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:

Options

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

D

Explanation

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)`