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) to N_(2) +3H_(2)`
Rate `= -(1)/(2) (d[NH_(3)])/(dt)= +(d[N_(2)])/(dt)= +(1)/(3)(d[H_(2)])/(dt)`
For zero ORDER reaction, rate = k
`therefore -(1)/(2) (d[NH_(3)])/(dt)=(d[N_(2)])/(dt)=(1)/(3)(d[H_(2)])/(dt)=2.5xx10^(-4)"mol L"^(-1)s^(-1)`
`therefore ` Rate of production of `N_(2)=(d[N_(2)])/(dt)=2.5xx10^(-4)"mol L"^(-1)s^(-1)`
Rate of production of `H_(2)=(d[H_(2)])/(dt)`
`=3xx(2.5xx10^(-4)"mol L"^(-1)s^(-1))=7.5xx10^(-4)"mol L"^(-1)s^(-1)`.


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