Consider the reaction:
`N_(2(g))+3H_(2(g))rarr 2NH_(3(g))`.
The equally relationship between `-(d[NH_(3)])/(dt)` and `-(d[H_(2)])/(dt)` is:

Options

A. `(d[NH_(3)])/dt=-(1)/(3) (d[H_(2)])/(dt)`
B. `+(d[NH_(3)])/(dt)=-(1)/(3) (d[H_(2)])/dt`
C. `+(d[NH_(3)])/(dt)=-(3)/(2) (d[H_(2)])/dt`
D. `+(d[NH_(3)])/(dt)=-(d[H_(2)])/dt`

Answer

B

Explanation

For the reaction `N_(2)(g)+3H_(2)(g)to 2NH_(3)(g)`
`-(d[N_(2)])/dt=-(1)/(3)(d)/(dt)([H_(2)])=(1)/(2)(d)/(dt)([NH_(3)])`
`therefore(d)/(dt)([NH_(3)])=-(2)/(3)(d)/(dt)([h_(2)])`