Question

For the reaction, 2 N2O5--> 4 NO2+O2, the rate of formation of NO2 is 0.004 mol^-1 s^-1.

A) Calculate the rate of disappearance of N2O5

B) Calculate the rate of appearance of O2

Answer 1

Rate of the reaction is given by

Rate = -(1/2) * {d[N₂O₅] / dt } = (1/4) * {d[NO₂] / dt } = {d[O₂] / dt }

where d[N₂O₅] / dt   is rate of change in concentration of N₂O₅; d[NO₂] / dt is rate of formation of NO₂

and d[O₂] / dt is rate of formation of oxygen.

Given, rate of formation of NO₂, d[NO₂] / dt = 0.004 mol^-1 s^-1

we have, -(1/2) * {d[N₂O₅] / dt } = (1/4) * {d[NO₂] / dt }

So, {d[N₂O₅] / dt } = -(2/4) * {d[NO₂] / dt } = -(2/4) * 0.004 mol^-1 s^-1 = - 0.002 mol^-1 s^-1

So, {d[N₂O₅] / dt } = - 0.002 mol^-1 s^-1

Hence, rate of change in concentration of N₂O₅ is - 0.002 mol^-1 s^-1

negative sign indicates decrease in concentration with increase in time

Hence, rate of disappearance of N₂O₅ is 0.002 mol^-1 s^-1

We also have, (1/4) * {d[NO₂] / dt } = {d[O₂] / dt }

So, (1/4) * 0.004 mol^-1 s^-1 = {d[O₂] / dt }

So, {d[O₂] / dt } = 0.001 mol^-1 s^-1

Hence, rate of formation of oxygen is 0.001 mol^-1 s^-1

Answers: A) 0.002 mol^-1 s^-1   B) 0.001 mol^-1 s^-1