Question

Iodine and bromine react to give iodine monobromide, IBr. Ilg) + Br2(g) 2 Br(g) What is the equilibrium composition of a mixture at 145 C that initially contained 0.0019 mol each of iodine and bromine in a 5.0-L vessel? The equilibrium constant K, for this reaction at 145 C is 108.

Answer 1

Answer: Equilibrium concentration of $I_2$ = 0.00006 M

equilibrium concentration of $Br_2$ = 0.00006 M

and equilibrium concentration of $IBr$  = 0.00064 M

Explanation:

Initial moles of  $I_2$ = 0.0019 mole

Volume of container = 5.0 L

Initial concentration of $I_2=\frac{moles}{volume}=\frac{0.0019moles}{5.0L}=0.00038M$  

initial concentration of $Br_2=\frac{moles}{volume}=\frac{0.0019mole}{5.0L}=0.00038M$

The chemical reaction for the decomposition of phosgene follows the equation:

                     $I_2(g)+Br_2(g)\rightleftharpoons 2IBr(g)$

Initial conc.        0.00038 M       0.00038 M       0

At eqm. conc.    ( 0.00038-x) M   (0.00038-x) M  (2x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[IBr]^2}{[I_2]\times [Br_2]}$

$108=\frac{[2x]^2}{(0.00038-x)^2}$

$x=0.00032$

Thus equilibrium concentration of $I_2$  = (0.00038-x) M =(0.00038-0.00032) M = 0.00006 M

equilibrium concentration of $Br_2$ = (0.00038-x) M =(0.00038-0.00032) M = 0.00006 M

equilibrium concentration of $IBr$ = 2x M = 2(0.00032) M=0.00064 M