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.
\[2A+3B\to {{A}_{2}}{{B}_{3}}\]
\[A+B\to AB(\text{fast})\]
\[AB+{{B}_{2}}\to A{{B}_{3}}(\text{slow})\] (Rate determining step)
\[A{{B}_{3}}+A\to {{A}_{2}}{{B}_{3}}(\text{fast})\]
(Here, the overall order of reaction is equal to two.)
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 |
\[\frac{{{2}^{n-1}}-1}{(n-1){{k}_{n}}{{(a)}^{n-1}}}\] |
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