The balanced equation below represents the reaction used in the Haber process to produce ammonia - NSC Physical Sciences - Question 6 - 2019 - Paper 2
Question 6
The balanced equation below represents the reaction used in the Haber process to produce ammonia.
N2(g) + 3H2(g) ⇌ 2NH3(g) ΔH < 0
In industry the product is remov... show full transcript
Worked Solution & Example Answer:The balanced equation below represents the reaction used in the Haber process to produce ammonia - NSC Physical Sciences - Question 6 - 2019 - Paper 2
Step 1
6.1 Write down the meaning of the double arrow used in the equation above.
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Answer
The double arrow in the equation indicates that the reaction is reversible, meaning that both the forward and reverse reactions can occur. Products can be converted back to reactants.
Step 2
6.2 Give ONE reason why ammonia is removed from the reaction vessel as quickly as it forms.
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Answer
Ammonia is removed quickly to favor the forward reaction and prevent the decomposition of NH3.
Step 3
6.3 Write down the percentage yield of ammonia at a temperature of 450 °C and a pressure of 200 atmospheres.
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Answer
The percentage yield of ammonia at a temperature of 450 °C and a pressure of 200 atmospheres is 20%.
Step 4
6.4.1 For a given pressure, the yield of ammonia at 500 °C is much lower than that at 350 °C.
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Answer
At 500 °C, the forward reaction is less favored because it is exothermic, which means an increase in temperature shifts the equilibrium to decrease the yield of ammonia.
Step 5
6.4.2 For a given temperature, the yield of ammonia at 350 atmospheres is much higher than at 150 atmospheres.
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Answer
Increasing the pressure at a constant temperature favors the reaction producing fewer moles of gas, which is the forward reaction in the production of ammonia.
Step 6
6.5.1 Calculate the maximum number of moles of NH3(g) that can be obtained in this reaction.
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Answer
Based on the stoichiometry of the reaction where 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3, starting with 6 moles of H2, the maximum number of NH3 produced would be 6 moles H2 x (2 moles NH3 / 3 moles H2) = 4 moles NH3.
Step 7
6.5.2 Use the graph above and calculate the equilibrium constant, Kc, for this reaction under these conditions.
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Answer
Using the graph, at equilibrium with given concentrations, Kc can be calculated as follows:
Let [NH3] = 2.8 M, [N2] = 0.002 M, and [H2] = 7.8 M.
The equilibrium constant expression is given by:
Kc=[N2][H2]3[NH3]2
Substituting in the concentrations:
Kc=(0.002)(7.8)3(2.8)2
