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To find the half-life of a first-order reaction, we can use the formula for half-life, which is given by:
\[
t_{1/2} = \frac{0.693}{k}
\]
where \(k\) is the rate constant of the reaction. First, we need to determine the rate constant \(k\) using the information provided in the question.
### Step 1: Determine the change in concentration
From the question, we know that 1 mole of reactant A produces 0.2 moles of B after 100 minutes. Therefore, the change in concentration of A can be calculated as follows:
\[
\text{Initial concentration of A} = 1 \, \text{mol}
\]
\[
\text{Final concentration of A} = 1 - 0.2 = 0.8 \, \text{mol}
\]
\[
\Delta [A] = [A]_0 - [A] = 1 - 0.8 = 0.2 \, \text{mol}
\]
### Step 2: Calculate the rate constant \(k\)
For a first-order reaction, the integrated rate law is given by:
\[
\ln \left( \frac{[A]_0}{[A]} \right) = kt
\]
Substituting the values we have:
\[
\ln \left( \frac{1}{0.8} \right) = k \cdot 100
\]
Calculating the left side:
\[
\ln(1.25) \approx 0.2231
\]
Now, substituting back into the equation:
\[
0.2231 = k \cdot 100
\]
Solving for \(k\):
\[
k = \frac{0.2231}{100} = 0.002231 \, \text{min}^{-1}
\]
### Step 3: Calculate the half-life \(t_{1/2}\)
Now that we have the rate constant \(k\), we can calculate the half-life:
\[
t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.002231} \approx 310.5 \, \text{min}
\]
### Step 4: Round off to the nearest integer
Finally, rounding off the half-life to the nearest integer gives us:
\[
t_{1/2} \approx 311 \, \text{min}
\]
Thus, the half-life of the reaction is approximately **311 minutes**.
