How Does First-Order Kinetics Affect Reactant Concentration Over Time?

[konartist]

Homework Statement

The initial concentration of reactant in a first-order reaction is 0.75M. The rate constant for the reaction is 2.75/s. What is the concentration (mol/L) of reactant after 6.5s?

Homework Equations

-kt=[A]initial/[A]anytime

The Attempt at a Solution

-(2.75/s)(6.5s)=[0.75M]/[A]@6.5seconds

-17.875/1=0.75M/x; x= -4.20x10^-2; The answer to this problem is not right, but I do not have any idea how to go about it another way.

Problem 2.

The usefulness of radiocarbon dating is limited to 50,000 years. Show mathematically why this is true. (Hint: Remember half life follows first order kinetics. The half life of C-14 is 5.73 x 10^3 years).

Half life formula : t1/2= 0.693/k
Common integrate rate law for first order reactions: ln[A]=-kt + ln[A@ initial]
5.73x10^3 = 0.693/k, k = 1.21x10^-4/s

How would I use k in my integrated formula to prove that the half life is limited to 50,000 years?

 

[Gokul43201]

#1. You have the wrong rate equation for a first-order reaction. Look at the rate equation you've used in #2. That's the one you need to use here.

#2. What is (roughly) the relative abundance of C-14 (w.r. to C-12) in the atmosphere? Assume you have a sample that has about 1kg of C. How much of that would have been C-14, when the organism was alive? How much of that C-14 will be present in the sample today (use the rate equation to find this)?

 

[Blueshift5]

For the first problem, the correct answer can be found by rearranging the integrated rate law for first-order reactions: [A] = [A]initial * e^(-kt). Plugging in the given values, we get [A] = 0.75M * e^(-2.75/s * 6.5s) = 0.75M * e^(-17.875) = 0.75M * 1.83 x 10^-8 = 1.38 x 10^-8 M. This is the concentration of reactant after 6.5 seconds.

For the second problem, we can use the same integrated rate law to show that after 50,000 years, the concentration of C-14 will be too low to accurately measure. The half-life of C-14 is 5.73 x 10^3 years, so after 50,000 years, the concentration will have decreased by a factor of (1/2)^50,000/5,730 = 0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000