Themes > Science > Chemistry > Miscellenous > Help file Index > kinetics > Reaction Rate The reaction rate is a quantity that defines how the concentration of a reactant or product changes with time. The rate is defined to be a positive number. If we consider a reaction such as A + B -> C + D we can ask how the concentration of A, B, C and D change over time. The average reaction rate is therefore the simple expression average reaction rate = change in concentration/change in time For example, in the above reaction the reaction rate expression can be written as rate = D[C] /DT Since the rate of production of C is the same as the rate of production of D (One molecule of D is produced when one molecule of C is produced), and is the negative of the change in A and B (For each molecule of C, one A and one B get used up.), we also have rate = D[C] /DT = - D[D] /DT = -D[A] /DT = -D[B] /DT For more complex reactions, we have to take into account the fact that some species are produced or used up faster than others: consider the reaction 2N2O5 -> 4NO2 + O2 In this reaction, NO2 is produced twice as fast as N2O5 is used up, and oxygen is produced only 1/2 as fast. Therefore the relationships between the rates of change for the various species is -D[N2O5] = D[NO2/2] = D[O2]/0.5 The rate expression for this reaction is therefore rate = -D[N2O5]/DT = D[NO2]/2DT = D[O2]/0.5DT The graph below shows the various concentrations as a function of time: Example: If we study the reaction of nitrogen and hydrogen gases to form ammonia, we find that the nitrogen is disappearing at a rate of 0.100 mol/L*min. What is the rate expression of the reaction and what is the rate of disappearance of hydrogen and appearence of ammonia? N2 + 3H2 -> 2NH3 Solution: The nitrogen in this reaction is disappearing at a rate of 0.100 mol/L*min. The rate is therefore rate = -D[N2]/DT = 0.100 mol/L*minFor the other species, 3 molecules of hydrogen are used for every molecule of nitrogen, and two molecules of ammonia are produced. rate = -D[N2]/DT = -D[H2]/3DT = D[NH3]/2DT =0.100 mol/L*minTherefore, the change in concentration for each species is -D[H2]/3DT = 0.100 mol/L*min ---> D[H2]/DT = -0.300 mol/L*min D[NH3]/2DT = 0.100 mol/L*min ---> D[NH3]/DT = 0.200 mol/L*min Information provided by: http://learn.chem.vt.edu