCN)]^{2+}(aq)}$ . A few drops of concentrated iron(III) chloride make it deeper red: the equilibrium has shifted right, using up some of the added $\mathrm{Fe^{3+}}$ .

Why, in terms of $Q$ and $K$ : adding $\mathrm{Fe^{3+}}$ increases a term on the bottom of

$Q = \frac{[\mathrm{Fe(SCN)^{2+}}]}{[\mathrm{Fe^{3+}}][\mathrm{SCN^{-}}]}$

so $Q$ falls below $K$ . Net forward reaction follows until $Q = K$ again. $K$ itself has not changed. Adding a product, or removing a reactant, makes $Q > K$ and shifts the equilibrium left.

2. Changing the pressure of a gas mixture

Figure 4.1 follows the $\mathrm{N_2O_4 \rightleftharpoons 2NO_2}$ mixture at 25 °C in a sealed syringe when the plunger is pushed in to halve the volume. The numbers come from $K_p = 14.8\,\mathrm{kPa}$ (Lesson 3).

Figure 4.1: Compressing an equilibrium mixture of $\mathrm{NO_2}$ and $\mathrm{N_2O_4}$ at 25 °C.

Halving the volume doubles every partial pressure, to 63.6 kPa of $\mathrm{NO_2}$ and 136.4 kPa of $\mathrm{N_2O_4}$ . Then

$Q_p = \frac{(63.6)^2}{136.4} = 29.7\,\mathrm{kPa} > K_p = 14.8\,\mathrm{kPa}$

so the net reaction goes backwards, $\mathrm{2NO_2 \rightarrow N_2O_4}$ , until $p(\mathrm{NO_2}) = 46.3$ and $p(\mathrm{N_2O_4}) = 145.0\,\mathrm{kPa}$ . Two gas molecules become one, so the total pressure drops from 200 to 191 kPa: the system has partly opposed the rise in pressure. In general:

Increasing the pressure (by compressing) shifts the equilibrium towards the side with fewer moles of gas; decreasing it shifts it towards more moles of gas.
If both sides have the same number of moles of gas ( $\Delta n = 0$ , e.g. $\mathrm{H_2 + I_2 \rightleftharpoons 2HI}$ ), pressure has no effect on the position.
Adding an unreactive gas such as argon at constant volume does not change any partial pressure, so it has no effect either.

3. Changing the temperature

Temperature is different: it changes the value of $K$ itself. Raising the temperature favours the endothermic direction, which absorbs the added heat. Table 4.1 shows the effect on $\mathrm{N_2O_4 \rightleftharpoons 2NO_2}$ , whose forward reaction is endothermic.

Temperature / °C	0	25	50	75	100
$K_{p}$ / kPa	1.80	14.8	88.5	408	1530

Table 4.1: $K_{p}$ for $\mathrm{N_2O_4 \rightleftharpoons 2NO_2}$ , calculated from thermodynamic data.

For an endothermic forward reaction, $K$ increases with temperature and the yield of products rises. For an exothermic forward reaction, such as ammonia synthesis, $K$ decreases as the temperature rises and the yield falls.

4. Adding a catalyst

A catalyst lowers the activation energy of the forward and the backward reactions equally. Equilibrium is reached sooner, but $K$ , $Q$ and the position of equilibrium are unchanged.

Change	Does $K$ change?	Effect on position
add a reactant or remove a product	no	shifts right ( $Q < K$ )
add a product or remove a reactant	no	shifts left ( $Q > K$ )
increase pressure by compression	no	towards fewer moles of gas
raise the temperature	yes	towards the endothermic direction
add a catalyst	no	none; equilibrium is reached faster

Worked example

Problem (testing the principle with numbers). An equilibrium mixture at 430 °C contains $[\mathrm{H_2}] = [\mathrm{I_2}] = 0.00214$ and $[\mathrm{HI}] = 0.0157\,\mathrm{mol\,dm^{-3}}$ (Lesson 1). Hydrogen is added until $[\mathrm{H_2}] = 0.0121\,\mathrm{mol\,dm^{-3}}$ . Calculate the new equilibrium concentrations ( $K_{c} = 54.3$ ).

