Question

carbonyl bromide, COBr2, decomposes to CO and Br2 with an Kc= .190 . A .015 mole sample of carbonyl bromide is place in a 2.5 L flask. what will be the equilibrium concentrations of the reaction?

Answer 1

Consider a equilibrium reaction, COBr ₂ CO + Br ₂

For above reaction, K c = [ CO ] [ Br ₂ ] / [ COBr ₂ ] = 0.190

Let's use ICE table.

M	COBr 2 CO + Br 2
I	0.006
C	-X	+X	+X
E	0.006 - X	X	X

( [ COBr ₂ ] = 0.015 mol / 2.5 L = 0.006 M )

K c = ( X) (X) / 0.006 - X = 0.190

X ² / 0.006 - X = 0.190

X ² = 0.006 - X ( 0.190 )

X ² = 0.00114 - 0.190 X

X ² + 0.190 X - 0.00114= 0

Comparing above equation with a X ² + b X +c = 0 , we get a = 1 , b = 0.190 and c = - 0.00114

We know that, X = - b +/- b ² -4 ac / 2a

X = - 0.190 +/- ( 0.190 ) ² - 4 (1) ( - 0.00114 ) / 2 (1)

X = - 0.1958 or X = 0.005822

We can choose X = 0.005822

Equilibrium concentration of CO = X = 0.005822 M

Equilibrium concentration of Br ₂ = X = 0.005822 M

Equilibrium concentration of COBr ₂ = 0.006 - X = 0.000178 M