Time for Concentration to Drop to One Third

Oct 13, 2025 by wbrum.org 137 views

Introduction

In chemistry, a first-order reaction is a type of chemical reaction where the rate of reaction is directly proportional to the concentration of one reactant. The rate constant (k) of a first-order reaction is a measure of how fast the reaction occurs. In this article, we will calculate the half-life and the time taken for the concentration to drop to one third of a first-order reaction with a rate constant of 2.39 × 10⁻³ s⁻¹.

Half-Life of a First-Order Reaction

The half-life (t1/2) of a first-order reaction is the time it takes for the concentration of the reactant to decrease by half. The half-life of a first-order reaction is related to the rate constant (k) by the following equation:

t1/2 = 0.693 / k

where k is the rate constant in s⁻¹.

Calculating Half-Life

To calculate the half-life of the reaction, we can plug in the value of the rate constant (k = 2.39 × 10⁻³ s⁻¹) into the equation:

t1/2 = 0.693 / (2.39 × 10⁻³ s⁻¹)

t1/2 = 289.1 s

Therefore, the half-life of the reaction is approximately 289.1 seconds.

Time for Concentration to Drop to One Third

The time taken for the concentration to drop to one third of the initial concentration can be calculated using the following equation:

t = (ln(3)) / k

where t is the time in seconds, ln(3) is the natural logarithm of 3, and k is the rate constant in s⁻¹.

Calculating Time for Concentration Drop

To calculate the time taken for the concentration to drop to one third, we can plug in the value of the rate constant (k = 2.39 × 10⁻³ s⁻¹) into the equation:

t = (ln(3)) / (2.39 × 10⁻³ s⁻¹)

t = 289.1 s / ln(3)

t = 103.4 s

Therefore, the time taken for the concentration to drop to one third of the initial concentration is approximately 103.4 seconds.

Conclusion

In conclusion, we have calculated the half-life and the time taken for the concentration to drop to one third of a first-order reaction with a rate constant of 2.39 × 10⁻³ s⁻¹. The half-life of the reaction is approximately 289.1 seconds, and the time taken for the concentration to drop to one third is approximately 103.4 seconds.

References

Atkins, P. W., & De Paula, J. (2010). Physical chemistry. Oxford University Press.
Levine, I. N. (2014). Physical chemistry. McGraw-Hill Education.

Mathematical Derivations

Half-Life of a First-Order Reaction

The half-life (t1/2) of a first-order reaction is related to the rate constant (k) by the following equation:

t1/2 = 0.693 / k

where k is the rate constant in s⁻¹.

Time for Concentration to Drop to One Third

The time taken for the concentration to drop to one third of the initial concentration can be calculated using the following equation:

t = (ln(3)) / k

where t is the time in seconds, ln(3) is the natural logarithm of 3, and k is the rate constant in s⁻¹.

Derivation of Half-Life Equation

The half-life of a first-order reaction can be derived from the integrated rate equation:

ln([A]t / [A]0) = -kt

where [A]t is the concentration of the reactant at time t, [A]0 is the initial concentration, k is the rate constant, and t is the time.

Rearranging the equation to solve for t, we get:

t = (ln([A]0 / [A]t)) / k

Since the half-life is the time it takes for the concentration to decrease by half, we can set [A]t = [A]0 / 2:

t1/2 = (ln([A]0 / ([A]0 / 2))) / k

Simplifying the equation, we get:

t1/2 = 0.693 / k

Derivation of Time for Concentration Drop Equation

The time taken for the concentration to drop to one third of the initial concentration can be derived from the integrated rate equation:

ln([A]t / [A]0) = -kt

where [A]t is the concentration of the reactant at time t, [A]0 is the initial concentration, k is the rate constant, and t is the time.

Rearranging the equation to solve for t, we get:

t = (ln([A]0 / [A]t)) / k

Since the concentration has dropped to one third, we can set [A]t = [A]0 / 3:

t = (ln([A]0 / ([A]0 / 3))) / k

Simplifying the equation, we get:

t = (ln(3)) / k