Question

Given the reaction 2NO2 1/202 N2O5, what is the relationship between the rates of formation of N,0, and disappearance of the two reactants?

Answer 1

Answer:

$r=\frac{r(NO_{2})}{2} =\frac{r(O_{2})}{1/2}=\frac{r(N_{2}O_{5})}{1}$

Explanation:

Let us consider the reaction:

2 NO₂ + 1/2 O₂ ⇄ N₂O₅

The rate of formation of a substance is equal to the change in concentration of the product divided the change in time:

$r(N_{2}O_{5})=\frac{\Delta [N_{2}O_{5}] }{\Delta t}$

The rate of disappearance of a reactant is equal to to the change in concentration of the reactant divided the change in time, with a negative sign so that the rate is always a positive variable.

$r(NO_{2})=-\frac{\Delta[NO_{2}] }{\Delta t}$

$r(O_{2})=-\frac{\Delta[O_{2}] }{\Delta t}$

The rate of the reaction is equal to the rate of any substance divided its stoichiometric coefficient. In this way, we can relate these expressions:

$r=\frac{r(NO_{2})}{2} =\frac{r(O_{2})}{1/2}=\frac{r(N_{2}O_{5})}{1}$