Question

Please show how X is derived

For the following reaction Kc = 2.20 x 102 at 74°C | CO(g) + Cl2(g) ←→ COCI2(g) Find the equilibrium concentrations of all chemical species starting with [CO]-0.161 M and [C12] = 0.205 M. [CO] = [C2] - [coci2)- COCI2]

Answer 1

First, let us define the equilibrium constant for any species:

The equilibrium constant will relate product and reactants distribution. It is similar to a ratio

The equilibrium is given by

rReactants -> pProducts

Keq = [products]^p / [reactants]^r

For a specific case:

aA + bB = cC + dD

Keq = [C]^c * [D]^d / ([A]^a * [B]^b)

now,

define Keq

Keq = [COCl2]/([CO][Cl2])

initially:

[CO] = 0.161

[Cl2] = 0.205

[COCl2] = 0

in equilibrium, due to stoichioemtry

[CO] = 0.161 - x

[Cl2] = 0.205 - x

[COCl2] = 0 + x

substitute in K

Keq = [COCl2]/([CO][Cl2])

220 = x / (0.161-x)(0.205-x)

220 *(0.161*0.205 - (0.161+0.205)x + x^2) = x

7.2611-80.52x+ 220x^2 = x

220x^2 -81.52x + 7.2611 =0

x = 0.1489

[CO] = 0.161 - 0.1489 = 0.013 M

[Cl2] = 0.205 - 0.1489 = 0.0561 M

[COCl2] = 0 + 0.1489 = 0.1489 M