Question

At a certain temperature, 0.740 mol SO 3 0.740 mol SO3 is placed in a 2.50 L 2.50 L container. 2 SO 3 ( g ) − ⇀ ↽ − 2 SO 2 ( g ) + O 2 ( g ) 2SO3(g)↽−−⇀2SO2(g)+O2(g) At equilibrium, 0.180 mol O 2 0.180 mol O2 is present. Calculate K c

Answer 1

Answer: $K_c$ = 0.0046

Explanation:

Moles of  $SO_3$ = 0.740 mole

Volume of solution = 2.50 L

Initial concentration of $SO_3$ = $\frac{moles}{volume}=\frac{0.740}{2.50}=0.296M$

The given balanced equilibrium reaction is,

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

 Initial conc.      0.296 M                             0M                      0M

 At eqm. conc.     (0.296-2x)M                     (2x) M          (x) M

The expression for equilibrium constant for this reaction will be,

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

Given : moles of $O_2$ at equilibrium = 0.180M

Concentration of $O_2$ at equilibrium= $\frac{moles}{volume}=\frac{0.180}{2.50}=0.072M$

x =  0.072 M

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

$K=\frac{(2x)^2\times (x)^2}{(0.296-2x)^2}$  

$K=\frac{(2\times 0.072)^2\times (0.072)^2}{(0.296-2\times 0.072)^2}$  

$K=0.0046$

Thus the value of $K_c$ is 0.0046