It takes 1 h for a first order recation to go to `50%` completion . The total time required for the same reaction to reach `87.5%` completion will be

Options

A. `1.75` h
B. `6.00` h
C. `3.50` h
D. `3.00` h

Explanation

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 1 hour to reach 50% completion.

### Step-by-step Solution:

1. **Understanding the Half-Life of a First-Order Reaction**:
The half-life (T₁/₂) of a first-order reaction is given as 1 hour. For first-order reactions, the half-life is constant and does not depend on the concentration of the reactants.

2. **Using the Half-Life to Find the Rate Constant (k)**:
The relationship between the half-life and the rate constant (k) for a first-order reaction is given by:
\[ T_{1/2} = \frac{0.693}{k} \]
Rearranging this formula to find k:
\[ k = \frac{0.693}{T_{1/2}} = \frac{0.693}{1 \text{ hour}} = 0.693 \text{ hour}^{-1} \]

3. **Using the Integrated Rate Law for First-Order Reactions**:
The integrated rate law for a first-order reaction is:
\[ \ln\left(\frac{[A_0]}{[A]}\right) = kt \]
Where:
- [A₀] = initial concentration
- [A] = concentration at time t
- k = rate constant
- t = time

4. **Calculating Time for 87.5% Completion**:
If the reaction reaches 87.5% completion, this means that 12.5% of the reactant remains. Therefore, we can express this as:
\[ [A] = 0.125 [A_0] \]
Substituting into the integrated rate law:
\[ \ln\left(\frac{[A_0]}{0.125 [A_0]}\right) = kt \]
Simplifying gives:
\[ \ln\left(\frac{1}{0.125}\right) = kt \]
\[ \ln(8) = kt \]

5. **Calculating \(\ln(8)\)**:
We know that:
\[ \ln(8) = \ln(2^3) = 3 \ln(2) \approx 3 \times 0.693 = 2.079 \]

6. **Substituting k and Solving for t**:
Now substituting k into the equation:
\[ 2.079 = (0.693) t \]
Solving for t:
\[ t = \frac{2.079}{0.693} \approx 3 \text{ hours} \]

### Final Answer:
The total time required for the reaction to reach 87.5% completion is approximately **3 hours**.

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