`80%` of a fisrt order reaction was completed in `70 min`. How much it will take for `90%` completion of a reaction?

Options

A. `114 min`
B. `140 min`
C. `100 min`
D. `200 min`

Explanation

To solve the problem of how long it will take for 90% completion of a first-order reaction given that 80% completion takes 70 minutes, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction**:
- We have a first-order reaction where the initial concentration of the reactant A is denoted as \( A_0 \).
- At \( t = 70 \) minutes, 80% of the reactant has been converted to product B, meaning that 20% of the reactant remains.

2. **Determine Remaining Concentration**:
- At 80% completion, the concentration of A left is:
\[ [A] = 0.20 A_0 \]
- At 90% completion, the concentration of A left will be:
\[ [A] = 0.10 A_0 \]

3. **Use the First-Order Kinetics Equation**:
- The integrated rate law for a first-order reaction is given by:
\[ t = \frac{2.303}{k} \log \left( \frac{[A_0]}{[A]} \right) \]
- For the 80% completion scenario:
\[ t_{80\%} = 70 \text{ minutes} = \frac{2.303}{k} \log \left( \frac{A_0}{0.20 A_0} \right) = \frac{2.303}{k} \log(5) \]

4. **Calculate the Rate Constant (k)**:
- From the equation for 80% completion, we can express \( k \):
\[ 70 = \frac{2.303}{k} \log(5) \]
- Rearranging gives:
\[ k = \frac{2.303 \log(5)}{70} \]

5. **Set Up for 90% Completion**:
- For the 90% completion scenario:
\[ t_{90\%} = \frac{2.303}{k} \log \left( \frac{A_0}{0.10 A_0} \right) = \frac{2.303}{k} \log(10) \]

6. **Relate \( t_{90\%} \) to \( t_{80\%} \)**:
- We can express \( t_{90\%} \) in terms of \( t_{80\%} \):
\[ \frac{t_{90\%}}{t_{80\%}} = \frac{\log(10)}{\log(5)} \]
- Plugging in the known value of \( t_{80\%} \):
\[ t_{90\%} = t_{80\%} \cdot \frac{\log(10)}{\log(5)} = 70 \cdot \frac{1}{\log(5)} \]

7. **Calculate the Logarithm Values**:
- Using logarithm values:
\[ \log(5) \approx 0.699 \]
- Therefore:
\[ t_{90\%} = 70 \cdot \frac{1}{0.699} \approx 100 \text{ minutes} \]

### Final Answer:
The time required for 90% completion of the reaction is approximately **100 minutes**.