A first order reaction is completed by 20% 2 minutes. How much further time is required for 64% of the initial concentration of the reactants to remain.

Explanation

To solve the problem step by step, we will use the first-order reaction kinetics formula and the information given in the question.

### Step 1: Understand the Problem
We know that a first-order reaction is 20% completed in 2 minutes. This means that 80% of the reactant remains after 2 minutes. We need to find out how much additional time is required for 64% of the initial concentration of the reactant to remain.

### Step 2: Define the Initial Concentration
Let's assume the initial concentration of the reactant is 100%. After 2 minutes, since the reaction is 20% completed, the remaining concentration is:
\[ \text{Remaining concentration} = 100\% - 20\% = 80\% \]

### Step 3: Calculate the Rate Constant (k)
Using the first-order kinetics formula:
\[ k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]} \right) \]
Where:
- \( [A]_0 \) = initial concentration = 100%
- \( [A] \) = concentration after 2 minutes = 80%
- \( t \) = time = 2 minutes

Substituting the values:
\[ k = \frac{2.303}{2} \log \left( \frac{100}{80} \right) \]

Calculating \( \log \left( \frac{100}{80} \right) \):
\[ \log \left( \frac{100}{80} \right) = \log(1.25) \approx 0.09691 \]

Now substituting this back into the equation for \( k \):
\[ k = \frac{2.303}{2} \times 0.09691 \approx 0.1117 \, \text{min}^{-1} \]

### Step 4: Calculate the Time for 64% Remaining
Now we need to find out how much time is required for the concentration to drop from 100% to 64%. Using the same formula:
\[ k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]} \right) \]
Where:
- \( [A]_0 = 100\% \)
- \( [A] = 64\% \)

Rearranging the formula to solve for \( t \):
\[ t = \frac{2.303}{k} \log \left( \frac{100}{64} \right) \]

Calculating \( \log \left( \frac{100}{64} \right) \):
\[ \log \left( \frac{100}{64} \right) = \log(1.5625) \approx 0.19382 \]

Now substituting \( k \) and the logarithm value:
\[ t = \frac{2.303}{0.1117} \times 0.19382 \approx 4.12 \, \text{minutes} \]

### Step 5: Calculate Additional Time Required
Since we already took 2 minutes to reach 80%, the additional time required to go from 80% to 64% is:
\[ \text{Additional time} = t - 2 \approx 4.12 - 2 = 2.12 \, \text{minutes} \]

### Final Answer
Thus, the total time required for 64% of the initial concentration of the reactants to remain is approximately **4 minutes**.

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