The decomposition of `NH_3` on platinum surface is a zero order reaction. What would be the rate of production of `N_2 and H_2` if `k=2.5xx10^(-4) "mol L"^(-1) s^(-1)` ?

Explanation

`2NH_3 to N_2 + 3H_2`
`-1/2(d[NH_3])/(dt)=(d[NH_2])/(dt)+1/3(d[H_2])/(dt)`
`(d[NH_3])/(dt)`=rate `=kxx[NH_3]^0`
`=2.5xx10^(-4) "mol L"^(-1) "sec"^(-1)`
`(d[N_2])/(dt)=-1/2 (d[NH_3])/(dt)`
`=1/2xx2.5xx10^(-4)"mol L"^(-1) "sec"^(-1)`
`d[H_2]=-3/2 (d[NH_3])/(dt)=3/2xx2.5xx10^(-4)`
`=3.75xx10^(-44) "mol L"^(-1) "sec"^(-1)`
Rate=`-(d[NH_3])/(dt)=kxx[NH_3]^0`
`=2.5xx10^(-4) "mol L"^(-1) "sec"^(-1)`
Rate of production of `N_2=2.5xx10^(-4) "mol L"^(-1) "sec"^(-1)`