A first order reaction takes 20 minutes for 25% decomposition. Calculate the time when 75% of the reaction will be completed. [Given : log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021]

Explanation

The equation for first order reaction is :
`t=(2.303)/(k)"log"([R]_(0))/(R )`
For `25%` of reaction :
`20=(2.303)/(k)"log"([R]_(0))/(0.75[R]_(0))=(2.303)/(k) "log"(4)/(3)" " …(i)`
For 75% of the reaction:
`t=(2.303)/(k) "log"([R]_(0))/(0.25[R]_(0))= (2.303)/(k)log 4" " ...(ii)`
Dividing (ii) by (i), we get
`(t)/(20)=(log 4)/("log"(4)/(3))=(log4)/(log 4-log3)=(0.6021)/(0.6021-0.4771)=(0.6021)/(0.1250)`
or ` t=(20xx0.6021)/(0.1250)=96.3` minutes