Respuesta :
Answer: [tex]K_c[/tex] = 0.0046
Explanation:
Moles of [tex]SO_3[/tex] = 0.740 mole
Volume of solution = 2.50 L
Initial concentration of [tex]SO_3[/tex] = [tex]\frac{moles}{volume}=\frac{0.740}{2.50}=0.296M[/tex]
The given balanced equilibrium reaction is,
[tex]2SO_3(g)\rightleftharpoons 2SO_2(g)+O_2(g)[/tex]
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,
[tex]K_c=\frac{[SO_2]^2\times [O_2]}{[SO_3]^2}[/tex]
Given : moles of [tex]O_2[/tex] at equilibrium = 0.180M
Concentration of [tex]O_2[/tex] at equilibrium= [tex]\frac{moles}{volume}=\frac{0.180}{2.50}=0.072M[/tex]
x = 0.072 M
Now put all the given values in this expression, we get :
[tex]K=\frac{(2x)^2\times (x)^2}{(0.296-2x)^2}[/tex]
[tex]K=\frac{(2\times 0.072)^2\times (0.072)^2}{(0.296-2\times 0.072)^2}[/tex]
[tex]K=0.0046[/tex]
Thus the value of [tex]K_c[/tex] is 0.0046
