Chemical Equilibrium

Lessons

Senior 6 · Chemistry

Chemical Equilibrium

5 lessons

Lesson 5 of 5

Equilibrium in Industry: the Haber and Contact Processes

25 min

Where does the nitrogen in a bag of fertiliser come from?

Rwanda's farmers use about 85 000 tonnes of fertiliser a year, and in 2023 all of it was imported. Since 20 December 2023, a plant in the Bugesera Special Economic Zone has blended imported fertilisers into mixtures suited to Rwandan soils and crops, with a capacity of 100 000 tonnes a year. The nitrogen in urea and NPK fertilisers, however, was first taken from the air by one reaction, carried out in large plants abroad:

$\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)} \qquad \Delta H = -92\,\mathrm{kJ\,mol^{-1}}$

This is the Haber process, first run on an industrial scale by BASF at Oppau, Germany, in 1913. It now uses 1–2 % of the world's energy supply. Lesson 4 predicts that a low temperature and a high pressure give the most ammonia. Yet real plants run at 400–500 °C and only about 200 atm, and convert only about 15 % of the gas on each pass. This lesson explains why those conditions are the best choice, and compares them with the Contact process for sulfuric acid.

A tall steel high-pressure reactor of the kind once used to synthesise ammonia by the Haber–Bosch process, now displayed outdoors in Germany. Photo: Harald Kucharek, CC BY-SA 2.0, via Wikimedia Commons

Definition

Equilibrium yield: the proportion of product present when the reaction mixture has reached equilibrium under the given conditions. Example: about 25 % ammonia by moles at 450 °C and 200 atm (Figure 5.1).

Conversion per pass: the fraction of the reactants turned into product during one passage through a reactor. It is lower than the equilibrium yield when the gas leaves before equilibrium is reached. Example: about 15 % in an ammonia converter.

Compromise conditions: operating conditions chosen to balance equilibrium yield against rate, energy cost, equipment cost and safety, so that the product is made as cheaply as possible.

Recycling: returning unreacted starting materials to the reactor, so that the overall conversion can be far higher than the conversion per pass.

The Haber process

The forward reaction is exothermic and turns 4 moles of gas into 2. Le Chatelier's principle therefore predicts the highest yield at low temperature and high pressure. Figure 5.1, calculated from the equilibrium constant at each temperature, confirms both predictions.

Figure 5.1: Equilibrium percentage of ammonia for a 1 : 3 mixture of $\mathrm{N_2}$ and $\mathrm{H_2}$ , calculated by Ganzaa assuming ideal gases. Measured values are a little higher at the highest pressures.

Condition chosen	Le Chatelier's prediction, and the problem with it
Temperature: 400–500 °C, a moderate yield reached quickly	Low temperature gives the highest yield, but below about 400 °C the reaction is too slow, even with the catalyst.
Pressure: about 200 atm (20 MPa)	High pressure gives a higher yield, but thick-walled vessels, pipes and compressors cost a lot to build and run, and leaks are more dangerous.
Catalyst: finely divided iron, so that equilibrium is approached in seconds	No effect on the yield; it only speeds up both reactions.
Removing product: the gas is cooled, ammonia condenses and is run off; unreacted $\mathrm{N_2}$ and $\mathrm{H_2}$ are recycled	Removing ammonia shifts the equilibrium to the right; there is no drawback.

The gas passes through the converter too quickly to reach equilibrium, so only about 15 % is converted on each pass. Because the unreacted gases are recycled again and again (Figure 5.2), the overall conversion reaches about 97 %.

Figure 5.2: The Haber process as a loop.

Worked example

Problem. At 450 °C and 200 atm, an equilibrium mixture made from $\mathrm{N_2}$ and $\mathrm{H_2}$ in the ratio 1 : 3 contains, by moles, 24.8 % $\mathrm{NH_3}$ , 18.8 % $\mathrm{N_2}$ and 56.4 % $\mathrm{H_2}$ . (a) Calculate $K_{p}$ . (b) At 400 °C, $K_p = 1.66 \times 10^{-4}\,\mathrm{atm^{-2}}$ . Show that this agrees with the sign of $\Delta H$ . (c) A converter at 450 °C and 200 atm gives 15 % ammonia. What fraction of the equilibrium yield is this, and why is it less?

Solution.

