Question

Find with respect to each reactant, order of the reaction:

& CO+C{{l}_{2}}\to COC{{l}_{2}} \\\ & 2NO+B{{r}_{2}}\to 2NOBr \\\ \end{aligned}$$

Answer 1

We have to find the slowest step in a given reaction. Because the slowest step is called rate determining step for a reaction. By using rate determining steps we can calculate the order of the reaction with respect to each reactant.

Complete step by step solution:
- In the question it is given to find the order of the reaction with respect to each reactant.
- The given reactions are as follows.

& CO+C{{l}_{2}}\to COC{{l}_{2}} \\\ & 2NO+B{{r}_{2}}\to 2NOBr \\\ \end{aligned}$$ \- Coming to first reaction $$CO+C{{l}_{2}}\to COC{{l}_{2}}$$ \- The above reaction does not happen in a single step and it happens in three steps and there are as follows. $$\begin{aligned} & i)~C{{l}_{2}}\underset{{{K}_{2}}}{\overset{{{K}_{1}}}{\longleftrightarrow}}2Cl~\left( fast \right) \\\ & ii)~Cl+CO\underset{K_{2}^{'}}{\overset{K_{1}^{'}}{\longleftrightarrow}}COCl~\left( fast \right) \\\ & iii)COCl+C{{l}_{2}}\xrightarrow{{{K}_{3}}}COC{{l}_{2}}+Cl~\left( slow \right) \\\ \end{aligned}$$ \- From Step -1we can write the rate expression as follows. $$\dfrac{{{K}_{1}}}{{{K}_{2}}}=\dfrac{{{\left[ Cl \right]}^{2}}}{\left[ C{{l}_{2}} \right]}\to (a)$$ \- From step-2 we can write the rate expression as follows. $$\dfrac{K_{1}^{'}}{K_{2}^{'}}=\dfrac{\left[ COCl \right]}{\left[ CO \right]\left[ Cl \right]}\to (b)$$ -From step-3 $$\dfrac{d[COC{{l}_{2}}]}{dt}={{K}_{3}}[COCl][C{{l}_{2}}]\to (c)$$ \- From the rate expressions (a) and (b) $$[COCl]=\dfrac{K_{1}^{'}}{K_{2}^{'}}[CO]\times \sqrt{\dfrac{{{K}_{1}}}{{{K}_{2}}}[C{{l}_{2}}]}\to (d)$$ \- From (c) and (d) $$\begin{aligned} & \dfrac{d[COC{{l}_{2}}]}{dt}={{K}_{3}}\dfrac{K_{1}^{'}}{K_{2}^{'}}{{\left( \dfrac{{{K}_{1}}}{{{K}_{2}}} \right)}^{1/2}}[CO]{{[C{{l}_{2}}]}^{1/2}}[C{{l}_{2}}] \\\ & \dfrac{d[COC{{l}_{2}}]}{dt}=K[CO]{{[C{{l}_{2}}]}^{3/2}} \\\ \end{aligned}$$ Therefore the order of the reaction with respect to CO is 1 and the order of the reaction with respect to chlorine is 3/2. \- Coming to equation 2 from the question. $$2NO+B{{r}_{2}}\to 2NOBr$$ \- The above reaction is also not a single step reaction. \- It takes place through two steps and they are as follows. $$\begin{aligned} & i)NO+B{{r}_{2}}\overset{fast}{\longleftrightarrow}NOB{{r}_{2}} \\\ & ii)NOB{{r}_{2}}+NO\xrightarrow{slow}2NOBr \\\ \end{aligned}$$ \- The rate of the above reaction can be determined by a slow step means from step-2. \- The rate of the reaction is as follows. Rate = $K[NOB{{r}_{2}}][NO]\to (a)$ \- The equilibrium constant ${{K}_{c}}$ is as follows. $$\begin{aligned} & {{K}_{c}}=\dfrac{[NOB{{r}_{2}}]}{[N{{O}_{2}}][B{{r}_{2}}]} \\\ & [NOB{{r}_{2}}]={{K}_{C}}[N{{O}_{2}}][B{{r}_{2}}]\to (b) \\\ \end{aligned}$$ From equations (a) and (b) Rate = $K{{[NO]}^{2}}[B{{r}_{2}}]$ -Therefore the overall order of the reaction is 3 and with respect to [NO] the order is two and with respect to bromine the order is 1. **Note:** The order of a reaction is the sum of the powers of the concentration terms involved in the rate of the reaction. Without writing the rate of the chemical reaction we cannot find the order of the particular chemical reaction.