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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...
Equilibrium Constant Expression:
The equilibrium constant expression for the reaction N₂ + 3H₂ ⇌ 2NH₃ can be written as:
Kc = [NH₃]² / [N₂][H₂]³
Given Information:
- 1 mole of N and 2 moles of H are allowed to react in a 1 dm³ vessel.
- At equilibrium, 0.8 mole of NH₃ is formed.
Calculating Concentration of NH₃:
The concentration of NH₃ can be calculated using the given moles and volume:
Volume of vessel = 1 dm³ = 1000 cm³
Concentration of NH₃ = (0.8 mole) / (1000 cm³)
Concentration of NH₃ = 0.0008 mole/cm³
Calculating Concentration of N₂ and H₂:
Since the moles of N and H are not given, we need to calculate their concentrations using the balanced equation.
At equilibrium, 1 mole of N₂ reacts to form 2 moles of NH₃.
Therefore, the concentration of N₂ = (1 - 0.8) mole / (1000 cm³)
Concentration of N₂ = 0.2 mole/cm³
At equilibrium, 2 moles of H₂ reacts to form 2 moles of NH₃.
Therefore, the concentration of H₂ = (2 - (2 * 0.8)) mole / (1000 cm³)
Concentration of H₂ = 0.4 mole/cm³
Answer:
The concentration of H₂ in the vessel is 0.4 mole/cm³. Hence, option 'B' is the correct answer.
The equilibrium constant expression for the reaction N₂ + 3H₂ ⇌ 2NH₃ can be written as:
Kc = [NH₃]² / [N₂][H₂]³
Given Information:
- 1 mole of N and 2 moles of H are allowed to react in a 1 dm³ vessel.
- At equilibrium, 0.8 mole of NH₃ is formed.
Calculating Concentration of NH₃:
The concentration of NH₃ can be calculated using the given moles and volume:
Volume of vessel = 1 dm³ = 1000 cm³
Concentration of NH₃ = (0.8 mole) / (1000 cm³)
Concentration of NH₃ = 0.0008 mole/cm³
Calculating Concentration of N₂ and H₂:
Since the moles of N and H are not given, we need to calculate their concentrations using the balanced equation.
At equilibrium, 1 mole of N₂ reacts to form 2 moles of NH₃.
Therefore, the concentration of N₂ = (1 - 0.8) mole / (1000 cm³)
Concentration of N₂ = 0.2 mole/cm³
At equilibrium, 2 moles of H₂ reacts to form 2 moles of NH₃.
Therefore, the concentration of H₂ = (2 - (2 * 0.8)) mole / (1000 cm³)
Concentration of H₂ = 0.4 mole/cm³
Answer:
The concentration of H₂ in the vessel is 0.4 mole/cm³. Hence, option 'B' is the correct 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?
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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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