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(D) The initial concentration of $SO_3$ is $1 \ mol/L$. The reaction is $2SO_3 \rightleftharpoons 2SO_2 + O_2$. At equilibrium,$0.6 \ mol$ of $SO_2$ i

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$0.675$

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(D) The initial concentration of $SO_3$ is $1 \ mol/L$. The reaction is $2SO_3 ightleftharpoons 2SO_2 + O_2$. At equilibrium,$0.6 \ mol$ of $SO_2$ is formed. According to the stoichiometry,if $2 \ moles$ of $SO_2$ are formed,$2 \ moles$ of $SO_3$ are consumed. Thus,$0.6 \ mol$ of $SO_2$ formation consumes $0.6 \ mol$ of $SO_3$. Equilibrium moles: $[SO_3] = 1 - 0.6 = 0.4 \ mol/L$ $[SO_2] = 0.6 \ mol/L$ $[O_2] = \frac{0.6}{2} = 0.3 \ mol/L$ $K_c = \frac{[SO_2]^2 [O_2]}{[SO_3]^2} = \frac{(0.6)^2 	imes (0.3)}{(0.4)^2} = \frac{0.36 	imes 0.3}{0.16} = \frac{0.108}{0.16} = 0.675$.

One mole of $SO_3$ was placed in a $1 \ L$ reaction vessel at a certain temperature. The following equilibrium was established: $2SO_3 \rightleftharpoons 2SO_2 + O_2$. At equilibrium,$0.6 \ moles$ of $SO_2$ were formed. The equilibrium constant $(K_c)$ of the reaction will be:

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