The decomposition of `NH_3` on platinum surface is zero· order with rate constant
`k = 2.5 xx 10^(-4) mol L^(-1) s^(-1)`. The rate of production of `H_2` will be

Options

A. `2.5 xx 10^(-4) mol L^(-1) s^(-1)`
B. `0.83 xx 10^(-4) mol L^(-1) s^(-1)`
C. `7.5 xx 10^(-4) mol L^(-1) s^(-1)`
D. `5.0 xx 10^(-4) mol L^(-1) s^(-1)`

Explanation

`2NH_3 overset(Pt)to N_2+3H_2`
`"Rate"= 1/2(d[NH_3])/(dt)=(d[N_2])/(dt)= 1/3(d[H_2])/(dt)`
`=1/3 (d[H_2])/(dt)=k = 2.5 xx 10^(-4)mol L^(-1)s^(-1)`
Rate of production of `H_2`.
`:. (d[H_2])/(dt) = 3 xx 2.5 xx 10^(-4)`
`=7.5 xx 10^(-4)mol L^(-1)s^(-1)`