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For the reaction, N(2)O(5) rarr 2NO(2)...

For the reaction,
`N_(2)O_(5) rarr 2NO_(2)+O_(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:

A

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

B

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

C

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

D

none of these

Text Solution

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The correct Answer is:
A
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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:

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) ?

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

For the reaction, N_2O_5rarr2NO_2+1/2O_2 Given : -(d[N_2O_5])/(dt)=K_1[N_2O_5] (d[NO_2])/(dt)=K_2[N_2O_5] and (d[O_2])/(dt)=K_3[N_2O_5] The relation 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:

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