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A first order reaction takes 40 min for 30% decomposition .Calculation t_((1)/(2))

Answer»

SOLUTION :The reaction is first ORDER,so it decomposes 30% decomposition so ,70% is remaining,
`therefore [R]_(t)70% of [R]_(0)`
`=(70)/(100)[R]_(0)=0.7 [R]_(0)`
Calculation of K:The reacton is first order ,so
`k=(2.303)/(t)` log `([R]_(0))/([R]_(t))`
`k=(2.303)/(t)` log `([R]_(0))/([R]_(t))`
`k=(2.303)/(t)` log `([R]_(0))/([R]_(t))` Where time t=40 MIN
`=(2.303)/(40 min `log `((10)/(7))` Calculation of `t_((1)/(2))`:Reaction is first order.
`therefore t_((t)/(2))=(0.693)/(k)=(0.693)/(8.9185xx10^(-3) min ^(-1))=77.7 min`


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