Question

Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene: \(3{{\rm{C}}_2}{{\rm{H}}_2}(g) \to {{\rm{C}}_6}{{\rm{H}}_6}(g)\). Which value of \({K_c}\) would make this reaction most useful commercially?

\({K_c} \approx 0.01,\;{K_c} \approx 1,\;\;{\rm{or}}\;{K_c} \approx 10.\) Explain your answer.

Short answer

The value that makes the reaction commercially useful is\({K_c} \approx 10\)_.

Step-by-step solution

Define equilibrium constant

The product of the reactant concentrations to the power of their reaction coefficient (the number before them in the reaction equation) divided by the product of the product concentrations to the power of their reaction coefficients, independent of the reaction, is the equilibrium constant.

Find the value of \({K_c}\)

The equilibrium constant for this reaction is expressed as:

\({K_c} = \frac{{c\left( {{C_6}{H_6},g} \right)}}{{c{{\left( {{C_2}{H_2},g} \right)}^3}}}\)

If our goal is to make as much benzene as feasible, the concentration of\({C_6}{H_6}\)should be as high as possible.

Thus, if \({K_c} \approx 10\), this reaction would be most commercially useful.