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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 ' = 2K : K'' =2K`

B

`k '=K : K'' =K`

C

`K' = 2K : k'' =K`

D

`k' = 2k : k'' = k//2`

Text Solution

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The correct Answer is:
D
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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 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:

2N_2O_5 rarr 4 NO_2 +O_2 If -(D[N_2O_5])/(dt) =k_1[N_2O_5] (d[NO_2])/(dt) =k_2[N_2O_5] ([O_2])/(dt) =k_3[N_2O_5] What is the relation between k_1, k_2 and k_3 ? .

Consider the following reaction, 2N_(2)O_(5)rarr4NO_(2)+O_(2),(d[NO_(2)] )/(dt)=k_(2)[N_(2)O_(5)] , (d[O_(2)])/(dt)=k_(3)[N_(2)O_(5)]" and "(d)/(dt)[N_(2)O_(5)]=k_(1) The relation between k_(1), k_(2) and k_(3) is

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