Carbonyl bromide, COBr2, decomposes to CO and Br2 with and equilibrium constant of Kc of 0.190...

Carbonyl bromide, COBr2, decomposes to CO and Br2 with and equilibrium constant of Kc of 0.190 at 73C.
COBr2(g) <--> CO (g) + Br2 (g)
if 0.015 mol COBr2 is in a 2.50 L flask at equilibrium, what are the concentrations of CO and Br2 at equilibrium?

Expert Solution

ans) COBr2 CO + Br2

given 0.015 /2.5 0 0

= 0.006M 0 0

at eqm conc 0.006 - x x x

Kc = [ CO ] [ Br2 ] / [ COBr2 ]

0.190 = x² / 0.006 -x

by solving x = 0.03376 M

[ CO ] =X = 0.03376M

[Br2 ] = X = 0.03376M

THANK U

queen_honey_blossom answered 2 years ago

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