Phosphorus pentachloride decomposes according to the chemical equation pcl5(g)↽−−⇀pcl3(g)+cl2(g)kc=1.80 at 250 ∘c a 0.157 mol sample of pcl5(g) is injected into an empty 2.50 l reaction vessel held at 250 ∘c. calculate the concentrations of pcl5(g) and pcl3(g) at equilibrium.

The concentration of PCl3 is 0.061 mol/L and for PCl5 is 0.002 mol/L

Explanation:

The initial concentration of PCl5 is

n/V = 0.157/2.5 = 0.063 mol/L

During the reaction x mol/L is consumed, and x mol/L is formed of each product (the stoichiometry is 1 mol : 1 mol : 1 mol).

So, doing a table of reaction

                   PCl5                  PCl3               Cl2

initial           0.063                    0                    0

reacted       -x                           +x                   +x

equilibrium  0.063 - x              x                     x

For a reaction aA + bB ↽−−⇀ cC +dD, the equilibrium constant is given by:

All these concentration are the concentration of equilibrium. So, for the reaction:

x² = 0.1134 - 1.8x

x² + 1.8x - 0.1134 = 0

Using Bhaskara, with a = 1, b = 1.8, and c = -0.1134

Δ = b² - 4ac = (1.8)² - 4*1*(-0.1134) = 3.6936

x is a molar concentration, so it must be positive, then:

x = (-1.8 + √3.6936)/2

x = 0.061 mol/L

Which is the concentration of PCl3. The concentration of PCl5 is

0.063 - 0.061 = 0.002 mol/L

c most likely

hop this let me know if it is wrong

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