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

Answer: A)
`0.002mol^(-1)s^(-1)`

B)
`0.001mol^(-1)s^(-1)`

Step-by-step explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

`2N_2O_5\rightarrow 4NO_2+O_2`

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of
`N_2O_5` =
`-(1d[N_2O_5])/(2dt)`

Rate in terms of appearance of
`NO_2` =
`(1d[NO_2])/(4dt)`

Rate in terms of appearance of
`O_2` =
`(1d[O_2])/(dt)`

Given :

Rate of formation of
`NO_2=+(1d[NO_2])/(dt)=0.004mol^(-1)s^(-1)`

As
`-(1d[N_2O_5])/(2dt)` =
`(1d[NO_2])/(4dt)` =
`(1d[O_2])/(dt)`

A) Rate of disappearance of
`N_2O_5=(2)/(4)* (1d[NO_2])/(dt)=(2)/(4)* 0.004=0.002mol^(-1)s^(-1)`

B) Rate of appearance of
`O_2` =
`(1)/(4)* (1d[NO_2])/(dt)=(1)/(4)* 0.004=0.001mol^(-1)s^(-1)`