Solution. Just after the addition, $Q = \frac{0.0157^2}{0.0121 \times 0.00214} = 9.5 < 54.3$ , so the reaction goes forward. Let $x$ mol dm⁻³ of $\mathrm{H_2}$ react:

$54.3 = \frac{(0.0157 + 2x)^2}{(0.0121 - x)(0.00214 - x)}$

Solving (by the quadratic formula or by trial) gives $x = 0.00153$ .

Answer: $[\mathrm{H_2}] = 0.0106$ , $[\mathrm{I_2}] = 0.00061$ , $[\mathrm{HI}] = 0.0188\,\mathrm{mol\,dm^{-3}}$ . The equilibrium shifted right and used up some of the added hydrogen, but only about 15 % of it: $[\mathrm{H_2}]$ stays far above its old value. Le Chatelier's principle says the system opposes the change; it never cancels it.

Worked example

Problem. For $\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}$ , $\Delta H = -197\,\mathrm{kJ\,mol^{-1}}$ , predict the effect on the equilibrium amount of $\mathrm{SO_3}$ , and on $K_{c}$ , of: (a) adding oxygen at constant volume; (b) compressing the mixture; (c) raising the temperature; (d) adding vanadium(V) oxide.

Solution.

(a) $[\mathrm{O_2}]$ rises, so $Q < K_{c}$ : more $\mathrm{SO_3}$ ; $K_{c}$ unchanged.
(b) 3 moles of gas on the left, 2 on the right. Compression shifts the equilibrium to the right: more $\mathrm{SO_3}$ ; $K_{c}$ unchanged.
(c) The forward reaction is exothermic, so the endothermic backward reaction is favoured: less $\mathrm{SO_3}$ ; $K_{c}$ decreases.
(d) $\mathrm{V_2O_5}$ is a catalyst: same amount of $\mathrm{SO_3}$ , reached sooner; $K_{c}$ unchanged.

Answer: more, more, less, no change; only (c) changes $K_{c}$ .

Le Chatelier at Lake Kivu

Lake Kivu, on Rwanda's border with the Democratic Republic of the Congo, holds an estimated 256 km³ of carbon dioxide and about 65 km³ of methane dissolved in its deep water. The gases stay in solution because the deep water is under high pressure: every 10 m of water adds roughly one atmosphere. Each dissolved gas is in a physical equilibrium such as

$\mathrm{CH_4(g) \rightleftharpoons CH_4(aq)}$

When the KivuWatt plant pumps gas-rich water up from the depths, the pressure falls, the equilibrium shifts towards the gas, and bubbles form, just as a fizzy drink fizzes when the bottle is opened. The methane is separated, cleaned and piped ashore to gas engines, which were generating 26 MW of electricity in 2016. The same principle explains the danger of the lake: if deep water were ever lifted suddenly, the dissolved carbon dioxide could come out of solution all at once.

A methane-extraction platform floating on Lake Kivu, Rwanda, photographed in 2009. Gas-rich water is pumped up from the depths, and the dissolved gases come out of solution as the pressure falls. Photo: U.S. Department of the Treasury, public domain, via Wikimedia Commons

Warning

Common mistakes

"Adding a reactant increases $K$ ." No: only temperature changes $K$ . Concentration and pressure changes alter $Q$ , and the system moves until $Q = K$ again.
"The equilibrium shifts to remove all the added substance." No: it removes only part of it, as the hydrogen example shows.
Applying the pressure rule to solids and liquids. Count moles of gas only: in $\mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)}$ , higher pressure shifts the equilibrium to the left.
"A catalyst increases the yield." No: it gives the same yield sooner.

Tip

In an explanation, examiners want two steps, not just "Le Chatelier": (1) the change ("increasing the pressure"), (2) the response, with its reason ("shifts the equilibrium to the right, because the right-hand side has fewer moles of gas, which lowers the pressure"). For a temperature change, always say which direction is endothermic.

