r = ∆ concentration   =  ∆[ ]
         ∆ time                   ∆t

Rate of Reaction

Rates of reactions are usually expressed in units of moles per liter per unit time. If we know the chemical equation for a reaction, its rate can be determined by following the change in concentration of any product or reactant that can be detected quantitatively.



To describe the rate of a reaction, we must determine the concentration of a reactant or product at various times as the reaction proceeds.

At some time, we observe that the reaction 2 N2O5 (g) → 4 NO2 (g) + O2 (g) is forming NO2 at the rate of 0.0072 mol / L∙s.

 (a) What is the rate of change of [O2],  ∆ [O2]/ ∆t in mol / L∙s?
(b) What is the rate of change of [N2O5], ∆ [N2O5]/ ∆t in mol / L∙s?

We can use the mole ratios from the balanced equation to determine the rates of change of other products and reactants. The rate of reaction can then be derived from any one of these individual rates.

(a) The balanced equation gives the reaction ratio 1 mol O2 : 4 mol NO2

rate of change of O2 =   ∆[O2]   =     0.0072 mol NO2    x    1 mol O2    = 0.0018 mol O2 / L ∙s
∆t L ∙s 4 mol NO2


(b) The balanced equation shows that 2 mol N2O5 is consumed for every 4 mol NO2 that is formed. Because [N2O5] is decreasing as [NO2] increases, we should write the reaction ratio as


rate of change of O2 =   ∆[O2]   =     0.0072 mol NO2    x   -2 mol N2O5    = - 0.0036 mol N2O5 / L ∙s 
∆t L ∙s 4 mol NO2


For the reaction:

C2H4O (g) -> CH4 (g) + CO (g)

the concentration of the ethanal (reactant) will decrease over time. If the concentration of ethanal was plotted versus time we would see that its concentration decreases:

The average rate over of ethanal decomposition over time can be determine by calculating the slope of a secant between those two time intervals.


(250,0.2375) (2250, 0.0375)

r = Δ[ethanal] / Δt

   = (0.0375 - 0.2375 mol/L) / (2250 - 250 s)

   = (- 0.2000 mol/L) / (2000 s)

   = - 1 x 10-4 mol L-1s-1

It is also possible to determine the instantaneous rate of ethanal deomposition at a particlar time by calculating the slope of a tangent.

(625,0.15) (1375, 0.05)

r = Δ[ethanal] / Δt

   = (0.05 - 0.15 mol/L) / (1375 - 625 s)

   = (-0.1 mol/L) / (750 s)

   = -1.3 x 10-4 mol L-1s-1