question_answer

  Following reaction takes place in one step: \[2NO(g)+{{O}_{2}}(g)\xrightarrow{\,}\,2\,N{{O}_{2}}(g)\] How will the rate of above reaction change if the volume of the reaction vessel is diminished to one third of its original volume? Will there be any change in the order of reaction with the reduced volume?

Answer:

                  Reaction rate \[=k{{[NO]}^{2}}\,[{{O}_{2}}]\] Since the volume of the vessel is reduced to one third (1/3) of the original volume, the concentration of the reactants per unit volume will become thrice. Rate with respect to NO will become nine times. Rate with respect to \[{{O}_{2}}\] will become thrice. Overall reaction rate will increase to 27 times. There will no change in the order of the reaction.