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Consider the reaction: N(2(g))+3H(2(g)...

Consider the reaction:
`N_(2(g))+3H_(2(g))rarr 2NH_(3(g))`.
The equally relationship between `-(d[NH_(3)])/(dt)` and `-(d[H_(2)])/(dt)` is:

A

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

B

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

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