A first order reaction is 20% complete in the 10 minutes. Calculate the time period for 75% completion of the reaction.

Explanation

Step 1: To Calculate k
`[R]_0=100(say),[R]=100-20=80`
t=10 min, k=?
`k=(2.303)/(t)log([R]_0)/([R])`
`=(2.303)/("10 min")log(100)/(80)`
`=(2.303)/("10 min")[log10-log8]`
`=(2.303)/("10 min")[1-0.9031]`
`=(2.303xx0.0969)/(10)min^(-1)`
`=0.0223" min"^(-1)`
Step II : To Calculate t.
`[R]_0=100,[R]=100-75=25`
`k=0.0223" min"^(-1)`
`t=(2.303)/(k)log([R]_0)/([R])`
`=(2.303)/(0.0223" min"^(-1))log(100)/(25)`
`=(2.303xxlog4)/(0.0223" min"^(-1))`
`=(2.303xx0.6021)/(0.0223)` min=62.18 min