Question

Which of the following is an example of a fractional order reaction?
A.$N{H_4}N{O_2}\xrightarrow{{}}{N_2} + 2{H_2}O$
B.$NO + {O_3}\xrightarrow{{}}N{O_2} + {O_2}$
C.$2NO + B{r_2}\xrightarrow{{}}2NOBr$
D.$C{H_3}CHO\xrightarrow{{}}C{H_4} + CO$

Answer 1

We can say in fractional order reactions, the order is a non-integer that indicates chemical chain reaction or other complex reaction mechanism.
The order of a chain reaction can be rationalized using the steady state approximation for the concentration of reactive intermediates such as free radicals.

Complete step by step answer:

We know that order of the chemical reaction is the sum of the power of concentration of reactants in the rate law expression.
For a reaction, $xA + yB\xrightarrow{{}}P$
Rate=$k{\left[ A \right]^x}{\left[ B \right]^y}$
By summing up the power of the concentration of reactants we give the order of the reaction.
Order=$x + y$
For the reaction, $N{H_4}N{O_2}\xrightarrow{{}}{N_2} + 2{H_2}O$
The rate of the reaction is $k = \left[ {N{H_4}N{O_2}} \right]$ and we can see the power of the concentration of reactant is one and the order of the reaction is 1.
Therefore, option (A) is incorrect.
For the reaction, $NO + {O_3}\xrightarrow{{}}N{O_2} + {O_2}$
The rate of the reaction is $k = \left[ {NO} \right]\left[ {{O_2}} \right]$ and we can see the power of concentration of $NO$ is one and power of concentration of ${O_2}$ is one. Adding both the powers $\left( {1 + 1 = 2} \right)$, order of the reaction is 2.
Therefore, option (B) is incorrect.
For the reaction, $2NO + B{r_2}\xrightarrow{{}}2NOBr$
The rate of the reaction is $k = {\left[ {NO} \right]^2}\left[ {B{r_2}} \right]$ and we can see the power of concentration of $NO$ is two and power of concentration of $B{r_2}$ is one. Adding both the powers $\left( {2 + 1 = 3} \right)$ the order of the reaction is 3.
Therefore, option (C) is incorrect.
For the reaction, $C{H_3}CHO\xrightarrow{{}}C{H_4} + CO$, the order of the reaction will be 1.5 with respect to aldehyde.
Initiation: $C{H_3}CHO\xrightarrow{{}} \bullet C{H_3} + \bullet CHO$
Propagation: $\bullet C{H_3} + C{H_3}CHO\xrightarrow{{}}C{H_3}CO \bullet + C{H_4}$
$C{H_3}CO \bullet \xrightarrow{{}} \bullet C{H_3} + CO$
Termination: $2 \bullet C{H_3}\xrightarrow{{}}{C_2}{H_6}$
In the steady state, the rates of formation and destruction of methyl radicals are equal, so that

$\dfrac{{d\left[ { \bullet C{H_3}} \right]}}{{dt}} = {k_i}\left[ {C{H_3}CHO} \right] - {k_t}{\left[ { \bullet C{H_3}} \right]^2} = 0$,
So that concentration of methyl radical satisfies,
$\left[ { \bullet C{H_3}} \right]\alpha {\left[ {C{H_3}CHO} \right]^{\dfrac{1}{2}}}$
The rate of the reaction is equal to the rate of the propagation steps which form the main reaction products methane and carbon monoxide:
$\dfrac{{d\left[ {C{H_4}} \right]}}{{dt}} = {k_p}\left[ { \bullet C{H_3}} \right]\left[ {C{H_3}CHO} \right]\alpha {\left[ {C{H_3}CHO} \right]^{\dfrac{3}{2}}}$
Here $\left[ { \bullet C{H_3}} \right]$ is free methyl radical.
Therefore, the order of the reaction is $\dfrac{3}{2}$.
Therefore, option (D) is correct.

Note:
We have to know that reaction order indicates the number of species whose concentration directly affects the rate of reaction.
The order of reaction does not depend on the stoichiometric coefficients corresponding to each species in the balanced reaction.
We have to know that the order of a chemical reaction is defined with the help of concentrations of reactants and not with concentrations of products.
The value of the order of reaction can be in the form of a whole number or a fraction. The order of the reaction could also be zero.