Answer:
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:
[tex]Kc = \frac{[C]^cx[D]^d}{[A]^ax[B]^b}[/tex]
All these concentration are the concentration of equilibrium. So, for the reaction:
[tex]Kc = \frac{[PCl3]x[Cl2]}{[PCl5]}[/tex]
[tex]1.8 = \frac{x*x}{0.063 - x}[/tex]
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
[tex]x = \frac{-b +/- \sqrt{delta} }{2a}[/tex]
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
