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The decomposition of NH_(3) on platinum surface is zero order reaction. What would be the rates of production of N_(2) and H_(2) if k=2.5xx10^(-4)"mol L"^(-1)s^(-1)? |
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Answer» Solution :For zero order reaction, rate of reaction = rate constant (k) `2NH_(3) to N_(2)+3H_(2)` Rate of reaction `=(d[N_(2)])/(dt)` HENCE, rate of production of `N_(2)=2.5xx10^(-4)"mol"^(-1)s^(-1)` Again, Rte of reaction `=(1)/(3)(d[H_(2)])/(dt)` or `(d[H_(2)])/(dt)=3xx` rate of reaction `=3xx2.5xx10^(-4)"mol L"^(-1)s^(-1)=7.5xx10^(-4)"mol L"^(-1)s^(-1)` |
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