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

`r=(r(NO_(2)))/(2) =(r(O_(2)))/(1/2)=(r(N_(2)O_(5)))/(1)`

Step-by-step 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))=(\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))=-(\Delta[NO_(2)] )/(\Delta t)`

`r(O_(2))=-(\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=(r(NO_(2)))/(2) =(r(O_(2)))/(1/2)=(r(N_(2)O_(5)))/(1)`