Question

A sealed 1.0 L flask is charged with 0.500 mol of I2 and 0.500 mol of Br2. An equilibrium reaction ensues: I2 (g) + Br2 (g) ↔ 2IBr (g) When the container contents achieve equilibrium, the flask contains 0.84 mol of IBr. The value of Keq is ________.

Answer 1

Answer:  Thus the value of $K_{eq}$ is 110.25

Explanation:

Initial moles of  $I_2$ = 0.500 mole

Initial moles of  $Br_2$ = 0.500 mole

Volume of container = 1 L

Initial concentration of $I_2=\frac{moles}{volume}=\frac{0.500moles}{1L}=0.500M$

Initial concentration of $Br_2=\frac{moles}{volume}=\frac{0.500moles}{1L}=0.500M$

equilibrium concentration of $IBr=\frac{moles}{volume}=\frac{0.84mole}{1L}=0.84M$ [/tex]

The given balanced equilibrium reaction is,

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

Initial conc.              0.500 M     0.500 M             0  M

At eqm. conc.    (0.500-x) M      (0.500-x) M     (2x) M

The expression for equilibrium constant for this reaction will be,

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

$K_c=\frac{(2x)^2}{(0.500-x)\times (0.500-x)}$

we are given : 2x = 0.84 M

x= 0.42

Now put all the given values in this expression, we get :

$K_c=\frac{(0.84)^2}{(0.500-0.42)\times (0.500-0.42)}$

$K_c=110.25$

Thus the value of the equilibrium constant is 110.25