Question

A reaction has a half-life of 10 min.
(i) Calculate rate constant for the first order reaction.
(ii) What fraction of the reactant will be left after an hour of the reaction?

Answer 1

A reaction has a half-life of 10 min.
(i) Calculate rate constant for the first order reaction.
(ii) What fraction of the reactant will be left after an hour of the reaction?
ans. The half-life of a radioactive isotope describes the amount of time that it takes half of the isotope in a sample to decay. In the case of radiocarbon dating, the half-life of carbon 14 is5,730 years.

Answer 2

i) The rate constant for the first order reaction is $0.0693min^{-1}$

ii) $\frac{1}{64}$ fraction of the reactant will be left after an hour of the reaction.

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

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

$k=\frac{0.693}{10}=0.0693min^{-1}$

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant =$0.0693min^{-1}$

t = time of decomposition = 1 hour = 60 min

a = let initial amount of the reactant  =100

a - x = amount left after decay process= ?

Putting in the values, we get:

$60=\frac{2.303}{0.0693}\log\frac{100}{(a-x)}$

$1.81=\log\frac{100}{(a-x)}$

$1.81=\log\frac{100}{(a-x)}$

$\frac{100}{(a-x)}=antilog(1.81)$

$\frac{100}{(a-x)}=64.6$

$(a-x)=1.5625$

Fraction of the reactant will be left after an hour of the reaction is $\frac{1.5}{100}=0.015=\frac{1}{64}$

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