1.

Half life of a first order reaction completes in 5 minutes. What persent of reactant reacts after 40 minutes ?

Answer»

Solution :`t_(½) = 15 min`
`K = (0.693)/(t_(½))`
`= (0.693)/(15)`
`K = 0.0462 min^(-1)`
For first order REACTION,
`K = (2.303)/(t) log_(10)((a)/(a-x))`
`a = 100 K =0.0462 min^(-1), x = ? T = 40 min`
`0.0462 = (2.303)/(40) log_(10)((100)/(a-x))`
`log_(10)((100)/(a-x)) = 0.8024`
`(100)/(a-x)` antilog (0.8024)
`a-x = (100)/(6.345)`
`a-x = 15.76`
`x = 100 - 15.76`
`x = 100 - 15.76`
`:. x = 84.24%`


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