The decomposition of `NH_(3)` on platinum surface is zero order reaction the rate of production of `H_(2)` is `(k=2.5xx10^(-4) M s^(-1))`

Options

A. `3.35xx10^(-4) and 1.25xx10^(-4),` respectively
B. `1.25xx10^(-4) and 3.75xx10^(-4),` respectively
C. `3.75xx10^(-3) and 2.45xx10^(-3),` respectively
D. `1.25xx10^(-3) and 3.25xx10^(-3),` respectively

Explanation

Given,`k=2.5xx10^(-4) molL^(-1) S^(-1)`
`2H_(3)(g)toN_(2)(g)+3H_(2)(g)`
On diving the Eq.by 2,
`NH_(3)(g)to(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)`
Rate=`(d[NH_(3)])/(dt)=+(2d[N_(2)])/(dt)=(2)/(3)(d[H_(2)])/(dt)`
For zero order reaction, rate of reaction =k
so,`(-d[NH_(3)])/(dt)=(2d[N_(2)])/(dt)=(2)/(3)(d[H_(2)])/(dt)=2.5xx10^(-4)" ""mol"L^(-1)S^(-1)`
`therefore "Rate of production of" N_(2)=(d[N_(2)])/(dt)=(k)/(2)`
`=2.5xx10^(-4)" ""mol"L^(-1)S^(-1)`
`therefore "Rate of production of" H_(2)=(d[H_(2)])/(dt)=(3)/(2)xx(2.5xx10^(-4)molL^(-1)S^(-1))`
`=3.75xx10^(-4)molL^(-1)S^(-1)`