At constant temperature & high pressure, the given reaction is of the zero order : `2NH_(3)(g)overset(pt)rarrN_(2)(g)+3H_(2)(g)` . If the rate constant of this reaction is `3xx10^(-4)"mol.L"^(-1).s^(-1)` , calculate the rate of formation of `N_(2)and H_(2)`.

Explanation

The rate of the given reaction is ,
rate `=-(1)/(2)(d[NH_(3)])/(dt)=(d[N_(2)])/(dt)=(1)/(2)(d[H_(2)])/(dt)=k[NH_(3)]^(0)=k`
`therefore([N_(2)])/(dt)=k=3xx10^(-4)"mol.L"^(-1).s^(-1)and`
`(d[H_(2)])/(dt)=3xxk=3xx3xx10^(-4)=9xx10^(-4)"mol. L"^(-1).s^(-1)`
Hence , the rate to formation of `N_(2)=3xx10^(-4)"mol.L"^(-1).s^(-1) "and that of" H_(2)=9xx10^(-4)"mol . L"^(-1).s^(-1)`