Dynamic Equilibrium and Le Chatelier's Principle

Many reactions never go to completion: as products form, they react to give back the reactants, and the mixture settles into a state where nothing seems to change. This note explains reversible reactions and dynamic equilibrium, states Le Chatelier's principle in the syllabus wording, and uses it to predict the effect of changing concentration, pressure and temperature or adding a catalyst. It finishes with the two industrial processes the syllabus names, the Haber process and the Contact process, where the conditions are a compromise between yield, rate and cost. "Explain the effect of increasing the pressure on the position of equilibrium" questions appear in almost every Paper 2.

Reversible reactions

A reversible reaction is one that can go in both directions: reactants form products, and products react to re-form the reactants. It is shown with the equilibrium sign, $\ce{<=>}$.

Heating blue hydrated copper(II) sulfate drives off water and leaves the white anhydrous salt; adding water to the white powder turns it blue again:

$$\ce{CuSO4.5H2O(s) <=> CuSO4(s) + 5H2O(l)}$$

The forward reaction (left to right) is endothermic; the reverse reaction is exothermic by exactly the same amount.

Dynamic equilibrium

Consider hydrogen and iodine heated in a sealed flask:

$$\ce{H2(g) + I2(g) <=> 2HI(g)}$$

At the start there is no HI, so only the forward reaction happens, quickly. As $\ce{H2}$ and $\ce{I2}$ are used up, the forward rate falls. As HI builds up, the reverse reaction starts and speeds up. Eventually the two rates become equal. From then on, HI is formed as fast as it decomposes, and the concentrations stop changing.

"Dynamic" means the reactions have not stopped: both are still happening, at the same rate. Individual molecules are constantly changing from reactants to products and back. Only the overall amounts stay the same.

The graph shows concentrations against time for a reaction that starts with reactant only. The falling curve is the reactant and the rising curve the product. They level off (become constant) but are not equal.

(Horizontal axis: time; vertical axis: concentration. Equilibrium is reached when both curves become horizontal, at about $t = 5$ on this sketch.)

The next sketch shows the rates. The forward rate (falling curve) decreases and the reverse rate (rising curve) increases until they are equal: that is the moment equilibrium is established.

Features of a dynamic equilibrium

Why must the system be closed? If a product escapes, for example $\ce{CO2}$ from heated limestone in an open container, the reverse reaction cannot happen, and the reaction goes to completion: $\ce{CaCO3 -> CaO + CO2}$. In a sealed container, $\ce{CaCO3 <=> CaO + CO2}$ reaches equilibrium.

The position of equilibrium describes the relative amounts of reactants and products. "The position of equilibrium lies to the right" means the equilibrium mixture contains mostly products; "moves to the left" means more reactants form.

Le Chatelier's principle

The principle predicts the direction in which the position shifts. Note that the system only partly cancels the change: it minimises it, it does not undo it.

Changing concentration

If the concentration of a reactant is increased, the position of equilibrium moves to the right to use up some of the added reactant, forming more product. If a product is removed, the position moves to the right to replace it.

Example: the chromate(VI)/dichromate(VI) equilibrium.

$$\ce{2CrO4^2-(aq) + 2H+(aq) <=> Cr2O7^2-(aq) + H2O(l)}$$

$\ce{CrO4^2-}$ is yellow and $\ce{Cr2O7^2-}$ orange. Adding acid (more $\ce{H+}$) moves the position to the right: the solution turns orange. Adding alkali removes $\ce{H+}$ (as water), so the position moves to the left: the solution turns yellow.

Changing pressure

Pressure affects equilibria involving gases, and only when the numbers of moles of gas on the two sides differ.

• Increasing pressure: the position moves to the side with fewer moles of gas, which reduces the pressure.
• Decreasing pressure: the position moves to the side with more moles of gas.

For $\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$ there are 4 moles of gas on the left and 2 on the right. Increasing pressure moves the position to the right, increasing the yield of ammonia.

For $\ce{H2(g) + I2(g) <=> 2HI(g)}$ there are 2 moles of gas on each side, so pressure has no effect on the position.

Changing temperature

Temperature is the only change that alters the value of the equilibrium constant (see the next note). Its effect depends on the sign of $\Delta H$ for the forward reaction.

• Increasing temperature: the position moves in the endothermic direction, which absorbs heat and so minimises the rise in temperature.
• Decreasing temperature: the position moves in the exothermic direction, which releases heat.

Example: $\ce{2NO2(g) <=> N2O4(g)}$, $\Delta H = -58\ \text{kJ mol}^{-1}$. $\ce{NO2}$ is brown; $\ce{N2O4}$ is colourless. The forward reaction is exothermic, so heating moves the position to the left (the endothermic direction): the mixture becomes darker brown. Cooling in ice makes it paler.

Adding a catalyst

A catalyst speeds up the forward and reverse reactions equally (it lowers the activation energy of both by the same amount). So it has no effect on the position of equilibrium or on the yield, but equilibrium is reached more quickly.

