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

Answer

C

Explanation

For the reaction, we have
`-(d[N_(2)])/(dt)=-(d[H_(2)])/(3dt)=(d[NH_(3)])/(2dt)`
Thus,
`(d[NH_(3)])/(dt)=-(2)/(3)(d([H_(2)])/(dt)`