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The half-life period of a first order reaction is 15 minutes. The amount of substance left after one hour will be :

Option: 1

1 /4 of the original amount


Option: 2

1/8 of the original amount


Option: 3

1/16 of the original amount


Option: 4

1/ 32 of the original amount


Answers (1)

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The half-life period of a first-order reaction-

t_\frac{1}{2}=\frac{0.693}{k}

The half-life period of a first-order reaction is 15 minutes. 

So, 

 15=\frac{0.693}{k}

\mathrm{k=\frac{ln2}{15}}

Now,  if the amount of substance left after one hour is x , (k will be the same) and if the initial amount is 1.

\mathrm{\frac{ln2}{15}=\frac{1}{60}\:ln\frac{1}{x}}

\mathrm{4ln2= ln\frac{1}{x}}

\mathrm{ln16= ln\frac{1}{x}}

x=\frac{1}{16} of the original amount.

Posted by

Suraj Bhandari

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