Initally, only H2S is present at a pressure of 0.224 bar in a closed container. What is the total pressure in the container at equilibrium?
H2S(g)<--->H2(g)+S(g)
Initally, only H2S is present at a pressure of 0.224 bar in a closed container. What is the total pressure in the container at equilibrium?
I got .428 Bar but got marked wrong. IDK what I am doing wrong.
the total pressure will be the sum of the partial pressure of all the gases at equilibrium.
To calculate the total pressure at equilibrium, you need to consider the stoichiometry of the reaction and establish the partial pressures of all the species involved.
The balanced equation for the reaction is:
H2S(g) <--> H2(g) + S(g)
Since the initial pressure is given as 0.224 bar, we can denote the initial pressure of H2S as P(H2S) = 0.224 bar.
At equilibrium, the total pressure will be the sum of the partial pressures of all the species. Let's denote the pressure of H2 as P(H2) and the pressure of S as P(S).
At equilibrium, the equilibrium constant expression for the reaction can be written as:
Kp = (P(H2) * P(S)) / P(H2S)
To solve for the equilibrium partial pressures, we need to know the value of the equilibrium constant (Kp) for this reaction. If the value of Kp is given, we can use it to calculate the partial pressures. However, since Kp is not provided in your question, we cannot calculate the partial pressures for H2 and S.
Therefore, without the value of the equilibrium constant, we cannot determine the total pressure in the container at equilibrium. Hence, your answer of 0.428 bar cannot be determined as correct or incorrect without additional information.