[PCl 5 (g)] = 0.203 mol /2.15 L = 0.094 M PCl 5 (g) <====> PCl 3 (g) + Cl 2 (g)

Question

[PCl5(g)] = 0.203 mol /2.15 L
                = 0.094 M

            PCl5(g)     <====>     PCl3(g)     +     Cl2(g)
IC:       0.094                             0                       0
C:         - x                              + x                   + x
EC:   0.094 – x                         x                       x

Kc = [PCl3] [Cl2] / [PCl5]
or, 1.80 = (x) (x) / (0.094 – x)
or, x2 = 1.80 (0.094 – x)
or, x2 = 0.1692 – 1.8x
or, x2 + 1.8x - 0.1692 = 0

Solving the quadratic equation, we get;
x = 0.090 and x = -1.890

Discarding the negative value of x, we get x = 0.090.

So, the equilibrium concentrations are
[PCl5(g)] = 0.094 – x = 0.094 – 0.090 = 0.004 M
[PCl3(g)] = x = 0.090 M
[[Cl2(g)] = x = 0.090 M

Explanation / Answer

Phosphorus pentachloride decomposes according to the chemical equation PCI, g) +Cl Kc = 1.80 at 250°C into an empty 2.15 L reaction vessel held at 250 C. A 0.203 mol sample of PCIs(g) is injected Calculate the concentrations of PCls(g) and PCla(g) at equilibrium.

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.