Question

40% of a first-order reaction is completed in 50 minutes. How much time will it take for the completion of 80% of this reaction?

Answer 1

For a first-order reaction

t = `2.303/"k"  log  "a"/("a"-"x")`

At  t_40%     a = 100

x = 40

t_40% = 50 minutes 

At t_80%         a = 100 

a = 80

t_80% = ? 

 

t_80% = `2.303/"k" log  100/(100-80)`

 

t_80% = `2.303/"k" log  100/20`

 

t_80% = `2.303/"k" log 5`

 

t_40% = `2.303/"k" log  100/(100-40)`

 

t_40% = `2.303/"K" log  100/60`

 

`= 2.303/"k" log  5/3`

 

`"t"_(80%)/"t"_(40%) = (2.303/"k" log 5)/(2.303/"k"  log  5/3`

 

`"t"_(80%)/50 log 5/(log5-log3)` 

 

`"t"_(80%)  = 50 [0.6990/(0.6990-0.4771)]`

 

`= (50xx0.6990)/0.2219`

 

t_80% = 157.5 minutes