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

Options

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

Answer

A

Explanation

(A) `-(1)/(2) (d[N_(2)])/(dt)=-(1)/(3)(d[H_(2)])/(dt)=(1)/(2) (d[NH_(3)])/(dt)`
or `(d[NH_(3)])/(dt) =-(2)/(3)(d[H_(2)])/(dt)`