`H_(2)O_(2)to H_(2)O+O_(2)` the rate most of this reaction is `2xx10^(-2)"min"^(-1)` (at certain temp.) & reaction is started with`0.5M` conc. of `H_(2)O_(2)`, after 23.03 min the reaction was suddenly stopped by lowering the temp & the remaining `H_(2)O_(2)` was completely reacted with `NaCl`. If no. of moles of `O_(2)` produced in second reaction is `1/a` find `a. (log1.5=0.2)`

Answer

3

Explanation

`H_(2)O_(2)to H_(2)O+O_(2)`
`NaOCl+H_(2)O_(2)to O_(2)+NaCl+H_(2)O`
In 1 min 0.02 mol/lt `H_(2)O_(2)` decomposes so in 20 min
`k=(2.303)/t "log" a/(a-x)`
`0.02="log" 0.5/(a-x)` (take `log 1.5=0.2)`
`1.5=0.5/(a-x)`
`a-x=0.5/1.5=1/3`
`1/3` mole `H_(2)O_(2)` is left so `1/3` mole `O_(2)` is produced in second reaction.