Industrial equilibria

In industry, the aim is the most product per hour at the lowest cost, so chemists balance yield (the position of equilibrium) against rate (how fast equilibrium is reached) and cost (energy, equipment, safety).

The Haber process

$$\ce{N2(g) + 3H2(g) <=> 2NH3(g)} \qquad \Delta H = -92\ \text{kJ mol}^{-1}$$

condition	used	equilibrium argument	rate and cost argument
temperature	about $400$ to $450\ ^\circ\text{C}$	forward reaction is exothermic, so a low temperature gives a higher yield	a low temperature makes the rate too slow; $450\ ^\circ\text{C}$ is a compromise between yield and rate
pressure	about $20\ \text{MPa}$ (200 atm)	4 moles of gas go to 2, so a high pressure gives a higher yield and also increases rate	very high pressures need expensive, thick-walled equipment and more energy for compression, and are more hazardous; 20 MPa is a compromise
catalyst	iron	no effect on yield	increases rate, so equilibrium is reached faster at a lower temperature

The conversion in one pass is only about 15%. The ammonia is removed by cooling (it liquefies), and the unreacted nitrogen and hydrogen are recycled, so the overall conversion is about 98%. Removing ammonia also prevents the reverse reaction.

The Contact process

The key step in the manufacture of sulfuric acid is the oxidation of sulfur dioxide:

$$\ce{2SO2(g) + O2(g) <=> 2SO3(g)} \qquad \Delta H = -197\ \text{kJ mol}^{-1}$$

condition	used	explanation
temperature	about $450\ ^\circ\text{C}$	forward reaction is exothermic, so a lower temperature gives a better yield, but the rate would be too slow; compromise
pressure	about $100$ to $200\ \text{kPa}$ (1 to 2 atm)	3 moles of gas go to 2, so high pressure would increase the yield, but the conversion is already about 99% at low pressure; extra pressure is not worth the cost
catalyst	vanadium(V) oxide, $\ce{V2O5}$	increases rate; no effect on yield
excess air	a slightly higher proportion of oxygen	increasing $[\ce{O2}]$ moves the position to the right, using more of the expensive $\ce{SO2}$

The sulfur trioxide is then absorbed in concentrated sulfuric acid (dissolving it directly in water produces an uncontrollable acid mist) and diluted to give more sulfuric acid.

Worked examples

Practice

Answers

1. A state in a reversible reaction in a closed system where the rate of the forward reaction equals the rate of the reverse reaction, so the concentrations of reactants and products remain constant; both reactions continue.
2. If the system is open, a product (especially a gas) can escape. The reverse reaction then cannot occur at the same rate as the forward reaction, so the concentrations keep changing and the reaction moves towards completion instead of reaching equilibrium.
3. High temperature: the forward reaction is endothermic, so increasing temperature moves the position to the right. Low pressure: 2 moles of gas on the left, 4 on the right; lowering the pressure moves the position to the side with more gas moles.
4. (a) No effect. (b) Decreased: the catalyst increases the rates of the forward and reverse reactions equally, so equilibrium is reached faster.
5. Orange to yellow. The hydroxide ions react with $\ce{H+}$ to form water, lowering $[\ce{H+}]$. The position of equilibrium moves to the left to replace some $\ce{H+}$, converting orange $\ce{Cr2O7^2-}$ into yellow $\ce{CrO4^2-}$.
6. Increasing the pressure (concentration) of chlorine moves the position to the right: more yellow solid forms and the brown liquid decreases. Removing chlorine moves the position to the left: the yellow solid decomposes to give brown liquid (and chlorine).
7. The position of equilibrium already lies far to the right at about 1 to 2 atm (about 99% conversion), so raising the pressure would increase the yield only slightly. The extra cost of compressors, energy and stronger equipment would not be justified.
8. Higher pressure requires stronger, thicker-walled and more expensive reactors and pipes; compressing gases uses large amounts of energy; there is a greater safety risk from leaks or explosions. (Any two.)
9. Heating makes the mixture darker, so more brown $\ce{NO2}$ forms: the position moves to the left. Increasing temperature moves the position in the endothermic direction, so the reverse reaction is endothermic and the forward reaction is exothermic: $\Delta H$ negative.
10. Temperature: the forward reaction is exothermic, so a low temperature favours the yield; but at low temperatures the rate is too slow. $250\ ^\circ\text{C}$ is a compromise that, with the catalyst, gives a reasonable rate and yield. Pressure: 3 moles of gas go to 1, so high pressure moves the position to the right and also increases rate; $5$ to $10\ \text{MPa}$ is high enough for a good yield without the excessive cost and danger of extreme pressures. Catalyst: speeds up the reaction so that a lower temperature can be used; no effect on yield. Methanol is condensed out and the unreacted CO and $\ce{H2}$ are recycled through the reactor, raising the overall conversion.