How much time in minutes passes between 99% to 99.9% completion of a first order reaction with half life `0.3010"min"^(-1)`?

Explanation

To solve the problem of how much time passes between 99% and 99.9% completion of a first-order reaction with a half-life of 0.3010 minutes, we can follow these steps:

### Step 1: Understand the relationship between half-life and rate constant
For a first-order reaction, the relationship between the half-life (t₁/₂) and the rate constant (k) is given by the formula:
\[ k = \frac{0.693}{t_{1/2}} \]
Given that \( t_{1/2} = 0.3010 \) minutes, we can calculate k.

### Step 2: Calculate the rate constant (k)
Using the formula:
\[ k = \frac{0.693}{0.3010} \]
Calculating this gives:
\[ k \approx 2.303 \, \text{min}^{-1} \]

### Step 3: Set up the equations for 99% and 99.9% completion
For a first-order reaction, the amount of reactant remaining can be expressed as:
\[ \frac{[A]_0}{[A]} = e^{kt} \]
Where:
- \([A]_0\) is the initial concentration,
- \([A]\) is the concentration at time \( t \),
- \( k \) is the rate constant,
- \( t \) is the time.

1. For 99% completion:
- Remaining concentration \([A] = 1\%\) of \([A]_0\)
- Therefore, \(\frac{[A]_0}{[A]} = 100\)
- The equation becomes:
\[ 100 = e^{kt_1} \]
Taking the natural logarithm:
\[ kt_1 = \ln(100) \]
\[ t_1 = \frac{\ln(100)}{k} \]

2. For 99.9% completion:
- Remaining concentration \([A] = 0.1\%\) of \([A]_0\)
- Therefore, \(\frac{[A]_0}{[A]} = 1000\)
- The equation becomes:
\[ 1000 = e^{kt_2} \]
Taking the natural logarithm:
\[ kt_2 = \ln(1000) \]
\[ t_2 = \frac{\ln(1000)}{k} \]

### Step 4: Calculate \( t_1 \) and \( t_2 \)
1. Calculate \( t_1 \):
\[ t_1 = \frac{\ln(100)}{2.303} \]
\[ t_1 \approx \frac{4.605}{2.303} \approx 2 \, \text{minutes} \]

2. Calculate \( t_2 \):
\[ t_2 = \frac{\ln(1000)}{2.303} \]
\[ t_2 \approx \frac{6.908}{2.303} \approx 3 \, \text{minutes} \]

### Step 5: Find the time difference between 99% and 99.9% completion
The time difference \( \Delta t \) is:
\[ \Delta t = t_2 - t_1 \]
\[ \Delta t = 3 \, \text{minutes} - 2 \, \text{minutes} = 1 \, \text{minute} \]

### Final Answer
The time that passes between 99% and 99.9% completion of the reaction is **1 minute**.

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