At a certain temperature, the Kp for the decomposition of H2S is 0.755
H2S (g) <-- --> H2 (g) + S (g)
Initially, only H2S is present at a pressue of 0.201 atm in a closed container. what is the total pressure in the container at equilibrium?
To calculate the total pressure at equilibrium, we first need to determine the equilibrium concentrations of H2 and S. Since we know the initial pressure of H2S, we can assume that at the start of the reaction, the pressure of H2 and S is zero.
Let's assume that the change in pressure for H2S is -x (as it is being consumed), and the change in pressure for H2 and S is +x (as they are being produced).
Using the ideal gas law, we can determine the equilibrium concentrations of the gases in terms of x:
PH2 = x
PS = x
Now, we can use the expression for Kp (the equilibrium constant in terms of partial pressures) to solve for x:
Kp = (PH2 * PS) / PH2S
Substituting the given values:
0.755 = (x * x) / (0.201 - x)
Now, solve this quadratic equation for x:
0.755(0.201 - x) = x^2
0.152055 - 0.755x = x^2
Rearranging the equation:
x^2 + 0.755x - 0.152055 = 0
Using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
where a = 1, b = 0.755, and c = -0.152055
Solving this equation gives us two possible values for x: x1 and x2.
Now, we need to determine which value of x is valid. In this case, since we are dealing with the decomposition of H2S, the concentration of H2S will decrease while the concentrations of H2 and S will increase. Therefore, we can discard the negative value of x (since it would imply negative concentrations), and keep the positive value.
Now that we have the value of x, we can calculate the equilibrium concentrations of H2 and S:
PH2 = x
PS = x
The total pressure at equilibrium can be calculated by adding the pressures of each gas:
Total pressure = PH2S + PH2 + PS
Substituting the values:
Total pressure = 0.201 atm (initial pressure of H2S) + x (pressure of H2) + x (pressure of S)
Finally, we can substitute the value of x into the equation to find the total pressure at equilibrium.