General Chemistry 2

Question

The equilibrium constant, K_c, for the following reaction is 1.20×10^-2 at 500 K.

PCl₅(g) PCl₃(g) + Cl₂(g)

Calculate the equilibrium concentrations of reactant and products when 0.388 moles of PCl₅(g) are introduced into a 1.00 L vessel at 500 K.

[PCl5]	=	M
[PCl3]	=	M
[Cl2]	=	M

Answer 1

PCl₅(g) <=======> PCl₃(g) +  Cl₂(g)

I 0.388M 0 0

Δ -x +x +x

Eq 0.388-x x x

Kc = [PCl3][Cl2]/[PCl5] = 1.2x10^-2

Kc = [PCl3][Cl2]/[PCl5] = (x)(x)/(0.388-x) = 1.20x10^-2 =

x² + 1.2x10^-2x - 4.66x10^-3 = 0

x = -b ± √b² - 4ac

2a

x = 0.0626

[PCl₅] = 0.388 - x = 0.388 - 0.0626 = 0.325 M

[PCl₃] = [Cl₂] = x = 0.063 M

Answer 2

PCl₅(g)  <==> PCl₃(g) + Cl₂(g) ... Kc = 1.20x10^-2

Set up an ICE table

PCl₅(g) <==> PCl₃(g) + Cl₂(g)

0.388..............0...............0.........Initial

-x...................+x.............+x........Change

0.388-x...........x................x.........Equilibrium

Kc = [PCl3][Cl2] / [PCl5]

1.20x10^-2 = (x)(x) / (0.388-x)

x² + 1.2x10^-2x - 4.66x10^-3 = 0

Solve for x using the quadratic equation

x = 0.0626

Final concentrations as equilibrium are as follows:

[PCl₅] = 0.388 - 0.0626 = 0.325 M

[PCl₃] = 0.063 M

[Cl₂] = 0.063 M