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1 mole of N₂ and 2 moles of H₂ are allowed to react in a 1 dm3 vessel. At equilibrium 0.8 mole of NH₃ is formed. The concentration of H₂ in the vessel is
- a)0.6 mole
- b)0.8 mole
- c)0.2 mole
- d)0.4 mole
Correct answer is option 'B'. Can you explain this answer?
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1 mole of N₂ and 2 moles of H₂ are allowed to react in a 1...
Given:
- 1 mole of N
- 2 moles of H
- 0.8 mole of NH formed at equilibrium
To find:
- Concentration of H in the vessel
Solution:
The balanced chemical equation for the reaction is:
N2(g) + 3H2(g) ⇌ 2NH3(g)
From the equation, we can see that:
- 1 mole of N reacts with 3 moles of H
- 2 moles of H react with 2/3 mole of N
Let x be the number of moles of H that reacted to form 0.8 mole of NH. Then, we have:
- 0.8 mole of NH formed from x mole of H
- 1 mole of N reacted with 3x/2 moles of H
At equilibrium, the number of moles of each substance present in the vessel can be calculated using the following table:
Substance | Initial moles | Change | Equilibrium moles
--- | --- | --- | ---
N2 | 1 | -1(3x/2) | 1-3x/2
H2 | 2 | -x | 2-x
NH3 | 0 | +0.8 | 0.8
Using the law of chemical equilibrium, we can write:
Kc = [NH3]^2/([N2][H2]^3)
At equilibrium, we have:
Kc = (0.8)^2/[(1-3x/2)(2-x)^3]
Simplifying the expression, we get:
Kc = 0.1067/(1.125-6.375x+10.125x^2-6x^3+x^4)
Assuming that x is small compared to 2, we can neglect the terms containing x^2 and x^3. Then, we have:
Kc ≈ 0.1067/(1.125-6.375x+x^4)
At equilibrium, the concentration of NH3 is 0.8/1 = 0.8 M. Using the stoichiometry of the reaction, we can write:
[H2] = (2-x)/1 = 2-x M
Plugging in these values into the expression for Kc, we get:
0.1067 = (0.8)^2/[(1-3x/2)(2-x)^3]
Solving for x, we get:
x = 0.4
Therefore, the concentration of H in the vessel is:
[H2] = 2-x = 2-0.4 = 1.6 M
Answer: Option B. 0.8 mole
- 1 mole of N
- 2 moles of H
- 0.8 mole of NH formed at equilibrium
To find:
- Concentration of H in the vessel
Solution:
The balanced chemical equation for the reaction is:
N2(g) + 3H2(g) ⇌ 2NH3(g)
From the equation, we can see that:
- 1 mole of N reacts with 3 moles of H
- 2 moles of H react with 2/3 mole of N
Let x be the number of moles of H that reacted to form 0.8 mole of NH. Then, we have:
- 0.8 mole of NH formed from x mole of H
- 1 mole of N reacted with 3x/2 moles of H
At equilibrium, the number of moles of each substance present in the vessel can be calculated using the following table:
Substance | Initial moles | Change | Equilibrium moles
--- | --- | --- | ---
N2 | 1 | -1(3x/2) | 1-3x/2
H2 | 2 | -x | 2-x
NH3 | 0 | +0.8 | 0.8
Using the law of chemical equilibrium, we can write:
Kc = [NH3]^2/([N2][H2]^3)
At equilibrium, we have:
Kc = (0.8)^2/[(1-3x/2)(2-x)^3]
Simplifying the expression, we get:
Kc = 0.1067/(1.125-6.375x+10.125x^2-6x^3+x^4)
Assuming that x is small compared to 2, we can neglect the terms containing x^2 and x^3. Then, we have:
Kc ≈ 0.1067/(1.125-6.375x+x^4)
At equilibrium, the concentration of NH3 is 0.8/1 = 0.8 M. Using the stoichiometry of the reaction, we can write:
[H2] = (2-x)/1 = 2-x M
Plugging in these values into the expression for Kc, we get:
0.1067 = (0.8)^2/[(1-3x/2)(2-x)^3]
Solving for x, we get:
x = 0.4
Therefore, the concentration of H in the vessel is:
[H2] = 2-x = 2-0.4 = 1.6 M
Answer: Option B. 0.8 mole
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1 mole of N₂ and 2 moles of H₂ are allowed to react in a 1 dm3 vessel. At equilibrium 0.8 mole of NH₃ is formed. The concentration of H₂ in the vessel isa)0.6 moleb)0.8 molec)0.2 moled)0.4 moleCorrect answer is option 'B'. Can you explain this answer?
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1 mole of N₂ and 2 moles of H₂ are allowed to react in a 1 dm3 vessel. At equilibrium 0.8 mole of NH₃ is formed. The concentration of H₂ in the vessel isa)0.6 moleb)0.8 molec)0.2 moled)0.4 moleCorrect answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 1 mole of N₂ and 2 moles of H₂ are allowed to react in a 1 dm3 vessel. At equilibrium 0.8 mole of NH₃ is formed. The concentration of H₂ in the vessel isa)0.6 moleb)0.8 molec)0.2 moled)0.4 moleCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1 mole of N₂ and 2 moles of H₂ are allowed to react in a 1 dm3 vessel. At equilibrium 0.8 mole of NH₃ is formed. The concentration of H₂ in the vessel isa)0.6 moleb)0.8 molec)0.2 moled)0.4 moleCorrect answer is option 'B'. Can you explain this answer?.
1 mole of N₂ and 2 moles of H₂ are allowed to react in a 1 dm3 vessel. At equilibrium 0.8 mole of NH₃ is formed. The concentration of H₂ in the vessel isa)0.6 moleb)0.8 molec)0.2 moled)0.4 moleCorrect answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 1 mole of N₂ and 2 moles of H₂ are allowed to react in a 1 dm3 vessel. At equilibrium 0.8 mole of NH₃ is formed. The concentration of H₂ in the vessel isa)0.6 moleb)0.8 molec)0.2 moled)0.4 moleCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1 mole of N₂ and 2 moles of H₂ are allowed to react in a 1 dm3 vessel. At equilibrium 0.8 mole of NH₃ is formed. The concentration of H₂ in the vessel isa)0.6 moleb)0.8 molec)0.2 moled)0.4 moleCorrect answer is option 'B'. Can you explain this answer?.
Solutions for 1 mole of N₂ and 2 moles of H₂ are allowed to react in a 1 dm3 vessel. At equilibrium 0.8 mole of NH₃ is formed. The concentration of H₂ in the vessel isa)0.6 moleb)0.8 molec)0.2 moled)0.4 moleCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for NEET.
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1 mole of N₂ and 2 moles of H₂ are allowed to react in a 1 dm3 vessel. At equilibrium 0.8 mole of NH₃ is formed. The concentration of H₂ in the vessel isa)0.6 moleb)0.8 molec)0.2 moled)0.4 moleCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for 1 mole of N₂ and 2 moles of H₂ are allowed to react in a 1 dm3 vessel. At equilibrium 0.8 mole of NH₃ is formed. The concentration of H₂ in the vessel isa)0.6 moleb)0.8 molec)0.2 moled)0.4 moleCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of 1 mole of N₂ and 2 moles of H₂ are allowed to react in a 1 dm3 vessel. At equilibrium 0.8 mole of NH₃ is formed. The concentration of H₂ in the vessel isa)0.6 moleb)0.8 molec)0.2 moled)0.4 moleCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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