1.

For the reaction, `N_(2) + 3H_(2) rarr 2NH_(3)`, if `(d[NH_(3)])/(d t) = 2 xx 10^(-4) "mol L"^(-1) s^(-1)`, the value of `(-d[H_(2)])/(d t)` would be:A. `4 xx 10^(-4) mol L^(-1)s^(-1)`B. `6 xx 10^(-4) mol L^(-1)s^(-1)`C. `1 xx 10^(-4) mol L^(-1)s^(-1)`D. `3 xx 10^(-4) molL^(-1)s^(-1)`

Answer» Correct Answer - D
d) Rate =`-1/3 (d[H_(2)])/(dt) = (-d[N_(2)])/(dt)`
`=-1/2(d[NH_(3)])/(dt)`
Given `(d[NH_(3)])/(dt) = 2 xx 10^(-4) mol L^(-1)s^(-1)`
`(-d[H_(2)])/(dt) = 3/2 (d[NH_(3)])/(dt) = 3/2 xx (2 xx 10^(-4))`
`3 xx 10^(-4) mol L^(-1)s^(-1)`


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