Question

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

Short answer

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

Step-by-step solution

Define Decomposed

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

Decomposition Reaction

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

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

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

Calculating equilibrium constant

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