![Kc=6.875x10^{-3}](/tpl/images/0619/6079/8354b.png)
Explanation:
Hello,
In this case, for the given chemical reaction at equilibrium:
![2 SO_3 ( g ) \rightleftharpoons 2 SO_ 2 ( g ) + O_ 2 ( g )](/tpl/images/0619/6079/a4a19.png)
The initial concentration of sulfur trioxide is:
![[SO_3]_0=\frac{0.660mol}{4.00L}=0.165M](/tpl/images/0619/6079/888d2.png)
Hence, the law of mass action to compute Kc results:
![Kc=\frac{[SO_2]^2[O_2]}{[SO_3]^2}](/tpl/images/0619/6079/e7b22.png)
In such a way, in terms of the change
due to the reaction extent, by using the ICE method, it is modified as:
![Kc=\frac{(2x)^2*x}{(0.165-2x)^2}](/tpl/images/0619/6079/6e43f.png)
In that case, as at equilibrium 0.11 moles of oxygen are present,
equals:
![x=[O_2]=\frac{0.110mol}{4.00L}=0.0275M](/tpl/images/0619/6079/54041.png)
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}](/tpl/images/0619/6079/d1257.png)
Best regards.