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For the reaction, 2NH(3)rarr N(2)+3H(2) ...

For the reaction, `2NH_(3)rarr N_(2)+3H_(2)`
if `(-d[NH_(3)])/(dt)=k_(1)[NH_(3)]`,
`(d[N_(2)])/(dt)=k_(2)[NH_(3)], (d[H_(2)])/(dt)=k_(3)[NH_(3)]`
then the relation between `k_(1), k_(2)` and `k_(3)` is

A

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

B

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

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
A
`+(1)/(2) (d[NH_(3)])/(dt) =-(1)/(1) (d[N_(2)])/(dt) =-(1)/(3)(d[H_(2)])/(dt) +(K_(1))/(2)=-K_(2)=(K_(3))/(3)`
`1.5 K_(1) =3K_(2)=K_(3)`
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