In a first order reaction the reacting substance has half-life period of ten minutes. What fraction of the substance will be left after an hour the reaction has occurred?

Question

In a first order reaction the reacting substance has half-life period of ten minutes. What fraction of the substance will be left after an hour the reaction has occurred? (A) 1/6 of initial concentration (B) 1/64 of initial concentration (C) 1/12 of initial concentration (D) 1/32 of initial concentration

Tardigrade Admin · Accepted Answer

t1/2=10 No. of half-life =(60/10)= 6 half-life Ct=(C0/(2)n)=(C0/(2)6)=((C0/64))

Q. In a first order reaction the reacting substance has half-life period of ten minutes. What fraction of the substance will be left after an hour the reaction has occurred?

$1/6$ of initial concentration

$1/64$ of initial concentration

$1/12$ of initial concentration

$1/32$ of initial concentration

$t_{1/2} = 10$ No. of half-life $= \frac{60}{10} = 6$ half-life
$C_{t} = \frac{C _{0}}{( 2 ) ^{n}} = \frac{C _{0}}{( 2 ) ^{6}} = (\frac{C _{0}}{64})$

Q. In a first order reaction the reacting substance has half-life period of ten minutes. What fraction of the substance will be left after an hour the reaction has occurred?

1/6 of initial concentration

1/64 of initial concentration

1/12 of initial concentration

1/32 of initial concentration

t1/2​=10 No. of half-life =1060​=6 half-life Ct​=(2)nC0​​=(2)6C0​​=(64C0​​)

$1/6$ of initial concentration

$1/64$ of initial concentration

$1/12$ of initial concentration

$1/32$ of initial concentration

$t_{1/2} = 10$ No. of half-life $= \frac{60}{10} = 6$ half-life
$C_{t} = \frac{C _{0}}{( 2 ) ^{n}} = \frac{C _{0}}{( 2 ) ^{6}} = (\frac{C _{0}}{64})$