1.

The ionization of H_(2)SO_(4) is completed in 2 steps .(i)H_(2)SO_(4)toH^(+)+HSO_(4)^(-1) k_(a)(1)=1.4xx10^(-2) mol L^(-1) s^(-1) (ii)HSO_(4)^(-)=3.5xx10^(-2) mol L^(-1)S^(-1) Then which of the following equation given below is true for rate?

Answer»

Rate=`K_(a)(1)[H_(2)SO_(4)]`
Rate =`K_(a)(2)[HSO_(4)^(-)]`
Rate =`K_(a)(1)[H_(2)SO_(4)]`
Rate=`K_(a)(2)[HSO_(4)^(-)]`

SOLUTION :The rate of the second reaction is very les.So the rate can be taken same as other LESS reaction.
(ii)`HSO_(4)^(-1)TOH^(+)+SO_(4)^(2-)`Ka (2) `3.5xx10^(-2)`
`therefore` Rate will be =`K_(a)(2)HSO_(4)^(-1)`
So,option (B) is correct.


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