1.

For the reaction, `N_(2)+3H_(2) rarr 2NH_(3)` the rate `(d[NH_(3)])/(dt)=2xx10^(-4) M s^(-1)`. Therefore the rate `-(d[N_(2)])/(dt)` is given as:A. `10^(-4) M sec^(-1)`B. `10^(4) M sec^(-1)`C. `10^(-2) M sec^(-1)`D. `10^(-4) sec^(-1) M^(-1)`

Answer» Correct Answer - a
`(-d[N_(2)])/(dt)=1/2(d[NH_(3)])/(dt)`


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