A first order reaction is 15% complete in 20 minutes. In what time will the reaction 60% complete ?

Explanation

To solve the problem step by step, we will use the first-order reaction kinetics formula.

### Step 1: Understand the first-order reaction formula
For a first-order reaction, the rate constant \( k \) can be expressed using the equation:
\[ k = \frac{1}{t} \ln\left(\frac{A_0}{A_t}\right) \]
where:
- \( A_0 \) is the initial concentration,
- \( A_t \) is the concentration at time \( t \),
- \( t \) is the time taken for the reaction.

### Step 2: Determine the values for the first part of the reaction
Given that the reaction is 15% complete in 20 minutes:
- If the reaction is 15% complete, then 85% of the reactant remains. Thus, we have:
- \( A_0 = 100 \) (assuming 100% initial concentration),
- \( A_t = 100 - 15 = 85 \).

Now, substituting these values into the formula:
\[ k = \frac{1}{20} \ln\left(\frac{100}{85}\right) \]

### Step 3: Calculate \( k \)
First, calculate \( \ln\left(\frac{100}{85}\right) \):
\[ \frac{100}{85} \approx 1.1765 \]
Now, find the natural logarithm:
\[ \ln(1.1765) \approx 0.1625 \]
Now substitute back to find \( k \):
\[ k = \frac{1}{20} \times 0.1625 = 0.008125 \text{ min}^{-1} \]

### Step 4: Determine the values for the second part of the reaction
Now we need to find the time when the reaction is 60% complete:
- If the reaction is 60% complete, then 40% of the reactant remains:
- \( A_t = 100 - 60 = 40 \).

Using the same formula:
\[ k = \frac{1}{t} \ln\left(\frac{100}{40}\right) \]

### Step 5: Calculate the new time \( t \)
Now calculate \( \ln\left(\frac{100}{40}\right) \):
\[ \frac{100}{40} = 2.5 \]
Now find the natural logarithm:
\[ \ln(2.5) \approx 0.9163 \]
Now substitute back to find \( t \):
\[ 0.008125 = \frac{1}{t} \times 0.9163 \]
Rearranging gives:
\[ t = \frac{0.9163}{0.008125} \approx 112.3 \text{ minutes} \]

### Final Answer:
The time required for the reaction to be 60% complete is approximately **112.3 minutes**.

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