A concentration of PCL5(g),4,0 mol/L, verifies that the process PCl5(g) = PCl3(g) + Cl2(g)

is achieved with a degree of balance of 20% Calculate the equilibrium of the reaction constant

This is a very oddly worded question and I'm not sure what the question is. I will restate what I think it is.

A concn of PCl5 at equilibrium is 4 mols/L. The reaction yield was 20%. Calculate Keq.

Let y = initial concentration and x = change as equilibrium is established.

.......PCl5 ==> PCl3 + Cl2
I......y.........0......0
C.....-x.........x......x
E.....y-x........x......x

The problem, as I've worded it, states that y-x = 4 and the 4 is 20% of the initial concentration. So y must be 20 (4/0.2 = 20)
Therefore y-x = 4 = (PCl5)
x = 16 = PCl3 = Cl2

Plug those values into the Keq constant expression and solve for Keq.

Thank you :D

To calculate the equilibrium constant (K) for the given reaction, we need to use the concentrations of the reactants and products at equilibrium. The equilibrium constant is determined using the following general expression:

K = [C]^c [D]^d / [A]^a [B]^b

Where A, B, C, and D are the reactants and products in the chemical equation, and a, b, c, and d are their respective coefficients.

For the given reaction: PCl5(g) = PCl3(g) + Cl2(g)

The concentrations at equilibrium can be written as follows:

[PCl5] = 4.0 mol/L (given)
[PCl3] = concentration at equilibrium (unknown)
[Cl2] = concentration at equilibrium (unknown)

Now, since the degree of balance is 20%, it means that only 20% of the initial PCl5 is converted into product at equilibrium. This implies that only 20% of [PCl5] is consumed, leaving 80% still present at equilibrium.

We can then calculate the concentrations at equilibrium as follows:

[PCl3] = 0.2 * [PCl5] = 0.2 * 4.0 mol/L = 0.8 mol/L
[Cl2] = 0.2 * [PCl5] = 0.2 * 4.0 mol/L = 0.8mol/L

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

K = [PCl3]^1 [Cl2]^1 / [PCl5]^1

K = (0.8)^1 * (0.8)^1 / (4.0)^1

K = 0.64 / 4.0

K ≈ 0.16 mol/L

Therefore, the equilibrium constant (K) for the given reaction is approximately 0.16 mol/L.