3/4 th of first order reaction was completed in 32 min 15/16 the part will be completed in

Options

A. 24 min
B. 64 min
C. 16 min
D. 32 min

Answer

To solve the problem step by step, we will follow the principles of first-order kinetics.

### Step 1: Understand the problem
We are given that 3/4 of a first-order reaction is completed in 32 minutes. We need to find out how long it will take for 15/16 of the reaction to be completed.

### Step 2: Determine the remaining concentration after 3/4 completion
Let the initial concentration of the reactant be 1 (A0 = 1). If 3/4 of the reaction is completed, the remaining concentration (A) is:
\[ A = A_0 - \frac{3}{4}A_0 = 1 - \frac{3}{4} = \frac{1}{4} \]

### Step 3: Use the first-order rate equation to find the rate constant (k)
The rate constant for a first-order reaction is given by the equation:
\[ k = \frac{2.303}{t} \log\left(\frac{A_0}{A}\right) \]
Substituting the values we have:
\[ k = \frac{2.303}{32} \log\left(\frac{1}{\frac{1}{4}}\right) = \frac{2.303}{32} \log(4) \]
Since \(\log(4) = 2 \log(2) \approx 0.602\), we can calculate:
\[ k = \frac{2.303}{32} \times 0.602 \approx 0.0431 \text{ min}^{-1} \]

### Step 4: Calculate the time for 15/16 completion
Now we need to find the time taken for 15/16 of the reaction to be completed. The remaining concentration when 15/16 is completed is:
\[ A = A_0 - \frac{15}{16}A_0 = 1 - \frac{15}{16} = \frac{1}{16} \]

### Step 5: Use the rate constant to find the new time (t)
Using the same rate equation:
\[ k = \frac{2.303}{t} \log\left(\frac{A_0}{A}\right) \]
Substituting the values:
\[ 0.0431 = \frac{2.303}{t} \log\left(\frac{1}{\frac{1}{16}}\right) = \frac{2.303}{t} \log(16) \]
Since \(\log(16) = 4 \log(2) \approx 1.204\):
\[ 0.0431 = \frac{2.303}{t} \times 1.204 \]
Rearranging gives:
\[ t = \frac{2.303 \times 1.204}{0.0431} \]
Calculating this:
\[ t \approx \frac{2.773}{0.0431} \approx 64 \text{ minutes} \]

### Final Answer
The time taken for 15/16 of the reaction to be completed is approximately **64 minutes**.

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