# JEE Main & Advanced Chemistry Chemical Kinetics Rate Law : Molecularity And Order Of A Reaction

Rate Law : Molecularity And Order Of A Reaction

Category : JEE Main & Advanced

Molecularity is the sum of the number of molecules of reactants involved in the balanced chemical equation. Molecularity of a complete reaction has no significance and overall kinetics of the reaction depends upon the rate determining step. Slowest step is the rate-determining step. This was proposed by Van't Hoff.

Example :     $N{{H}_{4}}N{{O}_{2}}\,\to {{N}_{2}}\,+\,2{{H}_{2}}O$     (Unimolecular)

$NO\,+\,{{O}_{3}}\,\to \,N{{O}_{2}}\,+\,{{O}_{2}}$            (Bimolecular)

$2NO\,+\,{{O}_{2}}\,\to \,2N{{O}_{2}}$               (Trimolecular)

The total number of molecules or atoms whose concentration determine the rate of reaction is known as order of reaction.

Order of reaction = Sum of exponents of the conc. terms in rate law

For the reaction $xA+yB\to \text{Products}$

The rate law is $\text{Rate}={{\text{ }\!\![\!\!\text{ A }\!\!]\!\!\text{ }}^{\text{x}}}[B{{[}^{y}}$

Then the overall order of reaction. $n=x+y$

where x and y are the orders with respect to individual reactants.

• If reaction is in the form of reaction mechanism then the order is determined by the slowest step of mechanism.

$2A+3B\to {{A}_{2}}{{B}_{3}}$

$A+B\to AB(\text{fast})$

$AB+{{B}_{2}}\to A{{B}_{3}}(\text{slow})$              (Rate determining step)

$A{{B}_{3}}+A\to {{A}_{2}}{{B}_{3}}(\text{fast})$

(Here, the overall order of reaction is equal to two.)

• Molecularity of a reaction is derived from the mechanism of the given reaction. Molecularity can not be greater than three because more than three molecules may not mutually collide with each other.
• Molecularity of a reaction can't be zero, negative  or fractional. order of a reaction may be zero, negative, positive or in fraction and greater than three. Infinite and imaginary values are not possible.
• When one of the reactants is present in the large excess, the second order reaction conforms to the first order and is known as pesudo unimolecular reaction. (Table 11.1)

#### Order and molecularity of some reaction

 S. No. Chemical equation Molecularity Rate law Order w.r.t. First reactant Second reactant Overall 1. $aA+bB\to$product a + b $\left( \frac{dx}{dt} \right)=k{{[A]}^{a}}{{[B]}^{b}}$ a b a + b 2. $aA+bB\to$product a + b $\left( \frac{dx}{dt} \right)=k{{[A]}^{2}}{{[B]}^{0}}$ 2 zero, if B is in excess 2 3. $2{{H}_{2}}{{O}_{2}}\xrightarrow{Pt,\Delta }2{{H}_{2}}O+{{O}_{2}}$ 2 (Bimolecular) $\left( \frac{dx}{dt} \right)=k[{{H}_{2}}{{O}_{2}}]$ 1* ­­----- 1 4. $C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}+{{H}_{2}}O\xrightarrow{{{H}^{+}}}$ $C{{H}_{3}}COOH+{{C}_{2}}{{H}_{5}}OH$ 2 (Bimolecular) $\left( \frac{dx}{dt} \right)=k[C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}]$ 1* Zero, if H2O is in excess 1 5. $\underset{\text{Sucrose}}{\mathop{{{C}_{12}}{{H}_{22}}{{O}_{11}}}}\,+{{H}_{2}}O\xrightarrow{{{H}^{+}}}$  $\underset{\text{Glucose}}{\mathop{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}\,+\underset{\text{Fructose}}{\mathop{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}\,$ 2 (Bimolecular) $\left( \frac{dx}{dt} \right)=k[{{C}_{12}}{{H}_{22}}{{O}_{11}}]$ 1* Zero, if H2O is in excess 1 6. ${{(C{{H}_{3}})}_{3}}CCl+O{{H}^{-}}\to$  ${{(C{{H}_{3}})}_{3}}COH+C{{l}^{-}}$ 2 (Bimolecular) $\left( \frac{dx}{dt} \right)=k[{{(C{{H}_{3}})}_{3}}CCl]$ 1* Zero, if OH– does not take part in slow step 1 7. $C{{H}_{3}}Cl+O{{H}^{^{-}}}\to$ $C{{H}_{3}}OH+C{{l}^{-}}$ 2 (Bimolecular) $\left( \frac{dx}{dt} \right)=k[C{{H}_{3}}Cl][O{{H}^{-}}]$ 1 1 2 8. ${{C}_{6}}{{H}_{5}}{{N}_{2}}Cl\xrightarrow{\Delta }$ ${{C}_{6}}{{H}_{5}}Cl+{{N}_{2}}$ 1 (Unimolecular) $\left( \frac{dx}{dt} \right)=k[{{C}_{6}}{{H}_{5}}{{N}_{2}}Cl]$ 1 ---- 1 9. $C{{H}_{3}}CHO\xrightarrow{\Delta }C{{H}_{4}}+CO$ 1 (Unimolecular) $\left( \frac{dx}{dt} \right)=k{{[C{{H}_{3}}CHO]}^{3/2}}$ 1.5 ---- 1.5 10. ${{H}_{2}}{{O}_{2}}+2{{I}^{-}}+2{{H}^{+}}\to$$2{{H}_{2}}O+{{I}_{2}}$ 5 $\left( \frac{dx}{dt} \right)=k[{{H}_{2}}{{O}_{2}}][{{I}^{-}}]$ 1 1 (H+is medium) 2 11. $2{{O}_{3}}\to 3{{O}_{2}}$ 2 (Bimolecular) $\left( \frac{dx}{dt} \right)=k{{[{{O}_{3}}]}^{2}}[{{O}_{2}}]$ 1 -1 with respect to O2 1

