At a certain temperature, the Kp for the decomposition of H2S is 0.832
H2S (g) <-- --> H2 (g) + S (g)
Initially, only H2S is present at a pressue of 0.121 atm in a closed container. what is the total pressure in the container at equilibrium?
Please tell me how to do it and confirm the right answer please. Thanks
To find the total pressure in the container at equilibrium, we need to determine the partial pressures of H2 and S at equilibrium and then add them together.
Given that the Kp for the decomposition of H2S is 0.832, we can use this value to set up an expression for the equilibrium partial pressures:
Kp = (P_H2 * P_S) / P_H2S
Since we initially have only H2S and no H2 or S, we can assume that the partial pressures of H2 and S at equilibrium will be equal. Let's represent this as "x":
Kp = (x * x) / (0.121 - x)
Simplifying the equation, we can rewrite it as:
0.832 = x^2 / (0.121 - x)
To solve this equation, we can use the quadratic formula. However, before proceeding, we need to check if the value of "x" we obtain is valid by making sure it is less than the initial pressure (0.121 atm).
Now, solving the equation using the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
For our equation, a = 1, b = 0, and c = -0.832:
x = (0 ± sqrt(0^2 - 4(1)(-0.832))) / 2(1)
= (0 ± sqrt(3.328)) / 2
= ±(sqrt(3.328) / 2)
Taking the positive root (as pressure cannot be negative), we have:
x ≈ 0.913
Since the value of "x" is less than the initial pressure, it is valid. Therefore, the equilibrium partial pressure of both H2 and S is approximately 0.913 atm.
Finally, we add the partial pressures of H2 and S to get the total pressure in the container at equilibrium:
Total pressure = Pressure_H2 + Pressure_S = 0.913 + 0.913 = 1.826 atm
Therefore, the total pressure in the container at equilibrium is approximately 1.826 atm.
Please note that this explanation assumes ideal gas behavior and no change in volume or temperature during the reaction.