Chapter 11, Problem 11.71PAE

Interpretation Introduction

Interpretation: Reaction of $N{O}_2\text{ and CO}$ occur in two steps. (a) Stoichiometric reaction for the whole mechanism, (b) molecularity of each step, (c) the experimental rate equation for the mechanism to be consistent with kinetic data, (d) intermediates in reaction should be stated.

Concept introduction:

Stoichiometric ratio represent the molar ratio of all the products and reactants and products, but not the intermediates participated in the reaction mechanism.

Molecularity is a theoretical concept and should not be negative, zero, fractional, infinite and imaginary. Molecularity is the total number of reactant molecules or atoms taking part in the chemical reaction.

Answer to Problem 11.71PAE

Solution:

(a) Stoichiometric equation

$N{O}_2{}_{(g)}+C{O}_{(g)} +N{O}_{3(g)}\to N{O}_{(g)}+N{O}_3{}_{(g)}+C{O}_2{}_{(g)}$

(b) Molecularity

Step 1- 2

Step 2 − 2

(c) Experimental rate equation

$\begin{array}{l}\begin{array}{l}Rate=\raisebox{1ex}{$d[NO]$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.=-k_1{\left[N{O}_2\right]}^2+k_2\left[N{O}_3\right]\left[CO\right]\\ \end{array}\hfill \\ \begin{array}{l}Here\\ k_1=rateconstantforthestep1\end{array}\hfill \\ k_2=rateconstantforthestep2\hfill \\ \hfill \end{array}$

(d) Intermediates

$NO{\text{ and NO}}_3$

Explanation of Solution

(a) The molar ratio of the reactant molecules, products in balance equation is called stoichiometry. Some reactions take so many steps to achieve their final products. There can be so many intermediates. But those are not involved in the balanced equation. If a reaction takes so many steps to give their final products, by adding them together we can their stoichiometric equation.

$\begin{array}{l}Step\text{ 1 }N{O}_{2(g)}+N{O}_{2(g)}\to N{O}_{(g)}+N{O}_{3(g)}\\ Step{\text{ 2 NO}}_{3(g0}+C{O}_{(g)}\to N{O}_{2(g)}+C{O}_{2(g)}\end{array}$

$\begin{array}{l} \hfill \\ 2N{O}_2+CO +N{O}_3\to N{O}_2+NO+N{O}_3+C{O}_2\hfill \end{array}$

Then we should cancel out same molecules present in both side, we can write overall reaction as follows,

$N{O}_2{}_{(g)}+C{O}_{(g)} +N{O}_{3(g)}\to N{O}_{(g)}+N{O}_3{}_{(g)}+C{O}_2{}_{(g)}$

(b) Molecularity is the total number of reactant molecules or atoms taking part in the chemical reaction.

Therefore, molecularity of step1 is 2 and molecularity of step 2 is also 2

(c) Rate equation can be written using either product or reactants. If take reactants, we should concern the reduction rate of reactants. If we take product, we should concern the growth rate of products.

$\begin{array}{l}\begin{array}{l}Rate=\raisebox{1ex}{$d[NO]$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.=-k_1{\left[N{O}_2\right]}^2+k_2\left[N{O}_3\right]\left[CO\right]\\ \end{array}\hfill \\ \begin{array}{l}Here\\ k_1=rateconstantforthestep1\end{array}\hfill \\ k_2=rateconstantforthestep2\hfill \\ \hfill \end{array}$

Some reactions should follow many steps to give their final product. In between these steps they will form intermediates. Intermediates don’t involve in the overall reaction. And they cannot be separated out. Here, $NO{\text{ and NO}}_3$

(d) are intermediates since they were produced by interaction of reactants, but they involved in further reactions to result in different final products.

Conclusion

In a reaction mechanism, more than 1 step is involved although only one could be rate determining. Intermediates are formed within the process and are not present in the final stoichiometric equation.