1.

The decomposition of NH_(3) on platinum surface is zero reaction. What are the rate of production of N_(2) and H_(2) is K=2.5xx10^(-4)"mol L"^(-1)s^(-1) ?

Answer»

Solution :The reaction is
`2NH_(3(g))overset("PT")rarr N_(2(g))+3H_(2(g))`
Here `K=2.5xx10^(-4)"MOL L"^(-1)s^(-1)`
The order of the reaction is zero i.e.,
Rate `=k ["Reactant"]^(@)`
Rate `= 2.5xx10^(-4)xx1=2.5xx10^(-4)"mol L"^(-1)s^(-1)`
`therefore (d)/(dt)[N_(2)]=(1)/(3)(d)/(dt)[H_(2)]`
The rate of formation of `N_(2)=2.5xx10^(-4)"mol L"^(-1)s^(-1)`
`therefore 2.5xx10^(-4)=(1)/(3)(d)/(dt)[H_(2)]`
`therefore (d)/(dt)[H_(2)]=7.5xx10^(-4)`
Therefore, rate of formation of `H_(2)`
`= 7.5xx10^(-4)"mol L"^(-1)s^(-1)`


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