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Decomposition of N(2)O(5) is expressed b...

Decomposition of `N_(2)O_(5)` is expressed by the equation ,
`N_(2)O_(5)rarr2NO_(2)+^(1//2)O_(2)`
If during a certain time interval, the rate of decomposition of `N_(2)O_(5) is 1.8xx10^(-3) "mol litre"^(-1) "min"^(-1)` , what will be the rates of formation of `MO_(2) and O_(2)` during the same intrval ?

Text Solution

Verified by Experts

The rate expression of the decomposition of `N_(2)O_(5)` is :
`-(Delta[N_(2)O_(5)])/(Deltat)=(1)/(2)(Delta[NO_(2)])/(Deltat)=2*(Delta[O_(2)])/(Deltat)`
So , `(Delta[NO_(2)])/(Deltat)=2(Delta[NO_(2)O_(5)])/(Deltat)=2xx1.8xx10^(-3)`
`=3.6xx10^(-3) "mol litre"^(-1) min"^(-1)`
and `(Delta[O_(2)])/(Deltat)=(1)/(2)(Delta[N_(2)O_(5)])/(Deltat)=(1)/(2)xx1.8xx10^(-3)`
`=09xx10^(-3) "m ol litre" ^(-1) "min" ^(-1)`
(Rate is always positive and hence `-(Delta[N_(2)O_(5)])/(Deltat)` is taken positive. )
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