A first order reaction is 15 % complete in 20 minutes .How long will it take to complete 60 % ? (log `1.1765 = 0.0705` log 2.5 = 0.3979)

Explanation

To solve the problem, we will use the first-order reaction kinetics formula. The first-order rate equation can be expressed as:

\[ k = \frac{2.303}{T} \log \left( \frac{[A]_0}{[A]} \right) \]

Where:
- \( k \) is the rate constant,
- \( T \) is the time,
- \([A]_0\) is the initial concentration,
- \([A]\) is the concentration at time \( T \).

### Step 1: Determine the initial and final concentrations for the first scenario (15% completion)

Given that the reaction is 15% complete, if we assume the initial concentration \([A]_0 = 100\):
- The concentration at 15% completion \([A] = 100 - 15 = 85\).

### Step 2: Calculate the rate constant \( k \) using the first scenario

Using the formula for first-order kinetics:
\[ k = \frac{2.303}{T_1} \log \left( \frac{[A]_0}{[A]} \right) \]

Substituting the values:
- \( T_1 = 20 \) minutes,
- \([A]_0 = 100\),
- \([A] = 85\).

So,
\[ k = \frac{2.303}{20} \log \left( \frac{100}{85} \right) \]

Calculating the logarithm:
\[ \log \left( \frac{100}{85} \right) = \log(1.1765) = 0.0705 \]

Now substituting this into the equation for \( k \):
\[ k = \frac{2.303}{20} \times 0.0705 \]

Calculating \( k \):
\[ k = \frac{2.303 \times 0.0705}{20} \]
\[ k = \frac{0.1623}{20} = 0.008115 \, \text{min}^{-1} \]

### Step 3: Determine the final concentrations for the second scenario (60% completion)

For the second scenario where the reaction is 60% complete:
- The concentration at 60% completion \([A] = 100 - 60 = 40\).

### Step 4: Use the rate constant \( k \) to find \( T_2 \)

Using the same formula for the second scenario:
\[ k = \frac{2.303}{T_2} \log \left( \frac{[A]_0}{[A]} \right) \]

Substituting the values:
\[ 0.008115 = \frac{2.303}{T_2} \log \left( \frac{100}{40} \right) \]

Calculating the logarithm:
\[ \log \left( \frac{100}{40} \right) = \log(2.5) = 0.3979 \]

Now substituting this into the equation for \( k \):
\[ 0.008115 = \frac{2.303}{T_2} \times 0.3979 \]

Rearranging to solve for \( T_2 \):
\[ T_2 = \frac{2.303 \times 0.3979}{0.008115} \]

Calculating \( T_2 \):
\[ T_2 = \frac{0.9168}{0.008115} \approx 113.5 \, \text{minutes} \]

### Final Answer
It will take approximately **113.5 minutes** to complete 60% of the reaction.

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