1.

The rate of a certain reaction is given by, rate = k[H^(+)]^(n). The rate increases 100 times when the pH changes from 3 to 1. The order (n) of the reaction is

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Solution : RATE (r) = `k[H^+]^N`
When pH = 3 , `[H^+]` = `10^(-3)`
and when pH = 1 , `[H^+]`=`10^(-1)`
`THEREFORE (r_1)/(r_2)=(k(10^(-3))^n)/(k(10^(-1))^n) implies (1)/(100) =((10^(-3))/(10^(-1)))^n (therefore r_2=100 r_1)`
`implies (10^(-2))^1=(10^(-2))^n implies n=1`


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