*Pseudo-unimolecular reactions.

#### Order

Rate constant

Unit of rate constant

Effect on rate by changing conc. to m times

(Half-life period) T50=

0

${{k}_{0}}=\frac{x}{t}$

conc. time–1

(mol L–1 s–1)

No change

$\frac{a}{2{{k}_{0}}}$

1

${{k}_{1}}=\frac{2.303}{t}{{\log }_{10}}\left( \frac{a}{a-x} \right)$, $C={{C}_{0}}{{e}^{-{{k}_{1}}t}}$

$N={{N}_{0}}{{e}^{-{{k}_{1}}t}}$, ${{k}_{1}}=\frac{2.303}{({{t}_{2}}-{{t}_{1}})}{{\log }_{10}}\frac{(a-{{x}_{1}})}{(a-{{x}_{2}})}$

time–1 (s–1)

m times

$\frac{0.693}{{{k}_{1}}}$

2

${{k}_{2}}=\frac{1}{t}\left[ \frac{1}{(a-x)}-\frac{1}{a} \right]$$=\frac{x}{ta(a-x)}$ (for the case when each reactant has equal concentration) ${{k}_{2}}=\frac{2.303}{t(a-b)}{{\log }_{10}}\left[ \frac{b(a-x)}{a(b-x)} \right]$(for the case when both reactants have different concentration)

conc–1 time–1

(mol L–1) s–1

L mol­–1 s–1

m2 times

$\frac{1}{{{k}_{2}}a}$

3

${{k}_{3}}=\frac{1}{2t}\left[ \frac{1}{{{(a-x)}^{2}}}-\frac{1}{{{a}^{2}}} \right]$

conc–2 time–1

(mol L–1)–2  s–1

L2 mol­–2 s–1

m3 times

$\frac{3}{2{{k}_{3}}{{a}^{2}}}$

n

${{k}_{n}}=\frac{1}{(n-1)t}\left[ \frac{1}{{{(a-x)}^{n-1}}}-\frac{1}{{{(a)}^{n-1}}} \right]$; $n\ge 2$

conc(1–n) time–1

(mol L–1)(1–n) s–1

L(n–1) mol­(1–n) s–1

mn times

$\frac{{{2}^{n-1}}-1}{(n-1){{k}_{n}}{{(a)}^{n-1}}}$

You need to login to perform this action.
You will be redirected in 3 sec