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. `1 xx 10^(-4) "mol" L^(-1) s^(-1)`B. `3 xx 10^(-4) "mol" L^(-1) s^(-1)`C. `4 xx 10^(-4) "mol" L^(-1) s^(-1)`D. `6 xx 10^(-4) "mol" L^(-1) s^(-1)`

Answer» Correct Answer - B
`(1)/(2)(d[NH_(3)])/(d t) = (1)/(3)(d[H_(2)])/(d t)`
`:. -(d[H_(2)])/(d t) = (3)/(2) xx (d[NH_(3)])/(d t)`
`= (3)/(2) xx 2 xx 10^(-4) = 3 xx 10^(-4)`


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