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SIZIF [17.4K]
3 years ago
8

The Haber Process is the main industrial procedure to produce ammonia. The reaction combines nitrogen from air with hydrogen mai

nly from natural gas (methane) and is reversible and exothermic. The enthalpy change for this reaction is - 92 kJ mol-1. In an experiment, 1.5 moles of N2 and 4.0 moles of H2 is mixed in a 1.50 dm3 reaction vessel at 450 °C. After reaching equilibrium, the mixture contained 0.9 mole of NH3.
A) With the above information, write the reaction equilibrium equation in the Haber process. t.
B) Calculate Kc for this reaction.
C) What is the equilibrium yield of ammonia in this reaction?
D) Referring to Le Chatelier's principle and above information, suggest two ways to increase the yield of ammonia in this reaction and explain.
Chemistry
1 answer:
Firdavs [7]3 years ago
7 0

Answer:

A) N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g).

B) Kc=0.0933.

C) 0.9 mol.

D) Increasing both temperature and pressure.

Explanation:

Hello,

In this case, given the information, we proceed as follows:

A)

N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)

B) For the calculation of Kc, we rate the equilibrium expression:

Kc=\frac{[NH_3]^2}{[N_2][H_2]^3}

Next, since at equilibrium the concentration of ammonia is 0.6 M (0.9 mol in 1.5 dm³ or L), in terms of the reaction extent x, we have:

[NH_3]=0.6M=2*x

x=\frac{0.6M}{2}=0.3M

Next, the concentrations of nitrogen and hydrogen at equilibrium are:

[N_2]=\frac{1.5mol}{1.5L}-x=1M-0.3M=0.7M

[H_2]=\frac{4mol}{1.5L}-3*x=2.67M-0.9M=1.77M

Therefore, the equilibrium constant is:

Kc=\frac{(0.6M)^2}{(0.7M)*(1.77M)^3}\\ \\Kc=0.0933

C) In this case, the equilibrium yield of ammonia is clearly 0.9 mol since is the yielded amount once equilibrium is established.

D) Here, since the reaction is endothermic (positive enthalpy change), one way to increase the yield of ammonia is increasing the temperature since heat is reactant for endothermic reactions. Moreover, since this reaction has less moles at the products, another way to increase the yield is increasing the pressure since when pressure is increased the side with fewer moles is favored.

Best regards.

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