Chemistry and Biochemistry Demo Lab at OSU: Effect of temperature on the conversion of NO₂ to N₂O₄. One tube is placed in hot water and one in ice water. Watch which tube turns darker, and use the sign of ΔH to explain it.

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Practice

State Le Chatelier's principle.
For $\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$ , $\Delta H = -92\,\mathrm{kJ\,mol^{-1}}$ , state the effect on the equilibrium yield of ammonia of: (a) raising the pressure; (b) raising the temperature; (c) removing ammonia as it forms; (d) using an iron catalyst.
Explain why changing the pressure has no effect on the position of $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$ , but compressing the mixture still makes its colour darker.
Use Table 4.1 to decide whether $\mathrm{N_2O_4 \rightleftharpoons 2NO_2}$ is exothermic or endothermic in the forward direction, and explain how the ampoule photograph agrees.
An equilibrium mixture of $\mathrm{N_2O_4}$ and $\mathrm{NO_2}$ at 25 °C has partial pressures of 68.2 kPa and 31.8 kPa. The volume is doubled at constant temperature. Calculate $Q_{p}$ just after the change, and state the direction of the shift.
For $\mathrm{CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(g)}$ , $\Delta H = -91\,\mathrm{kJ\,mol^{-1}}$ , choose the temperature and pressure conditions that would give the highest equilibrium yield of methanol, and explain why a manufacturer might not use them.
Dissolving carbon dioxide or methane in water is exothermic. Analyse whether a rise in the temperature of Lake Kivu's deep water would make it hold more or less dissolved gas, and what that would mean for the lake's safety.

Show answersHide answers

If a system at equilibrium is disturbed by a change in concentration, pressure or temperature, the position of equilibrium shifts in the direction that tends to oppose the change.
(a) increases (4 moles of gas → 2); (b) decreases (forward reaction exothermic, $K$ falls); (c) increases ( $Q < K$ , more forms); (d) no change in yield, equilibrium reached faster.
$\Delta n = 0$ , so $Q_{p}$ is unchanged by compression and stays equal to $K_{p}$ . The colour darkens because the same amount of violet iodine is squeezed into a smaller volume: its concentration rises, though the mole fractions do not change.
$K_{p}$ rises with temperature, so the forward reaction is endothermic. In the photograph the gas is darker (more $\mathrm{NO_2}$ ) at higher temperature, as expected.
Partial pressures halve to 34.1 and 15.9 kPa: $Q_p = \frac{15.9^2}{34.1} = 7.41\,\mathrm{kPa} < 14.8$ . Net forward reaction: more $\mathrm{NO_2}$ forms (towards more moles of gas), so the colour, after fading on expansion, darkens a little again.
Low temperature (exothermic forward reaction) and high pressure (3 moles of gas → 1) give the highest yield. But at low temperature the rate is too slow, and very high pressure is expensive and dangerous, so a compromise is used.
Dissolving is exothermic, so warming favours the endothermic reverse direction, $\mathrm{gas(aq) \rightarrow gas(g)}$ : warmer water holds less dissolved gas. Warming the deep water would therefore tend to release gas, adding to the danger from any fall in pressure.

Sources and credits

Lake Kivu gas amounts, the extraction method and the 26 MW output in 2016: Lake Kivu, Wikipedia contributors, Wikipedia, licensed under CC BY-SA 4.0, adapted by Ganzaa.
Table 4.1 and Figure 4.1 were calculated by Ganzaa from standard thermodynamic data in the CRC Handbook of Chemistry and Physics.
Image: Nitrogen dioxide at different temperatures.jpg by Eframgoldberg, via Wikimedia Commons, licensed under CC BY-SA 3.0.
Image: Methane extraction platform at Lake Kivu.jpg, U.S. Department of the Treasury, via Wikimedia Commons, public domain.
Video: Effect of Temperature on conversion of NO2 to N2O4 (Le Chatelier's Principle), Chemistry and Biochemistry Demo Lab at OSU, embedded from YouTube.
The text, syringe 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.

1One sealed ampoule, five colours

21. Changing a concentration

32. Changing the pressure of a gas mixture

43. Changing the temperature

54. Adding a catalyst

6Le Chatelier at Lake Kivu