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)Ls^(-1).

Answer»

Solution :`2NH_(3(G))overset(PT)rarr N_(2(g))+3H_(2(g))`
Rate `=-(1)/(2)(d[NH_(3)])/(DT)=(d[N_(2)])/(dtr)=(1)/(3)(d[N_(2)])/(dt)`
However, it given that the reaction is of zero ORDER. Therefore
`-(1)/(2)(d[NH_(3)])/(dt)=(d[N_(2)])/(dt)=(1)/(3)(d[H_(2)])/(dt)k`
`= 2.5xx10^(-4)"mol L"^(-1)s^(-1)`
The rate of production of `N_(2)` is
`(d[N_(2)])/(dt)=2.5xx10^(-4)"mol L"^(-1)s^(-1)`
The rate of production of `H_(2)` is
`(d[N_(2)])/(dt)=3xx2.5xx10^(-4)"mol L"^(-1)s^(-1)`
`= 7.5xx10^(-4)"mol L"^(-1)s^(-1)`.


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