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(D) The reaction is $A_{(g)} + B_{(g)} \rightleftharpoons 2C_{(g)}$.Initial moles: $A = 3$,$B = 1$,$C = 0$.At equilibrium,$1.5 \ mol$ of $C$ are forme

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$4$

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(D) The reaction is $A_{(g)} + B_{(g)} \rightleftharpoons 2C_{(g)}$.Initial moles: $A = 3$,$B = 1$,$C = 0$.At equilibrium,$1.5 \ mol$ of $C$ are formed. Since the stoichiometry of $C$ is $2$,the moles of $A$ and $B$ reacted are $1.5 / 2 = 0.75 \ mol$ each.Equilibrium moles: $A = 3 - 0.75 = 2.25 \ mol$,$B = 1 - 0.75 = 0.25 \ mol$,$C = 1.5 \ mol$.Since the volume is $1 \ L$,the concentrations are equal to the number of moles.$K_c = \frac{[C]^2}{[A][B]} = \frac{(1.5)^2}{(2.25)(0.25)} = \frac{2.25}{0.5625} = 4$.

When $3 \ mol$ of $A$ and $1 \ mol$ of $B$ are mixed in a $1 \ L$ vessel,the following reaction takes place: $A_{(g)} + B_{(g)} \rightleftharpoons 2C_{(g)}$. If $1.5 \ mol$ of $C$ are formed at equilibrium,the equilibrium constant $(K_c)$ for the reaction is:

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