Question

A mixture 0.500 mole of carbon monoxide and 0.400 mole of bromine was placed into a rigid 1.00 L container and the system was allowed to come to equilibrium. The equilibrium concentration of COBr2 was 0.233 M. What is the value of Kc for this reaction? CO(g) + Br2(g) ------>(equilb. arrows) COBr2(g)

Answer 1

Solution :-

Balanced reaction equation

Volume is 1 L so moles is nothing but the molarity for each

   CO    +     Br2 <----- > COBr2

0.500 M      0.400 M         0

-x                   -x               +x

0.500-x          0.400-x       0.233M

Value of x= 0.233 M

So lets calculate the equilibrium concentration of the CO and Br2

[CO]eq = 0.500 –x = 0.500 – 0.233 = 0.267 M

[Br2]eq = 0.400 –x = 0.400 – 0.233 = 0.167 M

Now lets calculate the equilibrium constant

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

Lets put the values in the formula

Kc = [0.233]/[0.267][0.167]

Kc= 5.23

So the equilibrium constant is 5.23