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 $3.3\times 10^{-4}$

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,

$K_c=\frac{[l]^2}{[I_2]}$  

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

$K_c=\frac{(0.0055)^2}{(0.095-0.00275)}$

$K_c=\frac{(0.0055)^2}{0.09225}=0.00033$

Thus the equilibrium constant is $3.3\times 10^{-4}$