Question

At a certain temperature, 0.900 mol SO 3 is placed in a 2.00 L container. 2 SO 3 ( g ) − ⇀ ↽ − 2 SO 2 ( g ) + O 2 ( g ) At equilibrium, 0.110 mol O 2 is present. Calculate K c .

Answer 1

Answer: The value of $K_c$ is 0.0057

Explanation:

Initial moles of  $SO_3$ = 0.900 mole

Volume of container = 2.00 L

Initial concentration of $SO_3=\frac{moles}{volume}=\frac{0.900moles}{2.00L}=0.450M$  

equilibrium concentration of $O_2=\frac{moles}{volume}=\frac{0.110mole}{2.00L}=0.055M$ [/tex]

The given balanced equilibrium reaction is,

                            $2SO_3(g)\rightleftharpoons 2SO_2(g)+O_2(g)$

Initial conc.              0.450 M               0        0

At eqm. conc.    (0.450 -2x) M         (2x) M   (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[O_2][SO_2]^2}{[SO_3]^2}$

$K_c=\frac{x\times (2x)^2}{0.450-2x)^2}$

we are given : x = 0.055

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

$K_c=\frac{0.055\times (2\times 0.055)^2}{0.450-2\times 0.055)^2}$

$K_c=0.0057$

Thus the value of the equilibrium constant is 0.0057