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