Answer:
At Equilibrium
[COCl₂] = 0.226 M
[CO] = 0.054 M
[Cl₂] = 0.054 M
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
