Question

The equilibrium constant, K, for the following reaction is 1.20×10-2 at 500 K. PCl5(g) PCl3(g) + Cl2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 500 K contains 0.218 M PCl5, 5.11×10-2 M PCl3 and 5.11×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 3.92×10-2 mol of PCl3(g) is added to the flask? [PCl5] = M

[PCl3] = M

[Cl2] = M

Answer 1

Sol.

As 3.92 × 10^-2 mol of PCl3(g) added to flask

Volume of flask = 1 L

So , Concentration of PCl3(g) added to flask

= 3.92 × 10^-2 / 1 = 0.0392 M

New Initial Concentration of PCl3(g)

= 5.11 × 10^-2 + 0.0392 = 0.0903 M

New Initial Concentration of Cl2(g)

= 5.11 × 10^-2 M = 0.0511 M

New Initial Concentration of PCl5(g)

= 0.218 M

Now ,

Reaction : PCl5(g) <-----> PCl3(g) + Cl2(g)

Initial 0.218 0.0903 0.0511

Change + x - x - x

Equilibrium 0.218 + x 0.0903 - x 0.0511 - x

So , Kc = [PCl3] × [Cl2] / [PCl5]

1.20 × 10^-2 = 0.012 = (0.0903 - x ) ( 0.0511 - x ) / (0.218 - x )

0.0046 - 0.1414x + x² = 0.0026 - 0.012x

x² - 0.1294x + 0.002 = 0

Solving this quadratic equation ,  

x = ( - ( - 0.1294 ) +- ( - 0.1294 × -0.1294 - 4 × 1 × 0.002 )^1/2 ) / 2  

x = ( 0.1294 +- 0.0935 ) / 2  

So , x = ( 0.1294 + 0.0935 ) / 2 = 0.1114

or , x = ( 0.1294 - 0.0935 ) / 2 = 0.0180

But x cannot be greater than 0.0903

Therefore x = 0.0180

And , Equilibrium Concentrations are :

[PCl3] = 0.0903 - x = 0.0903 -  0.0180 = 0.0723 M  

[Cl2] = 0.0511 - x = 0.0511 - 0.0180 =   0.0331 M  

[PCl5] = 0.218 + x = 0.218 + 0.0180 =   0.2360 M