Consider the following reversible chemical reactions:

\({A_2}\left( g \right)\; + \;{B_2}\left( g \right)\begin{array}{*{20}{c}} {{K_1}}\\ \rightleftharpoons \end{array}\;2AB\left( g \right)\)      …. (1)

\(6AB\left( g \right)\;\begin{array}{*{20}{c}} {{K_2}}\\ \rightleftharpoons \end{array}3{A_2}\left( g \right) + 3{B_2}\left( g \right)\)        …. (2)

The relation between K₁ and K₂ is:

Explanation:

A reversible reaction is a chemical reaction where the reactants form products that, in 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, K_c

\({{\rm{K}}_{\rm{c}}} = \frac{{{{\left[ {{\rm{product}}} \right]}^{\rm{m}}}}}{{{{\left[ {{\rm{reactant}}} \right]}^{\rm{n}}}}}\)

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:

\(6AB\left( g \right)\begin{array}{*{20}{c}} {{K_2}}\\ \rightleftharpoons \end{array}\;3{A_2}\left( g \right) + 3{B_2}\left( g \right)\)

\({K_2} = \frac{{{{\left[ {{A_2}} \right]}^3}{{\left[ {{B_2}} \right]}^3}}}{{{{\left[ {AB} \right]}^6}}}\)

\(= \frac{1}{{{{\left( {\frac{{{{\left[ {AB} \right]}^2}}}{{\left[ {{A_2}} \right]\left[ {{B_2}} \right]}}} \right)}^3}}}\)

\(= \frac{1}{{K_1^3}}\)

 → K2 = K1-3