The rate for the decomposition of `NH_(3)` on platinum surface is zero order. What are the rate of production of `N_(2)` and `H_(2)` if `K = 2.5 xx 10^(-4) mol litre^(-1)s^(-1)`?

Options

A. `3.75 xx 10^(-4), 1.25 xx 10^(-4)`
B. `1.25 xx 10^(-4), 3.75 xx 10^(-4)`
C. `1.25 xx 10^(-4), 3.75 xx 10^(4)`
D. `1.25 xx 10^(4), 3.75 xx 10^(-4)`

Answer

B

Explanation

`-1/2 (d[NH_(3)])/(dt) = (d[N_(2))/(dt))= 1/3(d[H_(2)])/(dt)`
Rate `=K[NH_(3)]^(0)`
`(d[NH_(3)])/(dt) = 2.5 xx 10^(-4) "mol.ltr"^(-1)s^(-1)`
`(d[N_(2)])/(dt) = 1/2 xx 2.5 xx 10^(-4) = 1.25 xx 10^(-4)`
`(d[H_(2)])/(dt) = 3/2 xx 2.5 xx 10^(-4) = 3.75 xx 10^(-4)`