Question

The equilibrium constant, Kc, for the following reaction is 1.29E-2 at 600 K. COCl2(g) CO(g) Cl2(g) Calculate the equilibrium concentrations of reactant and products when 0.280 moles of COCl2(g) are introduced into a 1.00 L vessel at 600 K. [COCl2]

Answer 1

At Equilibrium

[COCl₂] = 0.226 M

[CO] = 0.054 M

[Cl₂] = 0.054 M

Step-by-step explanation:

Given that;

equilibrium constant Kc = 1.29 × 10⁻² at 600k

the equilibrium concentrations of reactant and products = ?

when 0.280 moles of COCl2(g) are introduced into a 1.00 L vessel at 600 K. [COCl²]

Concentration of COCl₂ = 0.280 / 1.00 = 0.280 M

COCl₂(g) ----------> CO(g) + Cl₂(g)

0.280 0 0 ------------ Initial

-x x x

(0.280 - x) x x ----------- equilibrium

we know that; solid does not take part in equilibrium constant expression

so

KC = [CO][Cl₂] / COCl₂

we substitute

1.29 × 10⁻² = x² / (0.280 - x)

0.0129 (0.280 - x) = x²

x² = 0.003612 - 0.0129x

x² + 0.0129x - 0.003612 = 0

x = -b±√(b² - 4ac) / 2a

we substitute

x = [-(0.0129)±√((0.0129)² - 4×1×(-0.003612))] / [2 × 1 ]

x = [-0.0129 ± √( 0.00017 + 0.01445)] / 2

x = [-0.0129 ± 0.1209] / 2

Acceptable value of x =[ -0.0129 + 0.1209] / 2

x = 0.108 / 2

x = 0.054

At equilibrium

[COCl₂] = (0.280 - x) = 0.280 - 0.054 = 0.226 M

[CO] = 0.054 M

[Cl₂] = 0.054 M