Question:At 1 atm and \(2{5^o}C,N{O_2}\) with an initial concentration of \(1.00M\;is\;3.3 \times 1{0^{ - 3}}\% \)decomposed into \(NO\;and\;{O_2}\). Calculate the value of the equilibrium constant for the reaction. \(2N{O_2}(g) \rightleftharpoons 2NO(g) + {O_2}(g)\)

Decomposition or rot is the process by which dead organic substances are broken down into simpler organic or inorganic matter such as carbon dioxide, water, simple sugars and mineral salts

\(2N{O_2}(g) \rightleftharpoons 2NO(g) + {O_2}(g)\)

Calculate the change in concentration of\({\rm{N}}{{\rm{O}}_2}\)

\(\begin{array}{*{20}{c}}{\left( {{\rm{N}}{{\rm{O}}_2}} \right) - \left( {{\rm{N}}{{\rm{O}}_2}} \right)(eq) = \left( {{\rm{N}}{{\rm{O}}_2}} \right) \cdot 3.3 \cdot {{10}^{ - 3}}{\rm{\% }}}\\{ = 1.00{\rm{M}} \cdot 3.3 \cdot {{10}^{ - 3}}{\rm{\% }}}\\{ = 3.3 \cdot {{10}^{ - 5}}{\rm{M}}}\end{array}\)

The equilibrium constant for given reaction is

\(\begin{array}{*{20}{c}}{{K_c} = \frac{{{{(NO)}^2} \cdot \left( {{{\rm{O}}_2}} \right)}}{{{{\left( {{\rm{N}}{{\rm{O}}_2}} \right)}^2}}}}\\{ = \frac{{{{\left( {3.3 \cdot {{10}^{ - 5}}} \right)}^2} \cdot \left( {\frac{{3.3 \cdot {{10}^{ - 5}}}}{2}} \right)}}{{{{\left( {1.00 - 3.3 \cdot {{10}^{ - 5}}} \right)}^2}}}}\\{}\end{array}\)

The value of \({K_c} = 1.80 \cdot {10^{ - 14}}\)