Answer:
The appropriate solution is "1 M/s".
Explanation:
The equation is:
[tex]2N_2O_5 \rightarrow4NO_2+O_2[/tex]
or,
[tex]\frac{1}{2}(-\frac{\Delta(N_2O_5)}{dt} ) =\frac{1}{4}\frac{\Delta (NO_2)}{dt}[/tex]
then,
The rate of [tex]NO_2[/tex] formation:
= [tex]\frac{\Delta NO_2}{dt}[/tex]
= [tex]\frac{4}{2} (-\frac{\Delta(N_2O_5)}{dt})[/tex]
= [tex]2(-\frac{\Delta (N_2O_5)}{dt} )[/tex]
= [tex]2(-\frac{(0.45-0.5)}{0.1} )[/tex]
= [tex]1 \ M/s[/tex]
