Consider the reaction:

A(g)⇌2B(g)
Find the equilibrium partial pressures of A and B for each of the different values of Kp. Assume that the initial partial pressure of B in each case is 1.0 atm and that the initial partial pressure of A is 0.0 atm. Make any appropriate simplifying assumptions.

A) Kp= 3.0
B)Kp= 1.2×10^−4
C) Kp= 1.0×10^5

I know this is posted with different numbers on this site and that it is a quadratic but I can't get the right answer. Please help.

To solve this problem, we can start by writing the equilibrium expression for the given reaction:

Kp = (P_B)^2 / (P_A)

where Kp is the equilibrium constant, P_B is the partial pressure of B, and P_A is the partial pressure of A.

Now, let's solve for the equilibrium partial pressures of A and B for each case:

A) Kp = 3.0

Given: initial P_B = 1.0 atm, initial P_A = 0.0 atm

Substituting the given values into the equilibrium expression, we get:

3.0 = (1.0)^2 / P_A

Simplifying the equation, we find:

3.0 * P_A = 1.0

P_A = 1.0 / 3.0

P_A ≈ 0.333 atm

Since the initial partial pressure of A is 0.0 atm, and the reaction favors the formation of B, we can assume that the equilibrium partial pressure of B is nearly equal to the initial partial pressure of B:

P_B ≈ 1.0 atm

Therefore, the equilibrium partial pressures are approximately:

P_A ≈ 0.333 atm
P_B ≈ 1.0 atm

B) Kp = 1.2 × 10^(-4)

Given: initial P_B = 1.0 atm, initial P_A = 0.0 atm

Substituting the given values into the equilibrium expression, we get:

1.2 × 10^(-4) = (1.0)^2 / P_A

Simplifying the equation, we find:

1.2 × 10^(-4) * P_A = 1.0

P_A = 1.0 / (1.2 × 10^(-4))

P_A ≈ 8.33 × 10^3 atm

Since the initial partial pressure of A is 0.0 atm, and the reaction favors the formation of B, we can assume that the equilibrium partial pressure of B is nearly equal to the initial partial pressure of B:

P_B ≈ 1.0 atm

Therefore, the equilibrium partial pressures are approximately:

P_A ≈ 8.33 × 10^3 atm
P_B ≈ 1.0 atm

C) Kp = 1.0 × 10^(5)

Given: initial P_B = 1.0 atm, initial P_A = 0.0 atm

Substituting the given values into the equilibrium expression, we get:

1.0 × 10^(5) = (1.0)^2 / P_A

Simplifying the equation, we find:

1.0 × 10^(5) * P_A = 1.0

P_A = 1.0 / (1.0 × 10^(5))

P_A = 1.0 × 10^(-5) atm

Since the initial partial pressure of A is 0.0 atm, and the reaction favors the formation of B, we can assume that the equilibrium partial pressure of B is nearly equal to the initial partial pressure of B:

P_B ≈ 1.0 atm

Therefore, the equilibrium partial pressures are approximately:

P_A ≈ 1.0 × 10^(-5) atm
P_B ≈ 1.0 atm

In summary, the equilibrium partial pressures of A and B for each given value of Kp are:

A) Kp = 3.0: P_A ≈ 0.333 atm, P_B ≈ 1.0 atm
B) Kp = 1.2 × 10^(-4): P_A ≈ 8.33 × 10^3 atm, P_B ≈ 1.0 atm
C) Kp = 1.0 × 10^(5): P_A ≈ 1.0 × 10^(-5) atm, P_B ≈ 1.0 atm

To determine the equilibrium partial pressures of A and B for each value of Kp, we will assume that the initial partial pressure of B is 1.0 atm, and the initial partial pressure of A is 0.0 atm. We'll use the given reaction:

A(g) ⇌ 2B(g)

Let's assign variables to the equilibrium partial pressures:
P(A) = the partial pressure of A at equilibrium
P(B) = the partial pressure of B at equilibrium

Now, let's analyze each case separately:

A) Kp = 3.0
According to the given reaction, Kp = (P(B))^2 / P(A)
Since the initial partial pressure of A is 0.0 atm, we have:
3.0 = (1.0)^2 / P(A)
P(A) = (1.0)^2 / 3.0 = 0.333 atm
Therefore, at equilibrium, the partial pressures are:
P(A) = 0.333 atm
P(B) = 1.0 atm

B) Kp = 1.2 × 10^(-4)
Using the same approach as before:
1.2 × 10^(-4) = (1.0)^2 / P(A)
P(A) = (1.0)^2 / (1.2 × 10^(-4)) = 8.33 × 10^3 atm (approximately)
Therefore, at equilibrium, the partial pressures are:
P(A) = 8.33 × 10^3 atm (approximately)
P(B) = 1.0 atm

C) Kp = 1.0 × 10^5
Once again, using the same approach:
1.0 × 10^5 = (1.0)^2 / P(A)
P(A) = (1.0)^2 / (1.0 × 10^5) = 1.0 × 10^(-5) atm
Therefore, at equilibrium, the partial pressures are:
P(A) = 1.0 × 10^(-5) atm
P(B) = 1.0 atm

Remember to always double-check your calculations, as errors can occur during the process. If you're not achieving the correct answer, it's possible that there may be a calculation mistake.

If you're worked this several times with no luck I assume that means you are checking with a database. Why not show your work and let us find the error? It's much easier that way instead of us guessing what you have done right and what you have done wrong.