TURN, react together to give the reactants back. Reversible reactions will reach an equilibrium point where the concentrations of the reactants and products will no longer change.According to equilibrium CONSTANT, Kc\({{\rm{K}}_{\rm{c}}} = \frac{{{{\left[ {{\rm{product}}} \right]}^{\rm{m}}}}}{{{{\left[ {{\rm{REACTANT}}} \right]}^{\rm{n}}}}}\)Consider the given first equation:\({A_2}\left( g \right)\; + \;{B_2}\;\left( g \right)\begin{array}{*{20}{c}}{{K_1}}\\ \rightleftharpoons\end{array}\;2AB\left( g \right)\) \({K_1} = \frac{{{{\left[ {AB} \right]}^2}}}{{\left[ {{A_2}} \right]\left[ {{B_2}} \right]}}\)Consider the given second equation:\(6AB\left( g \right)\begin{array}{*{20}{c}}{{K_2}}\\ \rightleftharpoons\end{array}\;3{A_2}\left( g \right) + 3{B_2}\left( g \right)\) \({K_2} = \frac{{{{\left[ {{A_2}} \right]}^3}{{\left[ {{B_2}} \right]}^3}}}{{{{\left[ {AB} \right]}^6}}}\)\(= \frac{1}{{{{\left( {\frac{{{{\left[ {AB} \right]}^2}}}{{\left[ {{A_2}} \right]\left[ {{B_2}} \right]}}} \right)}^3}}}\) \(= \frac{1}{{K_1^3}}\) \(\Rightarrow {K_2} = K_1^{ - 3}\)