Question

1

The compound PCI, decomposes into Cl₂ and PCI3. The equilibrium of PCIs(g) = Cl₂(g) + PC3(g) has a Keq of 2.24 x 10-²
at 327°C. What is the equilibrium concentration of Cl₂ in a 1.00 liter vessel containing 0.235 mole of PCls and 0.174 mole
of PCI3? Remember to use the correct number of significant digits.
Cl₂ =
Are the products or reactants favored?

Answer 1

Step-by-step explanation:

The equilibrium constant expression for the given reaction is:

Keq = [Cl₂][PCI₃] / [PCl₃]

We are given the equilibrium constant (Keq) as 2.24 x 10^-2 at 327°C, as well as the initial concentrations of PCl₃ and PCI₃. We need to determine the equilibrium concentration of Cl₂.

Let's assume that the change in concentration of PCl₃ is -x, the change in concentration of Cl₂ is +x, and the change in concentration of PCI₃ is +x, at equilibrium. Then, the equilibrium concentrations are:

[PCl₃] = 0.235 - x

[Cl₂] = x

[PCI₃] = 0.174 + x

Substituting these expressions into the equilibrium constant expression, we get:

2.24 x 10^-2 = (x)(0.174 + x) / (0.235 - x)

Simplifying and rearranging this equation, we get a quadratic equation:

2.24 x 10^-2 (0.235 - x) = x (0.174 + x)

0.00526 - 2.24 x 10^-2 x = 0.174x + x^2

x^2 + 0.138x - 0.00526 = 0

We can solve this quadratic equation to find the value of x:

x = 0.046 M (rounded to 3 significant digits)

Therefore, the equilibrium concentration of Cl₂ is 0.046 M, which is the same as the change in concentration of Cl₂.

Since the equilibrium concentration of Cl₂ is greater than zero, this means that the products (Cl₂ and PCI₃) are favored at equilibrium.