Question

Consider the following equilibrium at 979 K for the dissociation of molecular iodine into atoms of iodine. I2(g) equilibrium reaction arrow 2 I(g); Kc = 1.60 ✕ 10−3 Suppose this reaction is initiated in a 3.9 L container with 0.072 mol I2 at 979 K. Calculate the concentrations of I2 and I at equilibrium.

Answer 1

Answer:  $I_2$=  0.0050 M

$I$ = 0.0155 M

Explanation:

Initial moles of  $I_2$ = 0.072 mole

Volume of container = 3.9 L

Initial concentration of $I_2=\frac{moles}{volume}=\frac{0.072moles}{3.9L}=0.018M$  

The given balanced equilibrium reaction is,

                 $I_2(g)\rightleftharpoons 2I(g)$

Initial conc.         0.018 M            0

At eqm. conc.    (0.018-x) M      (2x) M  

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[I]^2}{[I_2]}$

$K_c=\frac{(2x)^2}{0.2-x}$

we are given :  $K_c=1.60\times 10^{-3}$

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

$1.60\times 10^{-3}=\frac{(2x)^2}{(0.018-x)}$

$x=0.0025$

So, the concentrations for the components at equilibrium are:

$[I]=2\times x=2\times 0.0025=0.0050$

$[I_2]=0.018-x=0.018-0.0025=0.0155$

Hence, concentrations of $I_2$ and $I$ are 0.0050 M ad 0.0155 M respectively.