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A definite amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. NH4HS decomposes to give NH3 and H2S and at equilibrium total pressure in flask is 0.84 atm. The equilibrium constant for the reaction is :
- a)0.30
- b)0.18
- c)0.17
- d)0.11
Correct answer is option 'D'. Can you explain this answer?
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A definite amount of solid NH4HS is placed in a flask already containi...
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A definite amount of solid NH4HS is placed in a flask already containi...
Given:
Initial pressure of ammonia gas = 0.50 atm
Equilibrium pressure = 0.84 atm
Equilibrium constant (K) = ?
Reaction: NH4HS (s) ⇋ NH3 (g) + H2S (g)
To find the equilibrium constant (K) for the given reaction, we need to use the following equation:
Kp = (P(NH3) × P(H2S)) / P(NH4HS)
Where,
Kp = equilibrium constant in terms of partial pressures
P(NH3) = partial pressure of NH3 at equilibrium
P(H2S) = partial pressure of H2S at equilibrium
P(NH4HS) = partial pressure of NH4HS at equilibrium
Let x be the amount of NH4HS that decomposes at equilibrium.
The concentration of NH3 and H2S at equilibrium will be equal to x (since one mole of NH4HS gives one mole each of NH3 and H2S upon decomposition).
Therefore,
P(NH3) = x / (V - x)
P(H2S) = x / (V - x)
P(NH4HS) = (n - x) / V
Where,
V = volume of the flask
n = initial amount of NH4HS
At equilibrium, the total pressure in the flask is given by:
P(total) = P(NH3) + P(H2S) + P(NH4HS)
Substituting the above equations in the equation for Kp, we get:
Kp = [(x / (V - x))^2] / ((n - x) / V)
Simplifying,
Kp = (x^2 / (V - x)^2) × (V / (n - x))
Kp = x^2 / (n - x)(V - x)
Now, we know that at equilibrium,
P(total) = 0.84 atm
So,
P(NH3) + P(H2S) = 0.84 - P(NH4HS)
Substituting the values of P(NH3), P(H2S), and P(NH4HS) from above, we get:
(x / (V - x)) + (x / (V - x)) = 0.84 - ((n - x) / V)
2x / (V - x) = 0.84 - ((n - x) / V)
Simplifying,
2xV = 0.84V(V - x) - (n - x)(2x - V)
2xV = 0.84V^2 - 0.84Vx - 2x^2 + Vx + nx - nx + x^2
2xV = 0.84V^2 - 0.16Vx - x^2
Rearranging and simplifying,
x^2 + 0.16Vx - 0.84V^2 + 2xV = 0
Solving for x using the quadratic formula, we get:
x = 0.184 moles
Substituting this value in the equation for Kp, we get:
Kp
Initial pressure of ammonia gas = 0.50 atm
Equilibrium pressure = 0.84 atm
Equilibrium constant (K) = ?
Reaction: NH4HS (s) ⇋ NH3 (g) + H2S (g)
To find the equilibrium constant (K) for the given reaction, we need to use the following equation:
Kp = (P(NH3) × P(H2S)) / P(NH4HS)
Where,
Kp = equilibrium constant in terms of partial pressures
P(NH3) = partial pressure of NH3 at equilibrium
P(H2S) = partial pressure of H2S at equilibrium
P(NH4HS) = partial pressure of NH4HS at equilibrium
Let x be the amount of NH4HS that decomposes at equilibrium.
The concentration of NH3 and H2S at equilibrium will be equal to x (since one mole of NH4HS gives one mole each of NH3 and H2S upon decomposition).
Therefore,
P(NH3) = x / (V - x)
P(H2S) = x / (V - x)
P(NH4HS) = (n - x) / V
Where,
V = volume of the flask
n = initial amount of NH4HS
At equilibrium, the total pressure in the flask is given by:
P(total) = P(NH3) + P(H2S) + P(NH4HS)
Substituting the above equations in the equation for Kp, we get:
Kp = [(x / (V - x))^2] / ((n - x) / V)
Simplifying,
Kp = (x^2 / (V - x)^2) × (V / (n - x))
Kp = x^2 / (n - x)(V - x)
Now, we know that at equilibrium,
P(total) = 0.84 atm
So,
P(NH3) + P(H2S) = 0.84 - P(NH4HS)
Substituting the values of P(NH3), P(H2S), and P(NH4HS) from above, we get:
(x / (V - x)) + (x / (V - x)) = 0.84 - ((n - x) / V)
2x / (V - x) = 0.84 - ((n - x) / V)
Simplifying,
2xV = 0.84V(V - x) - (n - x)(2x - V)
2xV = 0.84V^2 - 0.84Vx - 2x^2 + Vx + nx - nx + x^2
2xV = 0.84V^2 - 0.16Vx - x^2
Rearranging and simplifying,
x^2 + 0.16Vx - 0.84V^2 + 2xV = 0
Solving for x using the quadratic formula, we get:
x = 0.184 moles
Substituting this value in the equation for Kp, we get:
Kp
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A definite amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. NH4HS decomposes to give NH3and H2S and at equilibrium total pressure in flask is 0.84 atm. The equilibrium constant for the reaction is :a)0.30b)0.18c)0.17d)0.11Correct answer is option 'D'. Can you explain this answer?
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A definite amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. NH4HS decomposes to give NH3and H2S and at equilibrium total pressure in flask is 0.84 atm. The equilibrium constant for the reaction is :a)0.30b)0.18c)0.17d)0.11Correct answer is option 'D'. 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 A definite amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. NH4HS decomposes to give NH3and H2S and at equilibrium total pressure in flask is 0.84 atm. The equilibrium constant for the reaction is :a)0.30b)0.18c)0.17d)0.11Correct answer is option 'D'. 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 A definite amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. NH4HS decomposes to give NH3and H2S and at equilibrium total pressure in flask is 0.84 atm. The equilibrium constant for the reaction is :a)0.30b)0.18c)0.17d)0.11Correct answer is option 'D'. Can you explain this answer?.
A definite amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. NH4HS decomposes to give NH3and H2S and at equilibrium total pressure in flask is 0.84 atm. The equilibrium constant for the reaction is :a)0.30b)0.18c)0.17d)0.11Correct answer is option 'D'. 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 A definite amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. NH4HS decomposes to give NH3and H2S and at equilibrium total pressure in flask is 0.84 atm. The equilibrium constant for the reaction is :a)0.30b)0.18c)0.17d)0.11Correct answer is option 'D'. 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 A definite amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. NH4HS decomposes to give NH3and H2S and at equilibrium total pressure in flask is 0.84 atm. The equilibrium constant for the reaction is :a)0.30b)0.18c)0.17d)0.11Correct answer is option 'D'. Can you explain this answer?.
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