Question

If the initial concentration of N2O5 was 0.500 M and the concentration of N2O5 was 0.450 M after 0.100 s, what is the average rate of NO2 formation during the first 100 milliseconds

Answer 1

Answer:

The appropriate solution is "1 M/s".

Explanation:

The equation is:

$2N_2O_5 \rightarrow4NO_2+O_2$

or,

$\frac{1}{2}(-\frac{\Delta(N_2O_5)}{dt} ) =\frac{1}{4}\frac{\Delta (NO_2)}{dt}$

then,

The rate of $NO_2$ formation:

= $\frac{\Delta NO_2}{dt}$

= $\frac{4}{2} (-\frac{\Delta(N_2O_5)}{dt})$

= $2(-\frac{\Delta (N_2O_5)}{dt} )$

= $2(-\frac{(0.45-0.5)}{0.1} )$

= $1 \ M/s$