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

Explanation

For the reaction, `2NH_(3) rarrN_(2)+3H_(2)`
`therefore " Rate" = - (1)/(2) (d[NH_(3)])/(dt)=(d[N_(2)])/(dt)=(1)/(3)(d[H_(2)])/(dt)`
For a zero order reaction , rate = k
`therefore` Rate of production of `N_(2)=(d[N_(2)])/(dt)`
`=2.5xx10^(-4)"mol.L"^(-1).s^(-1)`
and rate of production of `H_(2)=(d[H_(2)])/(dt)`
`=3k=3xx(2.5xx10^(-4))`
`=7.5xx10^(-4)"mol.L"^(-1).s^(-1)`