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

Step-by-step explanation:

Initial moles of
`I_2` = 0.072 mole

Volume of container = 3.9 L

Initial concentration of
`I_2=(moles)/(volume)=(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=([I]^2)/([I_2])`

`K_c=((2x)^2)/(0.2-x)`

we are given :
`K_c=1.60* 10^(-3)`

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

`1.60* 10^(-3)=((2x)^2)/((0.018-x))`

`x=0.0025`

So, the concentrations for the components at equilibrium are:

`[I]=2* x=2* 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.