Question

The concentration of Rn−222 in the basement of a house is 1.45 × 10−6 mol/L. Assume the air remains static and calculate the concentration of the radon after 3.00 days. The half-life of Rn−222 is 3.82 days.

Answer 1

Answer: The concentration of radon after the given time is $3.83\times 10^{-30}mol/L$

Explanation:

All the radioactive reactions follows first order kinetics.

The equation used to calculate half life for first order kinetics:

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

We are given:

$t_{1/2}=3.82days$

Putting values in above equation, we get:

$k=\frac{0.693}{3.82}=0.181days^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,  

k = rate constant = $0.181days^{-1}$

t = time taken for decay process = 3.00 days

$[A_o]$ = initial amount of the reactant = $1.45\times 10^{-6}mol/L$

[A] = amount left after decay process =  ?

Putting values in above equation, we get:

$0.181days^{-1}=\frac{2.303}{3.00days}\log\frac{1.45\times 10^{-6}}{[A]}$

$[A]=3.83\times 10^{-30}mol/L$

Hence, the concentration of radon after the given time is $3.83\times 10^{-30}mol/L$