The initial concentration of PCl5 was 0.1983 M. It was allowed to decompose at some temperature until equilibrium had been attained (see reaction below). What must be the equilibrium concentration of PCl3?

PCl5(g) ↔ PCl3(g) + Cl2(g) Kc = 0.0580

a. 0.1162 M

b. 0.0496 M

c. 0.0486 M

d. 0.0821 M

.......PCl5 --> PCl3 + Cl2

I...0.1983.......0......0
C.......-x.......x......x
E....0.1983-x....x......x

Substitute the E line into the Kc expression for PCl5 and solve for x.

To find the equilibrium concentration of PCl3, we need to use the equilibrium constant (Kc) and the initial concentration of PCl5.

The balanced equation for the reaction is:
PCl5(g) ↔ PCl3(g) + Cl2(g)

The equilibrium constant expression (Kc) for this reaction is:
Kc = [PCl3][Cl2] / [PCl5]

Given:
Initial concentration of PCl5 = 0.1983 M
Kc = 0.0580

Let's represent the equilibrium concentration of PCl3 as x (M).

At equilibrium, the concentration of Cl2 will also be x (M) since one mole of PCl5 produces one mole of Cl2.

The concentration of PCl5 at equilibrium will be the initial concentration minus x since x moles of PCl5 is consumed to form x moles of PCl3.

Using the equilibrium constant expression, we can write:
0.0580 = (x)(x) / (0.1983 - x)

Rearranging the equation:
0.0580 * (0.1983 - x) = x^2

Expanding and simplifying:
0.0113274 - 0.0580x = x^2

Rearranging further:
x^2 + 0.0580x - 0.0113274 = 0

Now we can solve this quadratic equation to find the value of x (equilibrium concentration of PCl3).

Using a quadratic solver or equation solver software, we find that x ≈ 0.0496 M.

Therefore, the equilibrium concentration of PCl3 is approximately 0.0496 M.

The correct answer is (b) 0.0496 M.