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In the Haber process the following chemical equilibrium is established using an iron catalyst - Leaving Cert Chemistry - Question 9 - 2021

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In the Haber process the following chemical equilibrium is established using an iron catalyst. $$N_2 (g) + 3H_2 (g) \rightleftharpoons 2NH_3 (g)$$ The graph shows ... show full transcript

Worked Solution & Example Answer:In the Haber process the following chemical equilibrium is established using an iron catalyst - Leaving Cert Chemistry - Question 9 - 2021

Step 1

Select the area of the graph corresponding to the conditions most closely associated with good yields and slower rates of reaction

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Answer

The area of the graph that corresponds to conditions of good yields and slower rates of reaction is box A. This area represents conditions with higher pressure and lower temperature, which are beneficial for maximizing ammonia production while also slowing down the reaction kinetics.

Step 2

Select the area of the graph corresponding to the conditions most closely associated with higher yields but more costly and less safe ammonia production

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Answer

The area of the graph that corresponds to conditions of higher yields but is more costly and less safe for ammonia production is box B. This area represents higher temperatures and pressures that lead to increased ammonia yield but may involve higher operational costs and safety risks.

Step 3

Deduce whether the production of ammonia is exothermic or endothermic and justify your answer

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Answer

The production of ammonia in the Haber process is exothermic. This is deduced from the graph, which shows that as the temperature increases, the equilibrium position shifts towards lower ammonia production (indicating lower yields). In exothermic reactions, increasing the temperature typically favors the reactants, thus confirming the process is exothermic.

Step 4

Calculate the equilibrium constant (Kc) for the reaction at temperature T

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Answer

To calculate the equilibrium constant (Kc), we start with concentrations at equilibrium. Initially, we have:

  • 9.0 mol N2 and 27.0 mol H2 in a 10.0 L container.
  • At equilibrium, we have 6.0 mol NH3.

Let x be the change in moles for N2. Since 2 moles of NH3 produce 1 mole of N2:

Initially:

  • N2 = 0.9 mol
  • H2 = 2.7 mol
  • NH3 = 0.6 mol

At equilibrium:

  • N2 = 0.9 - 0.3 = 0.6 mol
  • H2 = 2.7 - 0.9 = 1.8 mol

Now we relate these to equilibrium concentrations:

Kc=[NH3]2[N2][H2]3=(0.6)2(0.6)(1.8)3=0.360.6×5.832=0.1K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(0.6)^2}{(0.6)(1.8)^3} = \frac{0.36}{0.6 \times 5.832} = 0.1

Step 5

State the percentage yield of ammonia at equilibrium in this container at equilibrium

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Answer

The percentage yield of ammonia at equilibrium can be calculated using the formula:

Percentage Yield=(moles of NH3 at equilibriuminitial moles of NH3)×100\text{Percentage Yield} = \left( \frac{\text{moles of } NH_3 \text{ at equilibrium}}{\text{initial moles of } NH_3} \right) \times 100 Thus, the yield is 20%20\%.

Step 6

State the effect (and explain) of increasing the pressure at temperature T on the yield of ammonia production

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Answer

Increasing the pressure at temperature T generally leads to a higher yield of ammonia. This is because the reaction produces fewer gas molecules (from 4 to 2), so increasing pressure shifts the equilibrium to the right, favoring ammonia production.

Step 7

State the effect (and explain) of using a more efficient catalyst on the yield of ammonia production

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Answer

Using a more efficient catalyst has no effect on the yield of ammonia in terms of the amount produced. However, it will speed up the rate at which equilibrium is attained. Catalysts increase the reaction rate without altering the position of equilibrium or the yield.

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