Answer:
Average rate of reaction is 0.000565 M/min
Explanation:
Applying law of mass action for the given reaction:
Average rate = [tex]-\frac{1}{2}\frac{[N_{2}O_{5}]}{\Delta t}=\frac{1}{4}\frac{\Delta [NO_{2}]}{\Delta t}=\frac{\Delta [O_{2}]}{\Delta t}[/tex]
Where, [tex]-\frac{1}{2}\frac{[N_{2}O_{5}]}{\Delta t}[/tex] represents average rate of disappearance of [tex]N_{2}O_{5}[/tex], [tex]\frac{1}{4}\frac{[NO_{2}]}{\Delta t}[/tex] represents average rate of appearance of [tex]NO_{2}[/tex] and [tex]\frac{[O_{2}]}{\Delta t}[/tex] represents average rate of appearance of [tex]O_{2}[/tex]
Here,[tex]-\frac{[N_{2}O_{5}]}{\Delta t}[/tex] = [tex]-\frac{(2.16-2.36)}{(177-0)}M/min=0.00113M/min[/tex]
So average rate of reaction = [tex][tex]-\frac{1}{2}\frac{[N_{2}O_{5}]}{\Delta t}[/tex][/tex] = [tex]\frac{1}{2}\times (0.00113M/min)=0.000565M/min[/tex]