(a) Partial pressures: $p_{\mathrm{NH_3}} = 0.248 \times 200 = 49.6$ , $p_{\mathrm{N_2}} = 37.6$ , $p_{\mathrm{H_2}} = 112.8\,\mathrm{atm}$ .

$K_p = \frac{p_{\mathrm{NH_3}}^{2}}{p_{\mathrm{N_2}}\,p_{\mathrm{H_2}}^{3}} = \frac{(49.6)^2}{37.6 \times (112.8)^3} = 4.56 \times 10^{-5}\,\mathrm{atm^{-2}}$

(b) $K_{p}$ is larger at the lower temperature ( $1.66 \times 10^{-4} > 4.56 \times 10^{-5}$ ). Lowering the temperature favours the forward direction, so the forward reaction is exothermic, as $\Delta H = -92\,\mathrm{kJ\,mol^{-1}}$ says.

(c) $\frac{15}{24.8} = 0.60$ . The gas spends only seconds on the catalyst, which is not long enough to reach equilibrium. Waiting longer would give more ammonia per pass but less per hour, so it is cheaper to recycle.

Answer: (a) $K_p = 4.56 \times 10^{-5}\,\mathrm{atm^{-2}}$ ; (b) $K$ falls as $T$ rises, as expected for an exothermic reaction; (c) about 60 % of the equilibrium yield.

The Contact process

Sulfuric acid, used to make phosphate fertilisers, detergents, paints and batteries, is made by the Contact process. Sulfur is burned in air, and the sulfur dioxide is oxidised further:

$\mathrm{S(s) + O_2(g) \rightarrow SO_2(g)}$

$\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)} \qquad \Delta H = -197\,\mathrm{kJ\,mol^{-1}}$

The key step is again exothermic with fewer moles of gas on the right, so low temperature and high pressure would favour it. The choices differ from the Haber process for a good reason:

Condition chosen	Reason
Temperature: about 450 °C (range 420–620 °C)	a compromise between rate and yield; below about 420 °C the vanadium catalyst is deactivated
Pressure: 1–2 atm	$K_{p}$ is so large that the equilibrium conversion is already about 98 % at 1 atm; extra pressure would add cost but little yield
Catalyst: vanadium(V) oxide, $\mathrm{V_2O_5}$	speeds up the reaction; cheaper and less easily poisoned than platinum
Oxygen: excess air	raises $[\mathrm{O_2}]$ , so $Q < K$ and more $\mathrm{SO_2}$ is converted

The sulfur trioxide is not dissolved directly in water: that reaction is so exothermic that it forms a mist of acid droplets instead of a liquid. It is absorbed in 98 % sulfuric acid to give oleum, which is then diluted:

$\mathrm{SO_3(g) + H_2SO_4(l) \rightarrow H_2S_2O_7(l)} \qquad \mathrm{H_2S_2O_7(l) + H_2O(l) \rightarrow 2H_2SO_4(l)}$

Removing $\mathrm{SO_3}$ in a first absorber and passing the remaining gas over the catalyst again (the "double contact, double absorption" process) raises the overall conversion to as much as 99.8 % (Figure 5.3).

Figure 5.3: The Contact process.

Worked example

Problem (evaluating high pressure). Gas entering a converter at 450 °C contains, by moles, 10 % $\mathrm{SO_2}$ , 11 % $\mathrm{O_2}$ and 79 % $\mathrm{N_2}$ . With $K_p \approx 3 \times 10^{4}\,\mathrm{atm^{-1}}$ (calculated from thermodynamic data), the equilibrium conversion of $\mathrm{SO_2}$ works out at 97.8 % at 1 atm, 98.4 % at 2 atm and 99.3 % at 10 atm. (a) Show that the 1 atm figure fits $K_{p}$ . (b) Should a manufacturer run the converter at 10 atm?

Solution.

(a) Take 100 mol of gas and a conversion of 0.978. $\mathrm{SO_2}$ left $= 10 \times 0.022 = 0.22\,\mathrm{mol}$ ; $\mathrm{SO_3}$ formed $= 9.78\,\mathrm{mol}$ ; $\mathrm{O_2}$ left $= 11 - 4.89 = 6.11\,\mathrm{mol}$ ; total $= 100 - 4.89 = 95.11\,\mathrm{mol}$ . At 1 atm, partial pressures equal mole fractions:

$K_p = \frac{\left(\frac{9.78}{95.11}\right)^2}{\left(\frac{0.22}{95.11}\right)^2 \times \frac{6.11}{95.11}} = \frac{(9.78)^2 \times 95.11}{(0.22)^2 \times 6.11} = 3.1 \times 10^{4}\,\mathrm{atm^{-1}}$

(b) Going from 1 to 10 atm raises the conversion by only 1.5 percentage points, but needs compressors, stronger vessels and more energy. Double absorption already gives up to 99.8 % at 1–2 atm. So no: unlike ammonia, whose $K$ is small at operating temperatures, sulfur trioxide gains little from pressure.

