Question

A compound decomposes by a first-order process. What is the half-life of the compound if 25.0% of the compound decomposes in 60.0 minutes?

Answer 1

Answer : The half-life of the compound is, 145 years.

Explanation :

First we have to calculate the rate constant.

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

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

where,

k = rate constant  = ?

t = time passed by the sample  = 60.0 min

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process = 100 - 25 = 75 g

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

$k=\frac{2.303}{60.0}\log\frac{100g}{75g}$

$k=4.79\times 10^{-3}\text{ years}^{-1}$

Now we have to calculate the half-life of the compound.

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

$4.79\times 10^{-3}\text{ years}^{-1}=\frac{0.693}{t_{1/2}}$

$t_{1/2}=144.676\text{ years}\approx 145\text{ years}$

Therefore, the half-life of the compound is, 145 years.