Consider the following reaction:
`N_(2)(g) + 3H_(2)(g) to 2 NH_(3)(g)` the rate of this reaction in terms of `N_(2)` at T K is `=(-d[N_(2)])/(dt)= 0.02 mol L^(-1)s^(-1)`. What is the value of `(-d[N_(2)])/(dt)` (in units of ` mol L^(-1)s^(-1)`) at the same temeprature.

Options

A. 0.02
B. 50
C. 0.06
D. 0.04

Answer

D

Explanation

`N_(2)(g) + 3H_(2)(g) to 2NH_(3)(g)`
Rate of reaction `= - (-d[N_(2)])/(dt) =(-d[N_(2)])/(dt)= (-d[H_(2)])/(3dt)= (d[NH_(3)])/(2dt)`
` therefore (-d[H_(2)])/(dt)= 3 xx 0.02 = 0.06 mol^(-1)L^(-1)s^(-1)`
`[ because (-d[N_(2)])/(dt)=0.02 mol^(-1)s^(-1)]`