Determine if the following system is at equilibrium. If not, in which direction will the system need to shift to reach equilibrium?

\({\rm{S}}{{\rm{O}}_2}{\rm{C}}{{\rm{l}}_2}(g)\rightleftharpoons {\rm{S}}{{\rm{O}}_2}(g) + {\rm{C}}{{\rm{l}}_2}(g)\)

\(\left( {{\rm{S}}{{\rm{O}}_2}{\rm{C}}{{\rm{l}}_2}} \right) = 0.12\;{\rm{M}},\;\left( {{\rm{C}}{{\rm{l}}_2}} \right) = 0.16\;{\rm{M and }}\left( {{\rm{S}}{{\rm{O}}_2}} \right) = 0.050\;{\rm{M}}.\;{K_c}\) for the reaction is 0.078.

\({Q_c}\) is the product of the concentrations of all the reactants raised to the power of their respective coefficients in the reaction for any reversible reaction.

Thus, the reversible reaction of the form, \(aA + bB\rightleftharpoons cC + dD\)

\({Q_c} = \frac{{|C{|^c}|D{|^d}}}{{|A{|^a}|B{|^b}}}\)

The direction of a reaction at equilibrium can be predicted by comparing \({{\rm{Q}}_{\rm{c}}}\)and \({{\rm{K}}_{\rm{c}}}\).

When the numerator, or the amount of product existing at any one time, is more than the denominator, or the amount of reactant, this condition exists. When surplus product is present, the system tends to go in the direction of reducing the excess product/s by returning the reactant/s to achieve equilibrium (according to Le Chatlier's principle). As a result, the reaction occurs in reverse.

When the amount of reactant/s and product/s are constant, the reaction is already at equilibrium. This is because the forward reaction rate is equal to the reverse reaction rate, and the system has no inclination to create any more product/s or reactant/s. As a result, there is no change in the direction of the response.

When the numerator, or the amount of product existing at any one time, is less than the denominator, or the amount of reactant, this condition exists. When surplus reactant(s) are present, the system tends to proceed in the direction of reducing the excess reactant(s) by returning the product(s) in order to achieve equilibrium (as per Le Chatlier's principle). As a result, the reaction is moving forward.

For the reaction \({\rm{S}}{{\rm{O}}_2}{\rm{C}}{{\rm{l}}_2}(\;g)\rightleftharpoons {\rm{S}}{{\rm{O}}_2}(\;g) + {\rm{C}}{{\rm{l}}_2}(\;g)\)

\({{\rm{Q}}_{\rm{c}}} = \frac{{\left( {{\rm{S}}{{\rm{O}}_2}||{\rm{C}}{{\rm{l}}_2}} \right)}}{{\left( {{\rm{S}}{{\rm{O}}_2}{\rm{C}}{{\rm{l}}_2}} \right)}}\)

Given,

\(\begin{aligned}{}{{\rm{K}}_{\rm{c}}} &= 0.078,\\\left( {{\rm{S}}{{\rm{O}}_2}{\rm{C}}{{\rm{l}}_2}} \right) &= 0.12\,{\rm{M}},\\\left( {{\rm{C}}{{\rm{l}}_2}} \right) & = 0.16\,{\rm{M}},\\\left( {{\rm{S}}{{\rm{O}}_2}} \right) &= 0.050\,{\rm{M}}\\{{\rm{Q}}_{\rm{c}}} &= \frac{{(0.050\,{\rm{M}})(0.16\,{\rm{M}})}}{{(0.12\,{\rm{M}})}}\\ &= 0.066\end{aligned}\)

Therefore, the system will proceed in forward direction.