When nitrogen and hydrogen react to form ammonia, the reaction can reach a dynamic equilibrium - Edexcel - GCSE Chemistry - Question 4 - 2013 - Paper 1
Question 4
When nitrogen and hydrogen react to form ammonia, the reaction can reach a dynamic equilibrium.
N2(g) + 3H2(g) ⇌ 2NH3(g)
(a) Explain what is meant by a dynamic equ... show full transcript
Worked Solution & Example Answer:When nitrogen and hydrogen react to form ammonia, the reaction can reach a dynamic equilibrium - Edexcel - GCSE Chemistry - Question 4 - 2013 - Paper 1
Step 1
Explain what is meant by a dynamic equilibrium.
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Answer
A dynamic equilibrium occurs in a reversible reaction where the rate of the forward reaction equals the rate of the backward reaction. At this point, the concentrations of reactants and products remain constant over time, although both reactions continue to occur.
Step 2
Explain how the use of a higher pressure would affect the equilibrium yield of ammonia.
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Answer
Using a higher pressure favors the forward reaction, which produces ammonia, because it reduces the volume of gas present. Thus, the equilibrium shifts to the right, increasing the yield of ammonia, as there are fewer moles of gas on the product side.
Step 3
Explain how the use of a lower temperature would affect the equilibrium yield of ammonia.
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Answer
Lower temperatures favor the exothermic forward reaction, thus shifting the equilibrium to the right and increasing the yield of ammonia. However, lower temperatures may also slow down the rate of reaction.
Step 4
State what is used in industry to overcome this problem.
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Answer
A catalyst, typically iron, is used to increase the reaction rate without affecting the position of the equilibrium.
Step 5
Calculate the minimum volume of hydrogen required to completely convert 1000 dm³ of nitrogen into ammonia.
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Answer
The reaction ratio is 1:3 (N2:H2). Therefore:
Volume of hydrogen required = 3 x 1000 dm³ = 3000 dm³.
Step 6
Calculate the mass of ammonium nitrate produced by the complete reaction of 34 g of ammonia.
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Answer
Using the molar mass calculations:
Molar mass of NH3 = 14 + (3 x 1) = 17 g/mol.
Moles of NH3 in 34 g = 34 g / 17 g/mol = 2 moles.
The reaction produces 2 moles of NH4NO3 for every mole of NH3.
Molar mass of NH4NO3 = 14 + (4 x 1) + (3 x 16) = 80 g/mol.
Mass of NH4NO3 = 2 moles x 80 g/mol = 160 g.
Thus, the mass of ammonium nitrate produced is 160 g.