Question

3. Gold-198 decays with a half-life of 2.7 days according to the equation:198Au → 0−1β + 198Cd

a. Find the value of the rate constant, k. Include units.


b. How much of a 10.0 g sample of Au-198 remains after seven days? (Use the integrated rate law formula.)

Answer 1

a. k=0.256/day

b.sample of Au-198 remains after seven days : 1.67 g

Further explanation

1. A half-life of 2.7 days⇒t1/2=2.7 days

The half-life can be expressed in a decay constant( λ)

$\tt \displaystyle t_ {1/2} = {\dfrac {\ln (2)} {\lambda}}\\\\\lambda(or~k)=\dfrac{0.693}{2.7}=0.256/day$

2. We can use formula : (integrated rate law) :

$\tt ln[A]=-kt+ln[Ao]\\\\ln(\dfrac{Ao}{A})=kt$

Ao=10 g

t=7 days

k=0.256/day

$\tt ln(\dfrac{10}{A})=0.256\times 7\\\\ln(\dfrac{10}{A})=1.792\\\\e^{1.792}=\dfrac{10}{A}\\\\6=\dfrac{10}{A}\rightarrow A=1.67~g$

Answer 2

The mass of Au-198 that remains after seven days is 1.62 g.

We know that radioactive decay is a first order reaction hence;

k = 0.693/ t12

Where;

k = rate constant

t12 = half life

k =  0.693/ 2.7 days = 0.26 day-1

Using;

N =Noe^-kt

No = amount of radioactive material initially present

N = mass of radioactive substance present at time t

N = 10 e^-(0.26 × 7)

N = 1.62 g

The mass of Au-198 that remains after seven days is 1.62 g.

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