A first order reaction is 75% completed in 100 minutes. How long time will it take for its 87.5% completion?

Options

A. 125 min
B. 150 min
C. 175 min
D. 200 min

Answer

To solve the problem, we need to determine the time required for a first-order reaction to reach 87.5% completion, given that it takes 100 minutes to reach 75% completion.

### Step-by-Step Solution:

1. **Understanding the Reaction Completion:**
- For a first-order reaction, if 75% of the reactant has reacted, then 25% remains.
- If 87.5% has reacted, then 12.5% remains.

2. **Using the First-Order Kinetics Formula:**
- The formula for the time \( t \) for a first-order reaction is given by:
\[ t = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]_t} \right) \]
- Here, \([A]_0\) is the initial concentration, and \([A]_t\) is the concentration at time \( t \).

3. **Finding the Rate Constant \( k \):**
- From the problem, we know that 75% completion corresponds to 100 minutes:
\[ t_{75\%} = 100 \text{ minutes} \]
- Therefore, \([A]_0 = 100\%\) and \([A]_{75\%} = 25\%\).
- Plugging these values into the formula gives:
\[ 100 = \frac{2.303}{k} \log \left( \frac{100}{25} \right) \]
- Simplifying the logarithm:
\[ \log \left( \frac{100}{25} \right) = \log(4) \]
- Thus, we have:
\[ 100 = \frac{2.303}{k} \log(4) \]

4. **Calculating \( k \):**
- Rearranging the equation to solve for \( k \):
\[ k = \frac{2.303 \log(4)}{100} \]

5. **Finding Time for 87.5% Completion:**
- For 87.5% completion, we have:
\[ [A]_{87.5\%} = 12.5\% \]
- Using the first-order kinetics formula again:
\[ t_{87.5\%} = \frac{2.303}{k} \log \left( \frac{100}{12.5} \right) \]
- Simplifying the logarithm:
\[ \log \left( \frac{100}{12.5} \right) = \log(8) \]
- Thus, we have:
\[ t_{87.5\%} = \frac{2.303}{k} \log(8) \]

6. **Substituting \( k \) into the Equation:**
- Now substituting \( k \) from step 4 into the equation for \( t_{87.5\%} \):
\[ t_{87.5\%} = \frac{2.303}{\frac{2.303 \log(4)}{100}} \log(8) \]
- This simplifies to:
\[ t_{87.5\%} = \frac{100 \log(8)}{\log(4)} \]

7. **Calculating \( \log(8) \) and \( \log(4) \):**
- We know that:
\[ \log(8) = 3 \log(2) \quad \text{and} \quad \log(4) = 2 \log(2) \]
- Therefore:
\[ \frac{\log(8)}{\log(4)} = \frac{3 \log(2)}{2 \log(2)} = \frac{3}{2} \]

8. **Final Calculation:**
- Substituting this back gives:
\[ t_{87.5\%} = 100 \times \frac{3}{2} = 150 \text{ minutes} \]

### Conclusion:
The time required for the reaction to reach 87.5% completion is **150 minutes**.