The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of `N_2 and H_2 if k = 2.5 xx 10^(-4) mol L^(-1) s^(-1)`?

Explanation

`2 NH _(3) to N _(2) + 3 H _(2)`
Rate `=- (1)/(2) (d [ NH _(3) ] )/( dt ) =(d [ N _(2) ] )/( dt ) = (1)/(3) (d [ H _(2) ] )/( dt )`
For zro order reaction rate = K
` - (1)/(2) (d [ NH _(3)])/( dt ) = (d [ N _(2) ] )/( dt ) = (1)/(3) (d [ H _(2) ] )/( dt) = 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) mol L^(-1) s ^(-1)) = 7. 5 xx 10 ^(-4) mol L ^(-1) s ^(-1)`