To derive the integrated rate law for a first-order reaction and show that the half-life is independent of the reactant concentration, we start with the definition of a first-order reaction, which can be represented by the following equation:
\[ A \rightarrow \text{products} \]
The rate of reaction can be expressed as:
\[ -\frac{d[A]}{dt} = k[A] \]
where \( [A] \) is the concentration of reactant A at time \( t \), and \( k \) is the rate constant.
Step 1: Rearranging the Equation
We can rearrange the rate equation to separate variables:
\[ \frac{d[A]}{[A]} = -k , dt \]
Step 2: Integrating
Next, we integrate both sides. The left side integrates with respect to \( [A] \), and the right side integrates with respect to \( t \):
\[ \int \frac{d[A]}{[A]} = -k \int dt \]
This gives us:
\[ \ln[A] = -kt + C \]
where \( C \) is the integration constant.
Step 3: Solving for the Integration Constant
To find the constant \( C \), we can use the initial condition. Let \( [A]_0 \) be the initial concentration of \( A \) at \( t = 0 \):
\[ \ln[A]_0 = C \]
Step 4: Writing the Integrated Rate Law
Now, substituting \( C \) back into our equation, we have:
\[ \ln[A] = -kt + \ln[A]_0 \]
By exponentiating both sides, we obtain the integrated rate law for a first-order reaction:
\[ [A] = [A]_0 e^{-kt} \]
Step 5: Deriving the Half-Life
The half-life (\( t_{1/2} \)) of a reaction is the time required for the concentration of the reactant to decrease to half its initial value. We set \( [A] = \frac{1}{2}[A]_0 \) and substitute it into the integrated rate law:
\[ \frac{1}{2}[A]_0 = [A]0 e^{-kt{1/2}} \]
Dividing both sides by \( [A]_0 \) (assuming \( [A]_0 \neq 0 \)):
\[ \frac{1}{2} = e^{-kt_{1/2}} \]
Now, taking the natural logarithm of both sides:
\[ \ln\left(\frac{1}{2}\right) = -kt_{1/2} \]
Rearranging gives:
\[ t_{1/2} = -\frac{\ln\left(\frac{1}{2}\right)}{k} \]
Using the property of logarithms that \( \ln\left(\frac{1}{2}\right) = -\ln(2) \):
\[ t_{1/2} = \frac{\ln(2)}{k} \]
Conclusion: Independence from Initial Concentration
The final expression for the half-life of a first-order reaction is:
\[ t_{1/2} = \frac{\ln(2)}{k} \]
This shows that the half-life of a first-order reaction is independent of the initial concentration \( [A]_0 \). The only factors that influence the half-life are the rate constant \( k \) and the natural logarithm of 2, confirming the characteristics of first-order kinetics.
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