The decomposition of `NH_(3)` on platinum surface is zero order reaction. What would be the rates of production of `N_(2) and H_(2)` if `k=2.5xx10^(-4)"mol L"^(-1)s^(-1)`?

Explanation

For zero order reaction, rate of reaction = rate constant (k)
`2NH_(3) to N_(2)+3H_(2)`
Rate of reaction `=(d[N_(2)])/(dt)`
Hence, rate of production of `N_(2)=2.5xx10^(-4)"mol"^(-1)s^(-1)`
Again, Rte of reaction `=(1)/(3)(d[H_(2)])/(dt)`
or `(d[H_(2)])/(dt)=3xx` rate of reaction
`=3xx2.5xx10^(-4)"mol L"^(-1)s^(-1)=7.5xx10^(-4)"mol L"^(-1)s^(-1)`