I\'m not sure how to go about figuring out the estimate rock age. I found the fi

Question

I'm not sure how to go about figuring out the estimate rock age. I found the first part out just not sure where to begin to determine the age

Consider the composite reaction below where A decomposes into two possible products, X and Y (called a competitive or parallel reaction). Derive an expression for [A], [X], and [Y] as a function of time. Use this expression to show that at any time [X/[Y]-k,/k2 (Hint: Write down expression for d[AWdt, d|XI/dt, and d[YIdt. You should be able to solve one of these very quickly, and then substitute this result into the other expressions.) Potassium-argon dating is commonly used in geology to date sedimentary rocks. Potassium- 40 decavs via two different pathwavs 18K28Ca + e- (893% of products) (10.7% of products) The overall half life for the decay of potassium-40 is 13 x 10 years. Estimate the age of a 18K,:Ar +p" rock with argon-40 to potassium-40 ratio of 40 18-0.102 40 19

Explanation / Answer

1) The reaction is parallel reaction

We can say that for first conversion

-d[A] / dt = K1[A]

For second conversion

-d[A] / dt = K2[A]

Overall

-d[A] / dt = K1[A] + K2[A]

d[A] / dt = -[A] (K1+K2)

on inegration

[A] = [A0] e^-(K1+K2)t

The rate of formation of X will be = d[X] /dt

d[X] /dt = K1[A] = K1 [A0] e^-(k1+k2)t

Again on inegrating , with limits from A0 to A (lower and upper limits)

[X] = K1[A0] / K1+K2 (1-e^-(k1+k2)t )

hence similarly

[Y] = K2[A0] / K1+K2 (1-e^-(k1+k2)t )

so K1/K2 = [X] / [Y]

2) The half life is = 1.3 X 10^9 years

Here 10% is Ar-40 and 90% is K-40

Given: only 10.7% of K-40 decays into Ar-40 and 89.3 % will decay into Ca-40

So age of rock will be

Age = Half life of K-40 / 0.693 X ln (1 + ratio of argon and potassium X 1/0.107)

On plugging the values

AGe = 1.3 X 10^9 / 0.693 X ln ( 1+ 0.102 / 0.107)

Age = 1.875 X 10^9 X 2.332

Age = 4.37 X 10^9 years

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