Question

A student solved the following problem and found \(\left[ {{N_2}{O_4}} \right] = 0.16M\)at equilibrium. How could this student recognize that the answer was wrong without reworking the problem? The problem was: What is the equilibrium concentration of \(\left[ {{N_2}{O_4}} \right]\) in a mixture formed from a sample of \(N{O_2}\) with a concentration of \(0.10M\)?

\(2N{O_2}(g) \rightleftharpoons {N_2}{O_4}(g)\)

\({K_c} = 160\)

Short answer

\(\begin{array}{c}\left[ {{{\rm{N}}_2}{{\rm{O}}_4}} \right] = 0.0419{\rm{M}}\\\left[ {{\rm{N}}{{\rm{O}}_2}} \right] = 0.0162M\end{array}\)

The value obtained for \({{\rm{N}}_2}{{\rm{O}}_4}\) does not matches with the student’s result. So, the student’s results were wrong

Step-by-step solution

Find out how the student’s result was wrong:

\(2N{O_2}(g) \rightleftharpoons {N_2}{O_4}(g)\)

Initial(M)	\(0.10\)	\(0\)
Change(M)	\( - 2x\)	\( + x\)
Equilibrium(M)	\(0.10 - 2x\)	\(x\)

Find the value of x

\(\begin{array}{c}{K_C} = \frac{{\left[ {{{\rm{N}}_2}{{\rm{O}}_4}} \right]}}{{{{\left[ {2{\rm{N}}{{\rm{O}}_2}} \right]}^2}}}\\160 = \frac{x}{{{{\left( {0.10 - 2x} \right)}^2}}}\\x = 160\left( {0.01 - 0.4x + 4{x^2}} \right)\\640{x^2} - 65x + 1.6 = 0\\x = 0.0419M\end{array}\)

Therefore, the equilibrium concentration will be

\(\begin{array}{c}\left[ {{{\rm{N}}_2}{{\rm{O}}_4}} \right] = x = 0.0419{\rm{M}}\\\left[ {{\rm{N}}{{\rm{O}}_2}} \right] = 0.10 - 2x = 0.0162M\end{array}\)