The rate of decomposition of `NH_(3)` on platinum surface is zero order. What are rate of production of `N_(2)` and `H_(2)` if `k=2.5 xx 10^(-4)Ms^(-)`?

Explanation

The decomposition of `NH_(3)` on platinum surface is represented by the following equation .
`2 NH_(3 (g)) overset("Pt")(to) N_(2 (g)) + 3 H_(2 (g))`
Therefore ,
Rate = `-(1)/(2) (d[NH_(3)])/("dt") = (d [N_(2)])/("dt") = (1)/(3) (d [H_(2)])/("dt") = k`
`= 2.5 xx 10^(-4) "mol" L^(-1) s^(-1)`
Therefore , the rate of production of `N_(2)` is
`(d [N_(2)])/("dt") = 2.5 xx 10^(-4) "mol" L^(-1) s^(-1)`
And , the rate of production of `H_(2)` is
`(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)`