In a catalytic experiment involving Haber's process, `N_(2)(g)+3" H"_(2)(g)to2" NH"_(3)(g)`, the rate of reaction was measured as : Rate `=(d[NH_(3)])/(dt)=2.0xx10^(-4)"M s"^(-1)`. If there were no side reactions, what was the rate of reaction expressed in terms of `N_(2)` ?

Options

A. `1xx10^(-4)"M s"^(-1)`
B. `4xx10^(-4)"M s"^(-1)`
C. `5xx10^(-3)"M s"^(-1)`
D. `1xx10^(-3)"M s"^(-1)`

Answer

A

Explanation

Rate `= (1)/(2)(d[NH_(3)])/(dt) = -(1)/(3)(d[h_(2)])/(dt) = -(d[N_(2)])/(dt)`
`therefore `-(d[N_(2)])/(dt) = (1)/(2) (d[NH_(3)])/(dt) = (1)/(2) xx(2 xx 10^(-4) M s^(-1))`
`" " = Ms^(-1)`