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The half-life period of a 1^(st) order r...

The half-life period of a `1^(st)` order reaction is 60 minutes. What percentage will be left over after 240 minutes ?

A

`6.25%`

B

`4.25%`

C

`5%`

D

`6%`

Text Solution

Verified by Experts

The correct Answer is:
A
`t_(1//2) = (0.693)/(k) implies (0.693)/(t_(1//2) = k implies (0.693)/(60) = k`
k=0.01155` "min"^(-1)`
`k= (2.303)/t log""((a)/(a-x))`
Let the initial amount (a) be 100.
`0.01155"min"^(-1) =(2.303)/(240 "min") log ((100)/(a-x))`
`(0.01155 "min"^(-1)xx40"min")/(2.303) = log ((100)/(a-x))`
`1.204 = log 100 - log(a-x)`
`1.204 = 2- log(a-x)`
`log(a-x) = 2- 1.204`
`log(a-x) = 0.796`
`(a-x) = 6.25%`
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Knowledge Check

  • The half-life period of a 1st order reaction is 60 minutes .What percentage will be left over after 240 minutes?

    A
    0.05
    B
    0.0625
    C
    0.06
    D
    0.0425

  • The half - life period of a 1 order reaction is 60 mintus . What percentage will bee left over after 240 minutes ?

    A
    0.0625
    B
    0.0425
    C
    0.05
    D
    0.06

  • The half – life of a 1st order reaction is 60 minutes. What percentage will be left over after 240 minutes?

    A
    `5%`
    B
    `6 cdot 25%`
    C
    `6%`
    D
    `4 cdot 25%`

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