Calculate the equilibrium concentrations of NO, O₂, and NO₂ in a mixture at 250 °C that results from the reaction of 0.20 M NO and 0.10 M O₂. (Hint: K is large; assume the reaction goes to completion then comes back to equilibrium.)

\(2NO(g) + {O_2}(g) \rightleftharpoons 2N{O_2}(g)\quad {K_c} = 2.3 \times 1{0^5}\;at\;25{0^o}C\)

Given information:

\(2{\text{NO}}({\text{g}}) + {{\text{O}}_2}({\text{g}}) \rightleftharpoons 2{\text{N}}{{\text{O}}_2}({\text{g}})\)

• The concentration of NO is 0.20 M.
• The concentration of O₂ is 0.10 M.
• The equilibrium constant is \({K_c} = 2.3 \times {10^5}\).

We have to find the equilibrium concentrations of NO, O₂, and NO₂.

Since 2 moles of NO reacts with 1 mole of O₂, to produce 2 moles of NO₂, 0.20 mole of NO will exact with 0.10 mole of O₂, and produce 0.20 moles of NO₂(0.20 M).

We have to determine the value of x,

\(\begin{array}{*{20}{c}}{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{K_c} = \frac{{{{\left[ {{\rm{N}}{{\rm{O}}_2}} \right]}^2}}}{{{{[NO]}^2} \times \left[ {{O_2}} \right]}}}\\{2.3 \times {{10}^5} = \frac{{{{(0.2 - 2x)}^2}}}{{{{(2x)}^2} \times x}}}\end{array}\)

Since \({K_c}\)is too large, we will neglect the change in concentration of \({\rm{N}}{{\rm{O}}_2}\), and get

\(\begin{array}{*{20}{c}}{2.3 \times {{10}^5} = \frac{{{{(0.2)}^2}}}{{{{(2x)}^2} \times x}}}\\{\,\,\,\,\,\,\,\,\,\,\,4{x^3} = \frac{{{{(0.2)}^2}}}{{2.3 \times {{10}^5}}}}\\{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x^3} = 4.35 \times {{10}^{ - 8}}}\\{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 3.52 \times {{10}^{ - 3}}{\rm{M}}}\end{array}\)

The change in the concentration obtained \(x = 3.52 \times {10^{ - 3}}{\rm{M}}\)

Therefore, the equilibrium concentrations of NO, O₂, and NO₂ are

\(\begin{array}{*{20}{c}}{\,\,[{\rm{NO}}] = 2{\rm{x}} = 7.04 \times {{10}^{ - 3}}{\rm{M}}}\\{\left[ {{{\rm{O}}_2}} \right] = {\rm{x}} = 3.52 \times {{10}^{ - 3}}{\rm{M}}}\\{\,\,\,\,\,\,\,[{\rm{NO}}] = 0.2{\rm{M}} - 2{\rm{x}} = 0.193{\rm{M}}}\end{array}\).