At `25^(@)C`, the value of equilibrium `(K_(c))` for the reaction, `A(s)+4B(g)hArr 2C(l)+3D(g)` is 16. if the reaction is initiated with 0.2 mol of each of the component taken in a closed vessel of 1 L volume, then in which direction the reaction will occur at a higher rate to attain the equilibrium?

Explanation

Since the value of the container is 1L, the initial concentration of each of the component is 0.2 `mol*L^(-1)`.
Reaction quotient `(Q_(c))` at the beginning of the raction,
`Q_(c)=([D(g)]^(3))/([B(g)]^(4))[because a(s)]=1 and [C(l)]=1]`
`therefore Q_(c)=((0.2)^(3))/((0.2)^(4))=5`, Since `Q_(c) lt K_(c)`, so, the forward reaction will occur at a higher rate to attain the equilibrium.