The process for the manufacture of ammonia, NH3
, was originally devised at the start of the 20th century and was developed into a full-scale industrial process by Carl Bosch in 1913. The key step in the process is the reversible reaction of nitrogen and hydrogen in the presence of an iron catalyst.
N2(g)+3H2(g)→2NH3(g)
ΔH=–92kJmol(–1)
1. Give the name of this manufacturing process.
2. Give the change in oxidation numbers in the given reaction and explain what substance is oxidized and which substance is reduced.
1. The manufacturing process is called the Haber-Bosch process.
2. In the reaction given:
N2(g) + 3H2(g) → 2NH3(g)
The oxidation state of nitrogen in N2 is 0, and in NH3 it is -3. Therefore, nitrogen is reduced from oxidation state 0 to -3 in the reaction.
The oxidation state of hydrogen in H2 is 0, and in NH3 it is +1. Therefore, hydrogen is oxidized from oxidation state 0 to +1 in the reaction.
Overall, in this reaction, nitrogen is reduced and hydrogen is oxidized.
At high temperatures and in the presence of a catalyst, sulphur trioxide decomposes according to the following equation:
2SO3(g)⇌2SO2(g)+O2(g)
; ΔH=+196kJ⋅mol−1
In an experiment, 8,0mol
of sulphur trioxide is placed in a container of volume
12,0dm3
and allowed to come to equilibrium. At temperature T1 there is 1,4mol
of oxygen in the equilibrium mixture.
1. Define a “reversible” reaction
2. Calculate the value for the equilibrium constant, Kc
.
The experiment was repeated at the same temperature using the same amount of sulphur trioxide but in a larger container. How will it affect any of the following? Write only INCREASE, DECREASE or REMAINS THE SAME.
3. The amount, in moles, of oxygen in the new equilibrium mixture
4. The value of Kc
5. The experiment was repeated in the original container but at temperature T2. The value of Kc was smaller than the value at temperature T1
6. State which is the higher temperature, T1 or T2
7. Explain your answer for the previous question
1. A reversible reaction is a chemical reaction in which the products can react to produce the original reactants.
2. The equilibrium constant, Kc, is calculated as:
Kc = [SO2]^2 * [O2] / [SO3]^2
Given that at equilibrium, [O2] = 1,4 mol and [SO3] = 8,0 mol in a 12,0 dm3 container:
Kc = (1,4)^2 / (8,0)^2 = 0,03
3. INCREASE
4. REMAINS THE SAME
5. DECREASE
6. T1
7. Decrease in temperature causes an increase in equilibrium constant. A lower temperature shifts the equilibrium towards the exothermic direction to counteract the temperature decrease, resulting in a decrease in Kc. Thus, T1 is the higher temperature.