Question

The equilibrium constant, Kc, for the following reaction is 77.5 at 600 K. CO(g) Cl2(g) COCl2(g) Calculate the equilibrium concentrations of reactant and products when 0.555 moles of CO and 0.555 moles of Cl2 are introduced into a 1.00 L vessel at 600 K.

Answer 1

[CO] = 0.078M

[Cl2] = 0.078M

[COCl2] = 0.477M

Step-by-step explanation:

Based on the reaction:

CO(g) Cl2(g) ⇄ COCl2(g)

Where equilibrium constant, kc, is:

kc = 77.5 = [COCl2] / [CO] [Cl2]

[] represents the equilibrium concentration of each gas. The initial concentration of each gas is:

[CO] = 0.555mol/1.00L = 0.555M

[Cl2] = 0.555M

And equilibrium concentrations are:

[CO] = 0.555M - x

[Cl2] = 0.555M - x

[COCl2] = x

Where x is reaction coordinate

Replacing in kc expression:

77.5 = [x] / [0.555M - x] [0.555M - x]

77.5 = x / 0.308025 - 1.11 x + x²

23.8719 - 86.025 x + 77.5 x² = x

23.8719 - 87.025 x + 77.5 x² = 0

x = 0.477M. Right answer

x = 0.646M. False answer. Produce negative concentrations

Replacing:

[CO] = 0.555M - 0.477M = 0.078M

[Cl2] = 0.078M

[COCl2] = 0.477M

And those concentrations are the equilibrium concentrations