Kp=1.6
Kp=1.0*10^-4
A<->2B
Find the equilibrium partial pressures of A and B
B=1.0 atm
A=0.0 atm
This is what I have so far
A 2B
0 1.0
+p -p
p 1-p
[B]^2/A (1-p)^2/p = 1.6
then got a quadratic equation of p^2+1.6p-1 to get x=.48
To find the equilibrium partial pressures of A and B in the given reaction A <=> 2B, where B is at a partial pressure of 1.0 atm and A is at a partial pressure of 0.0 atm, you need to use the equilibrium constant expression for this reaction.
The equilibrium constant expression (Kp) for the given reaction is:
Kp = ([B]^2) / [A]
Given that Kp = 1.6, you can substitute the partial pressure values for B and A into the equation:
1.6 = (1.0)^2 / [A]
Simplifying the equation, you get:
1.6 = 1 / [A]
Multiply both sides of the equation by [A] to solve for [A]:
1.6 * [A] = 1
[A] = 1 / 1.6
[A] ≈ 0.625 atm
So the equilibrium partial pressure of A is approximately 0.625 atm.
To find the equilibrium partial pressure of B, you can use the stoichiometry of the reaction, which tells you that the ratio of the partial pressures of A and B is 1:2. Since the partial pressure of A is 0.625 atm, the partial pressure of B can be calculated as follows:
[B] = 2 * [A]
[B] = 2 * 0.625
[B] = 1.25 atm
Therefore, the equilibrium partial pressure of B is 1.25 atm.
To summarize, at equilibrium, the partial pressure of A is approximately 0.625 atm and the partial pressure of B is 1.25 atm.