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A reaction of first-order completed 90% ...

A reaction of first-order completed `90%` in 90 minutes, hence, it is completed `50%` in approximately :

Text Solution

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`k=(2.303)/(t_(90)) log ((a)/(a-x))`
`k=(2.303)/(90)log"" ((100)/(100-90))=(2.303)/(90) log 10 =(2.303)/(90)" m in"^(-1)`
Also, `k=(2.303)/(t_(50))log""(100)/(100-50)=(2.303)/(t_(50)) log2`
`(2.303)/(90)=(2.303)/(t_(50))" log 2 "rArr t_(50) =90 log 2 =27" m in"`
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