A first order reaction is 40% complete in 50 minutes. In what time will the reaction be 80% compelete ?

Explanation

Step I: To Calculate k
`[R]_0=100(say),[R]=100-40=60`
t=50 min,k=?
`k=(2.303)/(t)log([R]_0)/([R])`
`=(2.303)/("50 min")log(100)/(60)`
`=(2.303)/("50 min")log 1.6667`
`=(2.303xx0.2220)/("50 min")`
`=1.022xx10^(-2) min^(-1)`
Step II : To Calculate t.
`[R]_0=100,[R]=100-80=20`
`k=1.022xx10^(-2)min^(-1)`,t=?
`t=(2.303)/(k)log([R]_0)/([R])`
`=(2.303)/(1.022xx10^(-2)min^(-1))log(100)/(20)`
`=(2.303log2)/(1.022xx10^(-2)min^(-1))`
`=(2.303xx0.6990)/(1.022xx10^(-2))=(2.303xx69.9)/(1.022)` min
=157.51 min