Carbon-14 has a half-life of 5720 years and this is a first order reaction. If a piece of wood has converted 75% of the carbon-1

Question

Andreyy89

3 years ago

13

Carbon-14 has a half-life of 5720 years and this is a first order reaction. If a piece of wood has converted 75% of the carbon-1

4, then how old is it?

Chemistry

2 answers:

liraira [26] · Answer 1 · 2022-05-17T14:25:26+03:00

Answer: 1145.8 years

Explanation:

Half-life of carbon-14 = 5720 years

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{5720\text{years}}$

$k=1.21\times 10^{-4}\text{years}^{-1}$

Now we have to calculate the age of the sample:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $1.21\times 10^{-4}\text{years}^{-1}$

t = age of sample = ?

a = let initial amount of the reactant = 100 g

x = amount decayed = 75 g

a - x = amount left after decay process = 100 - 75 = 25g

Now put all the given values in above equation, we get

$t==\frac{2.303}{1.21\times 10^{-4}}\log\frac{100}{25}$

$t=1145.8years$

gizmo_the_mogwai [7] · Answer 2 · 2022-05-17T16:26:17+03:00

The age of the sample carbon-14 is $\boxed{\text{11459 years}}$ .

Further Explanation:

Radioactive decay is the process of stabilization of unstable atomic nucleus with the release of energy. This emission of energy can be in form of different particles like alpha, beta and gamma particles.

Half-life is the time period in which half of the radioactive species is consumed. It is represented by $t_{1/2}$ .

the expression for half-life is given as follows:

$t_{1/2}=\dfrac{0.693}{k}$ ...... (1)

Where,

$t_{1/2}$ is half-life period .

$k$ is the rate constant .

Rearrange equation (1) for $k$ .

$k=\dfrac{0.693}{t_{1/2}}$ ...... (2)

Substitute 5720 yr for $t_{1/2}$ in equation (2).

$\begin{aligned}k=&\dfrac{0.693}{5720\text{ yr}}\\=&1.21\times10^{-4} \text{yr}^{-1}\end{aligned}$

Since it is first-order reaction, the expression for rate its given as follows:

$k=\dfrac{2.303}{t}\text{log}\left(\dfrac{a}{a-x}\right)$ ...... (3)

Where,

$k$ is the rate constant.

$t$ is the time taken for decay process.

$a$ is the initial amount of sample.

$x$ is the amount of sample that has been decayed.

Rearrange equation (3) to calculate $t$ .

$t=\dfrac{2.303}{k}\text{log}\left(\dfrac{a}{a-x}\right)$ ...... (4)

Consider 100 g to be initial mass of sample. So mass of sample that is decayed becomes 75 g.

Substitute 100 g for $a$ ,75 g for $x$ and $1.21\times10^{-4} \text{yr}^{-1}$ for $k$ in equation (4).

$\begin{aligned}t=&\dfrac{2.303}{1.21\times10^{-4} \text{yr}^{-1}}\text{log}\left(\dfrac{100\text{ g}}{100\text{ g}-75\text{ g}}\right)\\=&11459\text{ years}\end{aligned}$