Class 11 Exam > Class 11 Questions > An amount of solid NH4HS is placed in a flask...
Start Learning for Free
An amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH3 and H2S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm ? The equilibrium constant for NH4HS decomposition at this temperature is [AIEEE-2005]
- a)0.18
- b)0.30
- c)0.11
- d)0.17
Correct answer is option 'C'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
An amount of solid NH4HS is placed in a flask already containing ammon...
Most Upvoted Answer
An amount of solid NH4HS is placed in a flask already containing ammon...
NH4HS decomposition reaction: NH4HS ⇌ NH3 + H2S
To solve this problem, we can use the ideal gas law and the concept of partial pressures to determine the equilibrium constant (K) for the reaction.
1. Write the balanced chemical equation:
NH4HS ⇌ NH3 + H2S
2. Define the initial and equilibrium pressures:
Initial pressure of NH4HS = 0.50 atm
Equilibrium pressure = 0.84 atm
3. Use the ideal gas law to calculate the initial and equilibrium moles of NH4HS:
PV = nRT
Initial moles of NH4HS:
n = (0.50 atm) / (R x T)
Equilibrium moles of NH4HS:
n = (0.84 atm) / (R x T)
4. Calculate the change in moles of NH4HS:
Δn = n(eq) - n(initial)
5. Use the balanced chemical equation to relate the change in moles of NH4HS to the moles of NH3 and H2S produced:
Δn(NH4HS) = -Δn(NH3) = -Δn(H2S)
6. Calculate the equilibrium pressures of NH3 and H2S:
P(NH3) = Δn(NH3) x (R x T) / V
P(H2S) = Δn(H2S) x (R x T) / V
7. Calculate the equilibrium constant (K) using the partial pressures:
K = (P(NH3) x P(H2S)) / P(NH4HS)
Now let's calculate the equilibrium constant (K):
- Assume the volume (V) of the flask is 1 L for simplicity.
- The value of the ideal gas constant (R) is 0.0821 L.atm/mol.K.
- The temperature (T) is constant.
1. Calculate the initial and equilibrium moles of NH4HS:
n(initial) = (0.50 atm) / (0.0821 L.atm/mol.K x T)
n(eq) = (0.84 atm) / (0.0821 L.atm/mol.K x T)
2. Calculate the change in moles of NH4HS:
Δn(NH4HS) = n(eq) - n(initial)
3. Calculate the equilibrium pressures of NH3 and H2S:
P(NH3) = -Δn(NH3) x (0.0821 L.atm/mol.K x T) / 1 L
P(H2S) = -Δn(H2S) x (0.0821 L.atm/mol.K x T) / 1 L
4. Calculate the equilibrium constant (K):
K = (P(NH3) x P(H2S)) / P(NH4HS)
By substituting the values into the equation and simplifying, we get the equilibrium constant:
K = (P(NH3) x P(H2S)) / P(NH4HS)
K = [(-Δn(NH3) x (0.0821 L.atm/mol.K x T) / 1 L) x (-Δn(H2S) x (0.0821 L.atm/mol.K x T) / 1 L)] /
To solve this problem, we can use the ideal gas law and the concept of partial pressures to determine the equilibrium constant (K) for the reaction.
1. Write the balanced chemical equation:
NH4HS ⇌ NH3 + H2S
2. Define the initial and equilibrium pressures:
Initial pressure of NH4HS = 0.50 atm
Equilibrium pressure = 0.84 atm
3. Use the ideal gas law to calculate the initial and equilibrium moles of NH4HS:
PV = nRT
Initial moles of NH4HS:
n = (0.50 atm) / (R x T)
Equilibrium moles of NH4HS:
n = (0.84 atm) / (R x T)
4. Calculate the change in moles of NH4HS:
Δn = n(eq) - n(initial)
5. Use the balanced chemical equation to relate the change in moles of NH4HS to the moles of NH3 and H2S produced:
Δn(NH4HS) = -Δn(NH3) = -Δn(H2S)
6. Calculate the equilibrium pressures of NH3 and H2S:
P(NH3) = Δn(NH3) x (R x T) / V
P(H2S) = Δn(H2S) x (R x T) / V
7. Calculate the equilibrium constant (K) using the partial pressures:
K = (P(NH3) x P(H2S)) / P(NH4HS)
Now let's calculate the equilibrium constant (K):
- Assume the volume (V) of the flask is 1 L for simplicity.
