At a coastal chemical works, gases from an air-separation unit and a hydrogen plant are fed into a high-pressure vessel. Fig. 1 shows the route used to make...
Question 1 Report
At a coastal chemical works, gases from an air-separation unit and a hydrogen plant are fed into a high-pressure vessel. Fig. 1 shows the route used to make ammonia for a solution fertiliser. The reaction mixture is cooled after leaving the reactor. Ammonia becomes a liquid and is removed, while nitrogen and hydrogen that have not reacted are returned to the compressor. Iron is used in the reactor.
In one pass through the reactor, 18% of the nitrogen reacts. Table 1 gives the amount of each gas entering the reactor.
| gas | amount entering reactor / mol |
|---|---|
| nitrogen, N2 | 200 |
| hydrogen, H2 | 600 |
(a) Complete the balanced symbol equation for the reversible reaction.
N2(g) + H2(g) ⇌ NH3(g) [2]
(b) Give one reason why the gases are compressed before entering the reactor. [1]
(c) Suggest two advantages of recycling the unreacted nitrogen and hydrogen shown in Fig. 1. [2]
(d) Use Table 1 and the equation to calculate the number of moles of ammonia made and the number of moles of nitrogen left after one pass. [3]
(e) What does the iron catalyst do to the energy needed for the reaction? [2]
Answer Details
(a) The balanced reversible equation is:
\[\text{N}_2(g)+3\text{H}_2(g)\rightleftharpoons2\text{NH}_3(g)\]
The coefficients give the required \(1:3:2\) ratio. [2]
(b) Compressing the gases increases the pressure, so gas particles collide more frequently. This increases the reaction rate. [1]
(c) Recycling the unreacted gases means less nitrogen and hydrogen are wasted and more ammonia can be made from the same starting materials. It also lowers raw-material costs. Any two of these advantages gain credit. [2]
(d) First calculate the nitrogen that reacts:
\[\frac{18}{100}\times200\text{ mol}=36\text{ mol N}_2\]
From \(\text{N}_2\rightarrow2\text{NH}_3\), each mole of nitrogen reacting produces two moles of ammonia:
\[36\times2=72\text{ mol NH}_3\]
Nitrogen left after one pass:
\[200-36=164\text{ mol N}_2\]
[3]
(e) Iron lowers the activation energy, meaning less energy is needed for successful collisions. It therefore increases the rate of reaction, but is not used up and does not change the equilibrium yield. [2]
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