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. `4xx10^(-4)mol L^(-1) s^(-1)`
B. `6xx10^(-4)mol L^(-1) s^(-1)`
C. `1xx10^(-4) mol L^(-1) s^(-1)`
D. `3xx10^(-4) mol L^(-1) s^(-1)`

Answer

D

Explanation

For the reaction
`N_(2)+3H_(2)rarr2NH_(3)`
we have
`-(d[N_(2)])/(dt)=-(d[H_(2)])/(3dt)=(d[NH_(3)])/(2dt)`
Thus,
`-(d[H_(2)])/(dt)=(3)/(2)(d[H_(2)])/(3dt)=(d[NH_(3)])/(dt)`
`=(3)/(2)(2xx10^(-4)molL^(-1)s^(-1)]`
`=3xx10^(-4)molL^(-1)s^(-1)`