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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

Text Solution

Verified by Experts

The rate of reaction is given by `-(1)/(2)(d[NH_(3)])/(dt)=+(d[N_(2)])/(dt)=+(1)/(3)(d[H_(2)])/(dt)`
Substituting the given values, we get `(1)/(2)k_(1)[NH_(3)]=k_(2)[NH_(3)]=(1)/(3)k_(3)[NH_(3)]" or "(1)/(2)k_(1)=k_(2)=(1)/(3)k_(3)`
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