Question

QUESTION 6 Consider the following reaction between the diatomic and monatomic forms of iodine: I2 (g) <-> 2I (g) When 0.095 M I2 is initially placed in a previously empty container and sealed, the system slowly reaches equilibrium. When equilibrium is reached, it is found that there is an equilibrium concentration of 0.0055 M of the monatomic form of iodine. Calculate the (unitless) equilibrium constant Kc. Round your answer to two sig figs, and express it in scientific notation.

Answer 1

Answer: The equilibrium constant is
`3.3* 10^(-4)`

Step-by-step explanation:

Initial concentration of
`I_2` = 0.095 M

The given balanced equilibrium reaction is,

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

Initial conc. 0.095 M 0 M

At eqm. conc. (0.095-x) M (2x) M

Given : 2x = 0.0055

x = 0.00275

The expression for equilibrium constant for this reaction will be,

`K_c=([l]^2)/([I_2])`

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

`K_c=((0.0055)^2)/((0.095-0.00275))`

`K_c=((0.0055)^2)/(0.09225)=0.00033`

Thus the equilibrium constant is
`3.3* 10^(-4)`