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Chapter 6: Problem 6

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

Short Answer

Expert verified
The relationship between the rate constant (k) and the half-life (t鈧嶁倎/鈧傗値) 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 by step solution

01

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.
02

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.
03

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: \[t_{1/2}=\frac{ln(1/2)}{-k}\]

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