Question

A 0.50 mol sample of COBr2 is transferred to a 9.50-L flask and heated until equilibrium is attained. Calculate the equilibrium concentrations of each species.

Answer 1

Equlibrium concentration for each species ae as follows:

[CO] = 0.043 mol/L

[Br₂] = 0.043 mol/L

[COBr₂] = 0.01 mol/L

Step-by-step explanation:

Let take a look at the chemical equation taking place at equilibrium

COBr2(g) ⇄ CO(g) + Br2(g)

The concentration of COBr2 i.e.

[COBr2] = no of moles/volume

= 0.50 mol/9.50 L

[COBr2] = 0.0530 mol/L

At standard conditions

Kc for COBr2 = 0.190

Now, the ICE table for the above reaction can be computed as follows:

COBr2(g) ⇄ CO(g) + Br2(g)

Initial 0.053 0 0

Change -x +x +x

Equilibrium (0.053 - x) x x

`\mathsf{K_c = ([CO][Br_2])/([COBr_2])}`

`K_c = ((x) (x))/((0,053 -x))`

`0.190= (x^2)/((0.053 -x))`

x² = 0.190(0.053 - x)

x² = 0.01007 - 0.190x

x² + 0.190x - 0.01007 = 0

Using quadratic formula:

x ≅ 0.043 mol/L

SInce: x = [CO][Br₂] = 0.043 mol/L

[COBr₂] = 0.053 - x

[COBr₂] = 0.053 - 0.043 mol/L

[COBr₂] = 0.01 mol/L