Question

At a certain temperature, 0.780 mol SO 3 0.780 mol SO3 is placed in a 4.00 L 4.00 L container. 2 SO 3 ( g ) − ⇀ ↽ − 2 SO 2 ( g ) + O 2 ( g ) 2SO3(g)↽−−⇀2SO2(g)+O2(g) At equilibrium, 0.100 mol O 2 0.100 mol O2 is present. Calculate K c .

Answer 1

Answer: 0.00297

Explanation:

Moles of  $SO_3$ = 0.780 mole

Volume of solution = 4.00 L

Initial concentration of $SO_2$ = $\frac{moles}{Volume}=\frac{0.780}{4.00L}=0.195M$

Equilibrium concentration of $O_2$ = $\frac{moles}{Volume}=\frac{0.100}{4.00L}=0.025M$

The given balanced equilibrium reaction is,

                            $2SO_3(g)\rightleftharpoons 2SO_2g)+O_2(g)$

Initial conc.         0.195 M            0 M       0 M

At eqm. conc.     (0.195-2x) M   (2x) M   (x) M

The expression for equilibrium constant for this reaction will be,

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

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

$K_c=\frac{(2x)^2\times (x)}{(0.195-2x)^2}$

Given $[O_2$ = x = 0.025 M

$K_c=\frac{(2\times 0.025)^2\times (0.025)}{(0.195-2\times 0.025)^2}$

$K_c=0.00297$

Thus $K_c$ is 0.00297