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?

Look under your other posts.

To find the total pressure in the container at equilibrium, we can use the concept of the equilibrium constant (Kp) and the given initial pressure of H2S.

The equilibrium constant expression for the given reaction is:

Kp = (P_H2 * P_S) / P_H2S

where P_H2, P_S, and P_H2S represent the partial pressures of H2, S, and H2S respectively.

Given that the equilibrium constant (Kp) is 0.832, and initially only H2S is present at a pressure of 0.121 atm, we can assign the pressure of H2S at equilibrium as P_H2S = 0.121 atm.

Since initially there is no H2 and S, we can assume their pressures to be zero. Therefore, P_H2 = 0 and P_S = 0.

Now, we can substitute these values into the equilibrium constant expression:

0.832 = (0 * 0) / 0.121

Since 0 multiplied by 0 is 0, the equation becomes:

0.832 = 0 / 0.121

Any number divided by zero is undefined. Therefore, the equilibrium constant is not defined in this case. This means that the given reaction does not reach equilibrium.

Since the reaction does not reach equilibrium, the total pressure in the container will still be the initial pressure of H2S, which is 0.121 atm.