1.

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

Answer»

Solution :Decomposition of `NH_(3)` is as follows:
`2NH_(3(G))toN_(2(g))+3H_(2(g))`
`k=2.5xx10^(-4) mol L^(-1)s^(-1)`
Reaction is of zero order
Rate of reaction =`-(1)/(2)(d[NH_(3)])/(dt)=(d[N_(2)])/(dt)=(1)/(3)-(d[H_(2)])/(dt)`
as the reaction is zero order
Rate of formation of `N_(2)=(d[N_(2)]^(0))/(dt)=k`
`2.5xx10^(-4)mol L^(-1)s^(-1)`
Rate of formation of `H_(2)`
`=k=(1)/(3) (d[H_(2)]^(0))/(dt)=2.5x10^(-4)=k`
`therefore (d[H_(2)])/(dt)=3xx2.5xx10^(-4)`


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