- The value of the ideal gas constant (R) is 0.0821 L.atm/mol.K.
- The temperature (T) is constant.
1. Calculate the initial and equilibrium moles of NH4HS:
n(initial) = (0.50 atm) / (0.0821 L.atm/mol.K x T)
n(eq) = (0.84 atm) / (0.0821 L.atm/mol.K x T)
2. Calculate the change in moles of NH4HS:
Δn(NH4HS) = n(eq) - n(initial)
3. Calculate the equilibrium pressures of NH3 and H2S:
P(NH3) = -Δn(NH3) x (0.0821 L.atm/mol.K x T) / 1 L
P(H2S) = -Δn(H2S) x (0.0821 L.atm/mol.K x T) / 1 L
4. Calculate the equilibrium constant (K):
K = (P(NH3) x P(H2S)) / P(NH4HS)
By substituting the values into the equation and simplifying, we get the equilibrium constant:
K = (P(NH3) x P(H2S)) / P(NH4HS)
K = [(-Δn(NH3) x (0.0821 L.atm/mol.K x T) / 1 L) x (-Δn(H2S) x (0.0821 L.atm/mol.K x T) / 1 L)] /
Attention Class 11 Students!
To make sure you are not studying endlessly, EduRev has designed Class 11 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 11.
|
Explore Courses for Class 11 exam
|
|
Similar Class 11 Doubts
Top Courses for Class 11View all
An amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH3 and H2S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm ? The equilibrium constant for NH4HS decomposition at this temperature is [AIEEE-2005] a)0.18b)0.30c)0.11d)0.17Correct answer is option 'C'. Can you explain this answer?
Question Description
An amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH3 and H2S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm ? The equilibrium constant for NH4HS decomposition at this temperature is [AIEEE-2005] a)0.18b)0.30c)0.11d)0.17Correct answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about An amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH3 and H2S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm ? The equilibrium constant for NH4HS decomposition at this temperature is [AIEEE-2005] a)0.18b)0.30c)0.11d)0.17Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH3 and H2S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm ? The equilibrium constant for NH4HS decomposition at this temperature is [AIEEE-2005] a)0.18b)0.30c)0.11d)0.17Correct answer is option 'C'. Can you explain this answer?.
An amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH3 and H2S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm ? The equilibrium constant for NH4HS decomposition at this temperature is [AIEEE-2005] a)0.18b)0.30c)0.11d)0.17Correct answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about An amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH3 and H2S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm ? The equilibrium constant for NH4HS decomposition at this temperature is [AIEEE-2005] a)0.18b)0.30c)0.11d)0.17Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH3 and H2S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm ? The equilibrium constant for NH4HS decomposition at this temperature is [AIEEE-2005] a)0.18b)0.30c)0.11d)0.17Correct answer is option 'C'. Can you explain this answer?.
Solutions for An amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH3 and H2S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm ? The equilibrium constant for NH4HS decomposition at this temperature is [AIEEE-2005] a)0.18b)0.30c)0.11d)0.17Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
Download more important topics, notes, lectures and mock test series for Class 11 Exam by signing up for free.
Here you can find the meaning of An amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH3 and H2S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm ? The equilibrium constant for NH4HS decomposition at this temperature is [AIEEE-2005] a)0.18b)0.30c)0.11d)0.17Correct answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
An amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH3 and H2S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm ? The equilibrium constant for NH4HS decomposition at this temperature is [AIEEE-2005] a)0.18b)0.30c)0.11d)0.17Correct answer is option 'C'. Can you explain this answer?, a detailed solution for An amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH3 and H2S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm ? The equilibrium constant for NH4HS decomposition at this temperature is [AIEEE-2005] a)0.18b)0.30c)0.11d)0.17Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of An amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH3 and H2S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm ? The equilibrium constant for NH4HS decomposition at this temperature is [AIEEE-2005] a)0.18b)0.30c)0.11d)0.17Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice An amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH3 and H2S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm ? The equilibrium constant for NH4HS decomposition at this temperature is [AIEEE-2005] a)0.18b)0.30c)0.11d)0.17Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice Class 11 tests.
|
|
Explore Courses for Class 11 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup