Question

48.
Half-life period of a first order reaction is 100 min. Afte
144.3 min, concentration of reactant is reduced to of th
original concentration
(a) 40%
(b) 30%
(c) 1/e
(d) 1/2​

Answer 1

Answer: (c) 1/e

Explanation:

Half-life of  first order reaction is = 100 min

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{100\text{min}}$

$k=0.00693\text{min}^{-1}$

Now we have to calculate the amount left after 144.3 min

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  = $0.00693\text{min}^{-1}$

t = time of decomposition = 144.3 min

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process  = ?

Now put all the given values in above equation, we get

$144.3=\frac{2.303}{0.00693}}\log\frac{100}{a-x}$

$(a-x)=36$

Thus the concentration of reactant is reduced to $\frac{1}{e}=\frac{1}{2.718}=0.36$ of original concentration.