Order of Reaction

The mathematical expression showing the dependence of rate on the concentration(s) of reactant(s) is known as rate-law or rate-expression of the reaction and sum of the indices (powers) of the concentration terms appearing in the rate law as observed experimentally is called order of reaction .

To understand what is order of reaction, consider the reaction:

2NO(g) + 2H2(g) –> N2(g) + 2H2O (g)

Kinetic experiment carried out at 1100 K upon this reaction has shown following rate data.

Expt. No.   [NO] (mole dm−3)   [H2] (mole dm−3)   Rate (mole dm−3s−1)

1.             5 × 10−3         2.5 × 10−3                     3 × 10−5

2.             1.0 × 10−2         2.5 × 10−3               1.2 × 10−4

3.             1.0 × 10−2         5.0 × 10−3               2.4 × 10−4

From the Expt. No.1 and 2, it is evident that rate increases 4 fold when concentration of NO is doubled keeping the concentration of H2 constant i.e.

Rate ∝ [NO]2 when [H2] is constant again from Expt. No.2 and 3, it is evident that when concentration of H2 is doubled keeping the concentration of NO constant, the rate is just doubled i.e.

Rate ∝ [H2] when [NO] is constant

From Expt. (1) and Expt. (3), the rate increases 8-fold when concentrations of both NO and H2 are doubled simultaneously i.e.

Rate ∝ [NO]2 [H2]

This is the rate-law of reaction as observed experimentally. In the rate law, the power of nitric oxide concentration is 2 while that of hydrogen concentration is 1. So, order of reaction w.r.t. NO is 2 and that w.r.t. H2 is 1 and overall order is 2 + 1 i.e. 3.

Note that the experimental rate law is not consistent with the stoichiometric coefficient of H2 in the chemical equation for the reaction. This fact immediately suggests that the reaction is complicated and it does not occur in single step as written. In order to explain the rate law, following mechanism has been proposed.

i) NO + NO <–> N2O2 (fast and reversible)

ii) N2O2 + H2 –> N2O + H2O (slow)

iii) N2O + H2 –> N2 + H2O (fast)

Let us see how this mechanism corresponds to the rate law as found experimentally and mentioned above.

The step II being the slowest step is the rate-determining step. Thus, rate of overall reaction or rate of formation of N2, will be equal to the rate of step II or rate of formation of H2O. So, we have according to Law of Mass Action.

Rate of overall reaction = Rate of step II = k [N2O2] [H2]

Where k = rate constant of step II.

N2O2 being intermediate for the overall reaction, its concentration has to be evaluated in terms of the concentration(s) of reactant(s) and this can be done by applying Law of Mass Action upon the equilibrium of Step I. Thus,

$\large K_C = \frac{[N_2O_2]}{[NO]^2}$

or, [N2O2] = KC[NO]2

where KC = equilibrium constant of Step II. Putting this value of concentration of N2O2 in the above rate expression, we get

Rate reaction = k.Kc . [NO]2 [H2]

or Rate of reaction = k1[NO]2 [H2]

Rate of reaction ∝ [NO]2[H2]

Where k.Kc = k1 = another constant, rate constant of overall reaction.

Note that from the knowledge of any two out of k , Kc and k1 , the rest one may be calculated.