A student solved the following problem and found the equilibrium concentrations to be \(\left[ {S{O_2}} \right] = 0.590M\), \(\left[ {{O_2}} \right] = 0.0450M\), and \(\left[ {S{O_3}} \right] = 0.260M\). How could this student check the work without reworking the problem? The problem was: For the following reaction at \(60{0^0}C\):

\(2S{O_2}(g) + {O_2}(g) \rightleftharpoons 2S{O_3}(g)\)

\({K_c} = 4.32\)

What are the equilibrium concentrations of all species in a mixture that was prepared with \(\left[ {S{O_3}} \right] = 0.500M\), \(\left[ {S{O_2}} \right] = 0M\)and \(\left[ {{O_2}} \right] = 0.350M\)?

We must find the reaction quotient \(\left( {{Q_c}} \right)\). If \(\left( {{Q_c}} \right) = \left( {{K_c}} \right)\)the system is already at equilibrium and the results are correct.

\(\begin{array}{c}{Q_C} = \frac{{{{\left[ {S{O_3}} \right]}^2}}}{{{{\left[ {S{O_2}} \right]}^2} \times \left[ {{O_2}} \right]}}\\ = \frac{{{{\left( {0.260} \right)}^2}}}{{{{\left( {0.590} \right)}^2} \times 0.0450}}\\ = 4.3155\\ \approx 4.32\end{array}\)

Since \(\left( {{Q_c}} \right) = \left( {{K_c}} \right)\) the student’s results are correct.

Let us assume that the amount of \({\rm{S}}{{\rm{O}}_2}\)be x

\(\begin{array}{c}{K_C} = \frac{{{{\left[ {S{O_3}} \right]}^2}}}{{{{\left[ {S{O_2}} \right]}^2} \times \left[ {{O_2}} \right]}}\\4.32 = \frac{{{{\left( {0.500 - 2x} \right)}^2}}}{{{{\left( {2x} \right)}^2} \times \left( {0.350 + x} \right)}}\\4.32 = \frac{{0.25 - 2x + 4{x^2}}}{{1.4{x^2} + 4{x^3}}}\\4{x^2} - 2x + 0.25 = 17.28{x^3} + 6.048{x^2}\\17.28{x^3} + 2.048{x^2} + 2x - 0.25 = 0\\x = 0.104\end{array}\)

So, equilibrium constant of all species in the mixture are

\(\begin{array}{c}\left[ {S{O_2}} \right] = 2x = 0.208M\\\left[ {{O_2}} \right] = 0.350 + x = 0.454M\\\left[ {S{O_3}} \right] = 0.500 - 2x = 0.292M\end{array}\)