Answer: (a) $K_p \approx 3.1 \times 10^{4}\,\mathrm{atm^{-1}}$ , consistent; (b) no: the small gain in yield would not pay for the cost of high pressure.

Warning

Rate and yield are different questions

"A high temperature is used to increase the yield of ammonia." Wrong: a high temperature lowers the yield of an exothermic reaction. It is used to increase the rate.
"The catalyst increases the yield." Wrong: it only lets the mixture approach equilibrium faster.
"The Haber process reaches equilibrium in the converter." Not quite: the gas leaves at about 15 % ammonia, before equilibrium is reached.

Tip

For any "explain the conditions" question, use a four-column habit: condition → what Le Chatelier predicts → what rate or cost demands → the compromise. Always mention the catalyst (rate only), and mention recycling or excess of a cheap reactant where it applies.

The Organic Chemistry Tutor: Haber Process and Le Chatelier's Principle of Chemical Equilibrium. Watch how each change in temperature, pressure and concentration shifts the ammonia equilibrium, and compare the reasoning with the compromise conditions in this lesson.

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Practice

State the temperature, pressure and catalyst used in (a) the Haber process; (b) the Contact process.
Use Figure 5.1 to read the equilibrium percentage of ammonia at 200 atm and (a) 400 °C, (b) 500 °C. Explain the difference.
Explain why ammonia is removed from the Haber loop by cooling, and why this helps the overall yield.
What mass of ammonia can be made from 1.00 tonne of nitrogen if the overall conversion is 97 %?
Write equations for the absorption of sulfur trioxide in sulfuric acid and for the dilution of oleum. Why is sulfur trioxide not simply dissolved in water?
At 400 °C, $K_p = 1.66 \times 10^{-4}\,\mathrm{atm^{-2}}$ for $\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}$ . A mixture at 400 °C has $p_{\mathrm{NH_3}} = 60$ , $p_{\mathrm{N_2}} = 35$ and $p_{\mathrm{H_2}} = 105\,\mathrm{atm}$ . Calculate $Q_{p}$ and state which way the reaction goes.
The Contact process uses 1–2 atm but the Haber process uses about 200 atm, although both reactions decrease the number of moles of gas. Evaluate the reasons for this difference.

Show answersHide answers

(a) 400–500 °C, about 200 atm, iron; (b) about 450 °C, 1–2 atm, vanadium(V) oxide.
(a) about 35 %; (b) about 17 %. The forward reaction is exothermic, so $K$ and the equilibrium yield fall as the temperature rises.
Ammonia condenses at a much higher temperature than nitrogen or hydrogen, so cooling turns it to liquid while the other gases stay gaseous and are recycled. Removing the product also keeps $Q < K$ when the gas returns to the converter.
$1.00\,\mathrm{t} \times \frac{2 \times 17.03}{28.02} \times 0.97 = 1.18\,\mathrm{t}$ .
$\mathrm{SO_3 + H_2SO_4 \rightarrow H_2S_2O_7}$ ; $\mathrm{H_2S_2O_7 + H_2O \rightarrow 2H_2SO_4}$ . Direct reaction with water is so exothermic that a mist of acid droplets forms, which is hard to collect.
$Q_p = \frac{60^2}{35 \times 105^3} = 8.9 \times 10^{-5}\,\mathrm{atm^{-2}} < K_p$ . Net forward reaction: more ammonia forms.
For $\mathrm{SO_3}$ , $K_{p}$ at 450 °C is so large that conversion is about 98 % even at 1 atm, so high pressure would cost more than it gains. For ammonia, $K_{p}$ at 400–500 °C is small, so high pressure is needed to get a worthwhile yield; even then only about 15 % is converted per pass. The extra cost of 200 atm is justified for ammonia but not for sulfur trioxide.

1Where does the nitrogen in a bag of fertiliser come from?

2The Haber process

3The Contact process

Where does the nitrogen in a bag of fertiliser come from?

The Haber process

The Contact process