Question

A volcanic eruption melts a large area of rock, and all gases are expelled. after cooling, 40ar (atomic number 18) accumulates from the ongoing decay of 40k (atomic number 19) in the rock (t1/2 = 1.25 × 109 yr). when a piece of rock is analyzed, it is found to contain 1.33 mmol of 40k and 1.50 mmol of 40ar. how long ago did the rock cool? enter your answer in scientific notation.

Answer 1

Decay of Ar-40 produces 1.33 mmol of K-40. Remaining number of moles of Ar-40 is 1.50 mmol. Initial mmol of Ar-40 present will be sum of number of moles of K-40 and remaining number of moles of Ar-40.

$n_{Ar-40}=(1.33+1.50)m mol=2.83 mmol$

now, half life time of the reaction is $1.25\times 10^{9} year$

For first order reaction, rate constant and half-life time are related to each other as follows:

$k=\frac{0.6932}{t_{1/2}}$

Putting the value of $t_{1/2}$,

$k=\frac{0.6932}{1.25\times 10^{9} year}=5.54\times 10^{-10} year^{-1}$

Rate equation for first order reaction is as follows:

$t=\frac{2.303}{k}log\frac{[A_{0}]}{[A_{t}]}}$

Here, $[A_{0}]$ is initial concentration and $[A_{t}]$ is concentration at time t.

$t=\frac{2.303}{(5.54\times 10^{-10} year^{-1})}log\frac{(2.83 mmol)}{(1.50 mmol)}}=1.146\times 10^{9} year$

Therefore, time required to cool the rock is $1.146\times 10^{9} year$.