Question

At a certain temperature, 0.660 mol SO 3 is placed in a 4.00 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:

$Kc=6.875x10^{-3}$

Explanation:

Hello,

In this case, for the given chemical reaction at equilibrium:

$2 SO_3 ( g ) \rightleftharpoons 2 SO_ 2 ( g ) + O_ 2 ( g )$

The initial concentration of sulfur trioxide is:

$[SO_3]_0=\frac{0.660mol}{4.00L}=0.165M$

Hence, the law of mass action to compute Kc results:

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

In such a way, in terms of the change $x$ due to the reaction extent, by using the ICE method, it is modified as:

$Kc=\frac{(2x)^2*x}{(0.165-2x)^2}$

In that case, as at equilibrium 0.11 moles of oxygen are present, $x$ equals:

$x=[O_2]=\frac{0.110mol}{4.00L}=0.0275M$

Therefore, the equilibrium constant finally turns out:

$Kc=\frac{(2*0.0275)^2*0.0275}{(0.165-2*0.0275)^2} \\\\Kc=6.875x10^{-3}$

Best regards.