Respuesta :
1. If 25% of the c-14 atoms in the sample decay, that means the final weight would be 100-25%= 75% of the initial weight. Then, the amount of time elapsed would be:
final weight= initial weight * 1/2^ (t/t1/2)
0.75* of the initial weight= initial weight * 2^-t/5730 ---->1/2 = 2^-1
0.75 = 2^-t/5730
log2 0.75 = log2 (2^-t/5730)
-0.415= -t/5730 ---->2^-0.415= 0.75
t=0.415*5730
t=2378 years
2. if a sample of c-14 initially contains 1.5 mmol of c-14, how many millimoles are left after 2255 years?
final weight= initial weight * 1/2^ (t/t1/2)
final weight= 1.5mol * 1/2^ (2255/5730)
final weight= 1.5mol* 1/2^0.39354
final weight= 1.5mol* 0.761= 1.14 mol
final weight= initial weight * 1/2^ (t/t1/2)
0.75* of the initial weight= initial weight * 2^-t/5730 ---->1/2 = 2^-1
0.75 = 2^-t/5730
log2 0.75 = log2 (2^-t/5730)
-0.415= -t/5730 ---->2^-0.415= 0.75
t=0.415*5730
t=2378 years
2. if a sample of c-14 initially contains 1.5 mmol of c-14, how many millimoles are left after 2255 years?
final weight= initial weight * 1/2^ (t/t1/2)
final weight= 1.5mol * 1/2^ (2255/5730)
final weight= 1.5mol* 1/2^0.39354
final weight= 1.5mol* 0.761= 1.14 mol
Answer :
(1) The time passed by the sample is [tex]2.4\times 10^3\text{ years}[/tex]
(2) The amount left after decay process is 1.14 mmol.
Explanation :
Part 1 :
Half-life = 5730 years
First we have to calculate the rate constant, we use the formula :
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
[tex]k=\frac{0.693}{5730\text{ years}}[/tex]
[tex]k=1.21\times 10^{-4}\text{ years}^{-1}[/tex]
Now we have to calculate the time passed.
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = [tex]1.21\times 10^{-4}\text{ years}^{-1}[/tex]
t = time passed by the sample = ?
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
[tex]t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{100}{75}[/tex]
[tex]t=2377.9\text{ years}=2.4\times 10^3\text{ years}[/tex]
Therefore, the time passed by the sample is [tex]2.4\times 10^3\text{ years}[/tex]
Part 2 :
Now we have to calculate the amount left.
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = [tex]1.21\times 10^{-4}\text{ years}^{-1}[/tex]
t = time passed by the sample = 2255 years
a = let initial amount of the reactant = 1.5 mmol
a - x = amount left after decay process = ?
Now put all the given values in above equation, we get
[tex]2255=\frac{2.303}{1.21\times 10^{-4}}\log\frac{1.5}{a-x}[/tex]
[tex]a-x=1.14mmol[/tex]
Therefore, the amount left after decay process is 1.14 mmol.
