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
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.