Problem 6 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The relationship between the rate constant (k) and the half-life (t₍₁/₂₎) of a first-order reaction is given by the equation: \[t_{1/2}=\frac{ln(1/2)}{-k}\] This equation shows that the half-life depends only on the rate constant, and they are inversely related.

Step 1: Write the rate law for a first-order reaction

The rate law for a first-order reaction is given by: \[\text{rate} = k[\text{A}]\] where "rate" is the rate of reaction, \(k\) is the rate constant, and [\(\text{A}\)] is the concentration of the reactant A at any given time.

Step 2: Integrate the rate law to find the integrated rate law

By integrating the rate law, we can find the relationship between the reactant concentration and time. For a first-order reaction, the integrated rate law is given by: \[ln([\text{A}])=-kt+ln([\text{A}]_0)\] where \(ln([\text{A}])\) is the natural logarithm of the reactant concentration at any time t, \(ln([\text{A}]_0)\) is the natural logarithm of the initial reactant concentration, and \(t\) is the time.

Step 3: Determine the relationship between the half-life and the rate constant

At half-life, the reactant concentration is half of its initial value (\([\text{A}]=\frac{1}{2}[\text{A}]_0\)). We can substitute this into the integrated rate law to obtain the relationship between the half-life (\(t_{1/2}\)) and the rate constant (\(k\)): \[ln\left(\frac{1}{2}[\text{A}]_0\right)=-kt_{1/2}+ln([\text{A}]_0)\] Then, using logarithm properties, we can simplify the equation: \[ln\left(\frac{1}{2}\right)=-kt_{1/2}\] Finally, we can isolate the half-life on one side of the equation: \[t_{1/2}=\frac{ln(1/2)}{-k}\] Notice that this equation shows that the half-life of a first-order reaction depends only on the rate constant. The relationship between the half-life and the rate constant (\(k\)) for a first-order reaction is: \[t_{1/2}=\frac{ln(1/2)}{-k}\]