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* 10^(-30)mol/L`

Step-by-step explanation:

All the radioactive reactions follows first order kinetics.

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

`t_(1/2)=(0.693)/(k)`

We are given:

`t_(1/2)=3.82days`

Putting values in above equation, we get:

`k=(0.693)/(3.82)=0.181days^(-1)`

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

`k=(2.303)/(t)\log([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* 10^(-6)mol/L`

[A] = amount left after decay process = ?

Putting values in above equation, we get:

`0.181days^(-1)=(2.303)/(3.00days)\log(1.45* 10^(-6))/([A])`

`[A]=3.83* 10^(-30)mol/L`

Hence, the concentration of radon after the given time is
`3.83* 10^(-30)mol/L`