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Sagot :
Answer: The value of [tex]K_c[/tex] for the given chemical equation is 0.0457.
Explanation:
Given values:
Initial moles of [tex]SO_3[/tex] = 0.700 moles
Volume of conatiner = 3.50 L
The given chemical equation follows:
[tex]2SO_3(g)\rightleftharpoons 2SO_2(g)+O_2(g)[/tex]
I: 0.700
C: -2x +2x x
E: 0.700-2x 2x x
Equilibrium moles of [tex]O_2[/tex] = x = 0.180 moles
Equilibrium moles of [tex]SO_2[/tex] = 2x = [tex](2\times 0.180)=0.360moles[/tex]
Equilibrium moles of [tex]SO_3[/tex] = 0.700 - 2x = [tex]0.700-(2\times 0.180)=0.340moles[/tex]
Molarity is calculated by using the equation:
[tex]Molarity=\frac{Moles}{Volume}[/tex]
So,
[tex][SO_3]_{eq}=\frac{0.340}{3.50}=0.0971M[/tex]
[tex][SO_2]_{eq}=\frac{0.360}{3.50}=0.103M[/tex]
[tex][O_2]_{eq}=\frac{0.180}{3.50}=0.0514M[/tex]
The expression of [tex]K_c[/tex] for above equation follows:
[tex]K_c=\frac{[SO_2]^2[O_2]}{[SO_3]^2}[/tex]
Plugging values in above expression:
[tex]K_c=\frac{(0.0971)^2\times 0.0514}{(0.103)^2}\\\\K_c=0.0457[/tex]
Hence, the value of [tex]K_c[/tex] for the given chemical equation is 0.0457.
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