Consider the decomposition of solid NH4HS in a flask containing NH3(g) at a pressure of 2 atm. What will be the partial pressure of NH3(g) and H2S(g) after the equilibrium has been attained? KP for the reaction is 3.(A)p(NH3) = 4 atm, p(H2S) = 2 atm (B) p(NH3) = 1.732 atm, p(H2S)= 1.732 atm(C)p(NH3) = 3 atm, p(H2S)= 1 atm (D)p(NH3)= 1 atm, p(H2S)= 1 atm?

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Consider the decomposition of solid NH4HS in a flask containing NH3(g) at a pressure of 2 atm. What will be the partial pressure of NH3(g) and H2S(g) after the equilibrium has been attained? KP for the reaction is 3.
(A)p(NH3) = 4 atm, p(H2S) = 2 atm
(B) p(NH3) = 1.732 atm, p(H2S)= 1.732 atm
(C)p(NH3) = 3 atm, p(H2S)= 1 atm
(D)p(NH3)= 1 atm, p(H2S)= 1 atm?

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Consider the decomposition of solid NH4HS in a flask containing NH3(g)...

Given:
- Decomposition of solid NH4HS in a flask containing NH3(g) at a pressure of 2 atm.
- KP for the reaction is 3.

To Find:
- Partial pressure of NH3(g) and H2S(g) after the equilibrium has been attained.

Solution:

The decomposition of solid NH4HS can be represented by the equation:
NH4HS(s) ⇌ NH3(g) + H2S(g)

According to the given information, the initial pressure of NH3(g) is 2 atm.
Let's assume that x atm of NH3(g) reacts to form x atm of H2S(g).

Step 1: Writing the equilibrium expression:
KP = (p(NH3) * p(H2S)) / p(NH4HS)

Given that KP = 3, we can substitute the initial pressure of NH3(g) (2 atm) into the equation:
3 = (2 - x) * x / (p(NH4HS))

Step 2: Solving the equation:
To solve the equation, we rearrange it:
3 = (2 - x) * x / (p(NH4HS))
3 * p(NH4HS) = x^2 - 2x
x^2 - 2x - 3 * p(NH4HS) = 0

Step 3: Finding the value of x:
Using the quadratic formula, we can solve for x:
x = [-(-2) ± √((-2)^2 - 4 * 1 * (-3 * p(NH4HS))))] / (2 * 1)
x = [2 ± √(4 + 12 * p(NH4HS))] / 2
x = 1 ± √(1 + 3 * p(NH4HS))

Step 4: Determining the partial pressures:
Since x represents the pressure of H2S(g), we can write:
p(H2S) = 1 ± √(1 + 3 * p(NH4HS))

Given that p(NH4HS) = 2 atm, we can substitute this value into the equation:
p(H2S) = 1 ± √(1 + 3 * 2)
p(H2S) = 1 ± √(1 + 6)
p(H2S) = 1 ± √7

Therefore, the partial pressure of NH3(g) is p(NH3) = 2 - x = 2 - (1 ± √7).

Answer:
The partial pressure of NH3(g) and H2S(g) after the equilibrium has been attained is:
(A) p(NH3) = 4 atm, p(H2S) = 2 atm

Answered by Infinity Academy · View profile

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Consider the decomposition of solid NH4HS in a flask containing NH3(g) at a pressure of 2 atm. What will be the partial pressure of NH3(g) and H2S(g) after the equilibrium has been attained? KP for the reaction is 3.(A)p(NH3) = 4 atm, p(H2S) = 2 atm (B) p(NH3) = 1.732 atm, p(H2S)= 1.732 atm(C)p(NH3) = 3 atm, p(H2S)= 1 atm (D)p(NH3)= 1 atm, p(H2S)= 1 atm?

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Consider the decomposition of solid NH4HS in a flask containing NH3(g) at a pressure of 2 atm. What will be the partial pressure of NH3(g) and H2S(g) after the equilibrium has been attained? KP for the reaction is 3.(A)p(NH3) = 4 atm, p(H2S) = 2 atm (B) p(NH3) = 1.732 atm, p(H2S)= 1.732 atm(C)p(NH3) = 3 atm, p(H2S)= 1 atm (D)p(NH3)= 1 atm, p(H2S)= 1 atm? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Consider the decomposition of solid NH4HS in a flask containing NH3(g) at a pressure of 2 atm. What will be the partial pressure of NH3(g) and H2S(g) after the equilibrium has been attained? KP for the reaction is 3.(A)p(NH3) = 4 atm, p(H2S) = 2 atm (B) p(NH3) = 1.732 atm, p(H2S)= 1.732 atm(C)p(NH3) = 3 atm, p(H2S)= 1 atm (D)p(NH3)= 1 atm, p(H2S)= 1 atm? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider the decomposition of solid NH4HS in a flask containing NH3(g) at a pressure of 2 atm. What will be the partial pressure of NH3(g) and H2S(g) after the equilibrium has been attained? KP for the reaction is 3.(A)p(NH3) = 4 atm, p(H2S) = 2 atm (B) p(NH3) = 1.732 atm, p(H2S)= 1.732 atm(C)p(NH3) = 3 atm, p(H2S)= 1 atm (D)p(NH3)= 1 atm, p(H2S)= 1 atm?.

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