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Chapter 6: Problem 6

How are the rate constant and the half-life related?

Short Answer

Expert verified
Question: Explain the relationship between the rate constant and half-life of a first-order reaction. Answer: For a first-order reaction, the rate constant (k) and half-life (t_1/2) are inversely proportional and related by the expression: k = 0.693/t_1/2. This relationship shows that as the rate constant increases, the half-life of the reaction decreases, indicating a faster reaction. Conversely, as the rate constant decreases, the half-life increases, indicating a slower reaction.

Step by step solution

01

Define the rate equation for a first-order reaction

A first-order reaction has the rate equation of the form: Rate = k[A] where Rate is the rate of the reaction, k is the rate constant, and [A] is the concentration of the reactant at a specific time.
02

Integrate the rate equation to find the relationship between the concentration of the reactant and time

Divide both sides of the rate equation by [A] and multiply by dt. Then, integrate both sides of the equation: $$ \int_{[A]_0}^{[A]} \frac{d[A]}{[A]} = k \int_{0}^{t} dt $$ where [A]_0 is the initial concentration of the reactant and t is the time taken for the concentration to change from [A]_0 to [A]. Integration yields: $$ ln\frac{[A]}{[A]_0}= -kt $$
03

Define half-life

Half-life (t_1/2) is the time it takes for the concentration of the reactant to reduce to half of its initial value. So, at t_1/2: $$ [A] = \frac{[A]_0}{2} $$
04

Find the relationship between the rate constant and half-life

Substitute the half-life condition into the integrated rate equation: $$ ln\frac{\frac{[A]_0}{2}}{[A]_0}= -kt_{1/2} $$ Simplify the equation: $$ ln\frac{1}{2} = -kt_{1/2} $$ The natural logarithm of 1/2 is -0.693, hence: $$ -0.693 = -kt_{1/2} $$ Divide both sides by -1 and then divide by t_1/2: $$ k = \frac{0.693}{t_{1/2}} $$
05

Final Conclusion

The rate constant (k) and half-life (t_1/2) are related by the following expression for first-order reactions: $$ k = \frac{0.693}{t_{1/2}} $$

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