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

For the reaction `N_(2) + 3H_(2)rarr 2NH_(3)` the rate of formation of ammonia was found to be `2.0xx10^(-4) mol dm^(-3)s^(-1)` The rate of consumption of `H_(2)` will be

A

`1.0 xx 10^(-4) mol dm^(-3) s^(-1)`

B

`2.0 xx 10^(-4) mol dm^(-3) s^(-1)`

C

`3.0 xx 10^(-4) mol dm^(-3) s^(-1)`

D

`4.0 xx 10^(-4) mol dm^(-3) s^(-1)`

Text Solution

Verified by Experts

We have `-(d[N_(2)])/(d t) = -(1)/(3)(d[H_(2)])/(d t) = (1)/(2)(d[NH_(3)])/(d t)`
`(d[NH_(3)])/(d t) = 2 xx 10^(-4)` mol `dm^(-3) s^(-1)`. Hence , `-(d[H_(2)])/(d t) = (3)/(2)(d[NH_(3)])/(d t) = 3 xx 10^(-4) mol dm^(-3) s^(-1)`.
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