Question

In the reaction CO + Cl2 => COCl2 (g), Kc = 1. 2 × 10³ at 395K. What is the Concentration equilibrium value [COCl2] if at equilibrium [CO] = 2 Cl2 = 1/2 [COCl2

Answer 1

Given the reaction CO + Cl2 => COCl2 (g), Kc = 1. 2 × 10³ at 395K.

We will find the Concentration equilibrium value [COCl2] if at equilibrium [CO] = 2 Cl2 = 1/2 [COCl2​ by :

at equilibrium :

CO + Cl2 => COCl2 (g)

x : 1/2x : 2x

• Kc = (COCl2(g) / {(CO(g)) * ( Cl2(g))}

= 2x / ( x* 1/2x)............

= 2x / (1/2) .....

replace Kc with 1.2X10^ 3

1.2X10 ^3 = 2x /x^2/2

therefore 4/x = 1.2x10^3

x = 0.0033

This means that x = 0.0033, the equilibrium concentration of [COCl2] is twice the value of x ,

so, [COCl2](g) = 2 *0.0033 = 0.0066M

( extra notes :we have 2x / x* 1/2x, this is the same as :

(2x )/ (x *x/2) = (2x) / ( x^2 /2)

2x * 2/x^2 ( the fraction in the denominators changes positions when we multiply with the numerator )

so 2x *2/x^2 = 4x /x^2 ( x's on the numerator and denominator cancels, leaving us with 1 x at the bottom)

= 4/x)