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The rate of reaction. 2N(2)O(5) to 4NO...

The rate of reaction.
`2N_(2)O_(5) to 4NO_(2) + O_(2)`
can be written in three ways.
`(-d[N_(2)O_(5)])/(dt) = k[N_(2)O_(5)]`
`(d[N_(2)O_(5)])/(dt) =( k^(')[N_(2)O_(5)])`
`(d[O_(2)])/(dt) = (k^(')[N_(2)O_(5)])`
The relation between k and `k^(')` are:

A

`k' = 2k, k'' = 2k`

B

`k'=k,k''=k`

C

`k'=2k,k''=k`

D

`k'=2k,k''=k/2`

Text Solution

Verified by Experts

The correct Answer is:
D
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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) ?

For the reaction, N_(2)O_(5) rarr 2NO_(2)+1/2O_(2), Given -(d[N_(2)O_(5)])/(dt)=K_(1)[NO_(2)O_(5)] (d[NO_(2)])/(dt)=K_(2)[N_(2)O_(5)] and (d[O_(2)])/(dt)=K_(3)[N_(2)O_(5)] The relation in between K_(1), K_(2) and K_(3) is:

The reaction 2N_(2)O_(5)hArr 2N_(2)O_(4)+O_(2) is

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