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For the reaction N(2) to 3H(2) to 2NH(3)...

For the reaction `N_(2) to 3H_(2) to 2NH_(3)`, how are the rate of reaction expressions inter-related `(d[H_(2)])/(dt)` and `(d[NH_(3)])/(dt)`?

A

`(d[NH_(3)])/dt=-(1)/(3) (d[H_(2)])/(dt)`

B

`+(d[NH_(3)])/(dt)=-(1)/(3) (d[H_(2)])/dt`

C

`+(d[NH_(3)])/(dt)=-(3)/(2) (d[H_(2)])/dt`

D

`+(d[NH_(3)])/(dt)=-(d[H_(2)])/dt`

Text Solution

Verified by Experts

The correct Answer is:
B
For the reaction `N_(2)(g)+3H_(2)(g)to 2NH_(3)(g)`
`-(d[N_(2)])/dt=-(1)/(3)(d)/(dt)([H_(2)])=(1)/(2)(d)/(dt)([NH_(3)])`
`therefore(d)/(dt)([NH_(3)])=-(2)/(3)(d)/(dt)([h_(2)])`
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