If a first order reaction takes 32 minutes for 75% completion, then time required for `50%` completion is :
If a first order reaction takes 32 minutes for 75% completion, then time required for `50%` completion is :
A. 32 min
B. 16 min
C. 8 min
D. 4 min
4 views
1 Answers
`k=(2.303)/(t)log((a)/(a-x))`
When , a=100, x=75, t=32
`k=(2.303)/(32)log_(10)((100)/(25))=0.0433 "min"^(-1)`
Time for `50%` completion `(t_(1//2))=(0.693)/(k)=(0.693)/(0.0433)~~16"min"^(-1)`
4 views
Answered