Question

At a certain temperature, 0.700 mol SO3 is placed in a 3.50 L container. 2SO3(g)↽−−⇀2SO2(g)+O2(g) At equilibrium, 0.180 mol O2 is present. Calculate c.

Answer 1

Answer: The value of $K_c$ for the given chemical equation is 0.0457.

Explanation:

Given values:

Initial moles of $SO_3$ = 0.700 moles

Volume of conatiner = 3.50 L

The given chemical equation follows:

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

I:             0.700

C:              -2x           +2x           x

E:           0.700-2x      2x            x

Equilibrium moles of $O_2$ = x = 0.180 moles

Equilibrium moles of $SO_2$ = 2x = $(2\times 0.180)=0.360moles$

Equilibrium moles of $SO_3$ = 0.700 - 2x = $0.700-(2\times 0.180)=0.340moles$

Molarity is calculated by using the equation:

$Molarity=\frac{Moles}{Volume}$

So,

$[SO_3]_{eq}=\frac{0.340}{3.50}=0.0971M$

$[SO_2]_{eq}=\frac{0.360}{3.50}=0.103M$

$[O_2]_{eq}=\frac{0.180}{3.50}=0.0514M$

The expression of $K_c$ for above equation follows:

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

Plugging values in above expression:

$K_c=\frac{(0.0971)^2\times 0.0514}{(0.103)^2}\\\\K_c=0.0457$

Hence, the value of $K_c$ for the given chemical equation is 0.0457.