Question

What is an elementary reaction? 3.5. Given the reaction 2NO, + 1/202 = N,Os, what is the relationship between the rates of formation of 1,0, and disappearance of the two reactants?

Answer 1

Answer:

a) Elementary reactions are those which proceed in single step only.

b) $-\dfrac{1}{2}\dfrac{d\left [ NO_{2} \right ]}{dt}\ =\ -2\dfrac{d\left [ O_{2} \right ]}{dt}\ =\ \dfrac{1}{2}\dfrac{d\left [ N_{2}O_{5} \right ]}{dt}$

Explanation:

a) In case of elementary reactions, there is only one transition sate formed during the reactions.

There are no intermediates formed in elementary reactions as they occur in one step.

b) For the reaction

$2NO_{2}\ +\ \dfrac{1}{2}O_{2}\ \rightarrow N_{2}O_{5}$

Relationship between rate of formation of products and rate of disappearance of reactants is given by rate law expression.

$-\dfrac{1}{2}\dfrac{d\left [ NO_{2} \right ]}{dt}\ =\ -2\dfrac{d\left [ O_{2} \right ]}{dt}\ =\ \dfrac{1}{2}\dfrac{d\left [ N_{2}O_{5} \right ]}{dt}$

Coefficients of the reactants and products are written as the reciprocals in the rate law expression.

It clearly means that rate of formation of $N_{2}O_{5}$ is double of the rate of disappearance of oxygen.

Rate of formation of $N_{2}O_{5}$ is half of the rate of disappearance of $NO_2$.