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

Options

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)`

Answer

C

Explanation

We hage `(d[NH_(3)])/(dt)=2xx10^(-4) mol dm^(-3) s^(-1)` hence `-(d[H_(2)])/(dt)=3/2(d[NH_(3)])/(dt)=3xx10^(-4) mol dm^(-3) s^(-1)`