Answer:
Equlibrium concentration for each species ae as follows:
[CO] = 0.043 mol/L
[Br₂] = 0.043 mol/L
[COBr₂] = 0.01 mol/L
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
[tex]\mathsf{K_c = \dfrac{[CO][Br_2]}{[COBr_2]}}[/tex]
[tex]K_c = \dfrac{(x) (x)}{(0,053 -x)}[/tex]
[tex]0.190= \dfrac{x^2}{(0.053 -x)}[/tex]
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
