Answer :
Answer:
2.8 mol/min is the rate of appearance of [tex]NO_2[/tex] at that moment.
Explanation:
[tex]2N_2O_5\rightarrow 4NO_2+O_2[/tex]
Rate of the reaction = R
[tex]R=\frac{-1}{2}\frac{d[N_2O_5]}{dt}=\frac{1}{4}\frac{d[NO_2]}{dt}=\frac{1}{1}\frac{d[O_2]}{dt}[/tex]
Rate of disappearance of [tex]N_2O_5[/tex] is [tex]=-\frac{d[N_2O_5]}{dt}= 1.40 mol/min[/tex]
[tex]R=\frac{-1}{2}\frac{d[NO_2]}{dt}[/tex]
[tex]R=\frac{1}{2}\times 1.40 mol/min=0.70 mol/min[/tex]
Rate of appearance of [tex]NO_2[/tex] is [tex]=\frac{1}{4}\frac{d[NO_2]}{dt}:[/tex]
[tex]R=\frac{1}{4}\frac{d[NO_2]}{dt}[/tex]
[tex]\frac{d[NO_2]}{dt}=4\times 0.70 mol/min=2.8 mol/min[/tex]
2.8 mol/min is the rate of appearance of [tex]NO_2[/tex] at that moment.