75% of a first order reaction was completed in 32 min. When was 50% of the reaction completed ?

Options

A. 16 min
B. 24 min
C. 8 min
D. 4 min

Answer

A

Explanation

For the reaction,
`k=(2.303)/(t).log.(a)/(a-x)=(2.303)/(32).log4 = 0.04332`
Hence, time required for 50% completion,
`t = (2.303)/(0.04332).log.(a)/(a-(a)/(2))`
`=(2.303)/(0.04332)xx0.3010=16min`