The rate for the decomposition of `NH_(3)` on platinum surface is zero order. What are the rate of production of `N_(2)` and `H_(2)` if `K = 2.5 xx 10^(-4) mol litre^(-1)s^(-1)`?

Answer

`(d[N_(2)])/(dt)=1.25xx10^(-4)Ms^(-1)`
`(d[H_(2)])/(dt)=3.75xx10^(-4)Ms^(-1)`

Explanation

`2NH_(3)rarr N_(2)+3H_(2)`
Rate`=k[NH_(3)]^(0)impliesk=2.5xx10^(-4)Ms^(-1)`
Rate `(-d[NH_(3)])/(dt)=(1)/(2)((-d[NH_(3)])/(dt))`
`=(1)/(2)xx2.5xx10^(-4)Ms^(-1)`
`=1.25xx10^(-4)Ms^(-1)`
`:.(d[H_(2)])/(dt)=(3)/(2)((-d[NH_(3)])/(dt))`
`=(3)/(2)xx2.5xx10^(-4)Ms^(-1)=3.75xx10^(-4)Ms^(-1)`