Show that for a first order reaction the time required to complete 75% of reaction is about 2 times more than that required to complete 50% of the reaction.

Explanation

As we know, rate constant for a first order reaction is given by,
`k=((2.303)/(t)) "log"_(10) ([R]^(0))/([R])`
For completion of 50% reaction, time required is `t_(50)` and for 75% completion time required is `t_(15)`.
Writing the equations and taking their ratios :
`k=(2.303)/(t)"log"_(10)(a)/(a-x)`
or `t_(75%) =(2.303)/(k)"log"(100)/(25)`
or `t_(75%)=(2.303)/(k)log xx 0.602`
`t_(75%) =(1.386)/(k)`
`t_(50%)=(2.303)/(k)"log"(100)/(50)`
or `t_(50%)=(2.303)/(k)log 2`
or `=(2.303)/(k) xx 0.3010`
or `=(0.693)/(k)`
`:.t_(75)=2xx t_(50)`
Hence, it is proved that time required for 75% completion of reaction is 2 times the time required for 50% completion of the reaction.