Help with Half-Life and a First-Order Process

1. Jan 16, 2008

rneely01

1. The problem statement, all variables and given/known data

½ Life and a First-Order Process

ln [R]t = -(0.18 d-1 ) (30.d)= -5.5
4.0 x 10 13atom/L

[R]t
4.0 x 10 13atom/L = e -5.5 = 0.0042

[R]t = 1.7 x 10 11 atom/L

2. Relevant equations

Arrhenhius Equation
k = Ae –Ea
RT

3. The attempt at a solution

=> -(0.18 d-1 ) (30.d)= -5.5 (should really be 5.4 not book answer of 5.5)
=> ln of 5.5 is 1.70 not .0042
=> no idea how book came up with 1.7 x 10 11 atom/L, since .0042/4.0 x 10 13atom/L is 1.05 x 10 -16 atom/L

2. Jan 16, 2008

rocomath

What is the original problem?

3. Jan 16, 2008

rneely01

Problem: Radioactive radon-222 gas ( 222 Rn) from natural sources can seep into the basement of a home. The half-life of 222 Rn is 3.8 days. If a basement has 4.0 x 10 13
atoms of 222 Rn per liter of air, and the radon gas is trapped in the basement, how many atoms of 222 Rn will remain after one month (30 days)?

Solution: Rate constant (k) is
K = 0.693 = 0.693 = 0.18 d-1
t ½ 3.8 d

This equation is relevant: ln [R]t = -kT
[R] 0

4. Jan 16, 2008

rocomath

Equations:

$$\ln{\frac{A_0}{A_t}=kt$$

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

So our givens are ...

$$A_0=4\times10^{13}L^{-1}$$

$$t_{\frac{1}{2}}=3.8d$$

What we want is ...

$$A_t$$ when $$t=30d$$

From our half-life equation, we can figure out what our constant k is. From our first-order equation, solve for our final concentration and plug in the value k that is found.

Last edited: Jan 16, 2008