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2N(2)O(5) rarr 4NO(2) + O(2) If (-d[N(...

`2N_(2)O_(5) rarr 4NO_(2) + O_(2)`
If `(-d[N_(2)O_(5)])/(dt) = k_(1)[N_(2)O_(5)]`
`(d[NO_(2)])/(dt) = k_(2)[N_(2)O_(5)]`
`(d[O_(2)])/(dt) = k_(3)[N_(2)O_(5)]`
What is the relation between `k_(1), k_(2)`, and `k_(3)`?

A

`k_(1)=k_(2)=k_(3)`

B

`k_(1)=2k_(2)=4k_(3)`

C

`2k_(1)=k_(2)=4k_(3)`

D

`4k_(1)=k_(2)=2k_(3)`

Text Solution

Verified by Experts

The correct Answer is:
C
The rate law of the given reaction is Rate =
`-(1)/(2)((d[N_(2)O_(5)])/(dt))=(1)/(4)((d[NO_(2)]]/(dt))=(d[O_(2)])/(dt)=K[N_(2)O_(5)]`
i.e. `K_(2)=4K` or `K=k_(2)//2`
`(d[NO_(2)])/(dt)=4K[N_(2)O_(5)]=K_(2)[N_(2)O_(5)]`
i.e. `K_(2)=4K` or `K=K_(2)//4`
`(d[O_(2)])/(dt)=K[N_(2)O_(5)]=K_(3)[N_(2)O_(5)]`
Combining all these results we have
`K=K_(1)//2=K_(2)//4=K_(3)`
Multiplying by `4` we get
`2K_(1)=K_(2)=4K_(3)`
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