Question

10% of a substance decays in 5 seconds in a first order reaction. The approximate amount left after 25 seconds

Answer 1

Answer:

As we learnt in concept

Number of nuclei after disintegration -

N=N_{0}e^{-\lambda t} or A=A_{0}e^{-\lambda t}

- wherein

Number of nucleor activity at a time is exponentional function

Initially \: N_1=0.90N_0 = N_0.e^{-\lambda t_1}

t_1= 5days

\Rightarrow e^{-5\lambda }=0.90 ............(1)

at \: t=t_2 =20 \:days

N=N_0.e^{-20\lambda }=N_0(e^{-5\lambda })^4

N=N_0.({0.90 })^4=0.65 N_0

\therefore 65 % original matter will left after 20 days.

Option 1)

60%

Incorrect

Option 2)

65%

Correct

Option 3)

70%

Incorrect

Option 4)

75%

Answer 2

Thus 59.52% of the substance is left after 25 seconds.

Given:

10% of a substance decays in 5 seconds in a first-order reaction.

To find:

The approximate amount left after 25 seconds.

Solution:

This follows 1st order rate law and according to it-

$k = \frac{1}{t} ln( \frac{x}{y} )$

Where k is the rate constant.

t is time =5 seconds.

x is the initial concentration of the substance=100%.

y is the final concentration of the substance equal to 100-10=90%.

$k = \frac{1}{5} ln( \frac{100}{90} ) \\ k = 0.02$Now t=25 seconds.

So, we can write-

$0.02 = \frac{1}{25} ln( \frac{100}{y} ) \\ 0.5 = ln( \frac{100}{y} ) \\ \frac{100}{y} = 1.68 \\ y = \frac{100}{1.68} \\ y = 59.52$

Thus 59.52% of the substance is left after 25 seconds.

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