A chemistry student wants to determine the effect of adding more reactant or product on the equilibrium position of a chemical reaction. The student has a 0.50 M solution of N2O4 and NO2 at equilibrium at 25°C, with the equation N2O4(g) <--> 2NO2(g). If the student adds more N2O4 to the reaction mixture, what will happen to the equilibrium concentration of NO2? Will it increase, decrease or stay the same? Justify your answer with a chemical equation and relevant calculations.

Question

A chemistry student wants to determine the effect of adding more reactant or product on the equilibrium position of a chemical reaction. The student has a 0.50 M solution of N2O4 and NO2 at equilibrium at 25°C, with the equation N2O4(g) <--> 2NO2(g). If the student adds more N2O4 to the reaction mixture, what will happen to the equilibrium concentration of NO2?

Answer 1

When the student adds more N2O4 to the reaction mixture, the equilibrium will shift to counteract the change according to Le Chatelier's principle. In this case, the equilibrium will shift to the right to consume the added N2O4 and produce more NO2. This will result in an increase in the equilibrium concentration of NO2.To justify this with a chemical equation and relevant calculations, let's assume the student adds x moles of N2O4 to the reaction mixture:Initial concentrations:[N2O4] = 0.50 M[NO2] = 0.50 MAfter adding x moles of N2O4:[N2O4] = 0.50 + x M[NO2] = 0.50 MAt the new equilibrium:[N2O4] =  0.50 + x  - y M  where y is the moles of N2O4 consumed [NO2] = 0.50 + 2y M  since 2 moles of NO2 are produced for every mole of N2O4 consumed The reaction equation is:N2O4 g  <--> 2NO2 g Since the equilibrium will shift to the right, y moles of N2O4 will be consumed, and 2y moles of NO2 will be produced. Therefore, the new equilibrium concentration of NO2 will be higher than the initial concentration:[NO2]new = 0.50 + 2y MIn conclusion, adding more N2O4 to the reaction mixture will cause the equilibrium concentration of NO2 to increase.