To solve the problem step by step, we will use the first-order kinetics formula and the information provided.
### Step 1: Understand the reaction and initial conditions
The reaction is given as:
\[ A \rightarrow B \]
From the problem, we know:
- Initial moles of \( A \) in the first condition: \( 0.8 \, \text{mol} \)
- Moles of \( B \) produced after 1 hour: \( 0.6 \, \text{mol} \)
- Therefore, moles of \( A \) remaining after 1 hour:
\[ 0.8 \, \text{mol} - 0.6 \, \text{mol} = 0.2 \, \text{mol} \]
### Step 2: Set up the first-order kinetics equation
For a first-order reaction, the rate constant \( k \) can be expressed using the formula:
\[ k = \frac{2.303}{t} \log_{10} \left( \frac{[A]_0}{[A]} \right) \]
Where:
- \( [A]_0 \) is the initial concentration of \( A \)
- \( [A] \) is the concentration of \( A \) at time \( t \)
### Step 3: Calculate the rate constant \( k \) using the first condition
Using the first condition:
- \( [A]_0 = 0.8 \, \text{mol} \)
- \( [A] = 0.2 \, \text{mol} \)
- \( t = 1 \, \text{hr} \)
Substituting these values into the equation:
\[ k = \frac{2.303}{1} \log_{10} \left( \frac{0.8}{0.2} \right) \]
\[ k = 2.303 \log_{10} (4) \]
Since \( \log_{10} (4) = 0.602 \):
\[ k = 2.303 \times 0.602 \]
\[ k \approx 1.384 \, \text{hr}^{-1} \]
### Step 4: Set up the second condition
For the second condition:
- Initial moles of \( A \): \( 0.9 \, \text{mol} \)
- Moles of \( B \) produced: \( 0.675 \, \text{mol} \)
- Therefore, moles of \( A \) remaining:
\[ 0.9 \, \text{mol} - 0.675 \, \text{mol} = 0.225 \, \text{mol} \]
### Step 5: Calculate the time \( t \) for the second condition
Using the first-order kinetics equation again:
\[ k = \frac{2.303}{t} \log_{10} \left( \frac{[A]_0}{[A]} \right) \]
Substituting the values for the second condition:
\[ 1.384 = \frac{2.303}{t} \log_{10} \left( \frac{0.9}{0.225} \right) \]
Calculating \( \log_{10} \left( \frac{0.9}{0.225} \right) \):
\[ \frac{0.9}{0.225} = 4 \]
So,
\[ \log_{10} (4) = 0.602 \]
Now substituting back into the equation:
\[ 1.384 = \frac{2.303}{t} \times 0.602 \]
Rearranging to find \( t \):
\[ t = \frac{2.303 \times 0.602}{1.384} \]
Calculating:
\[ t \approx \frac{1.384}{1.384} \]
\[ t \approx 1 \, \text{hr} \]
### Final Answer
The time taken for the conversion of \( 9.0 \, \text{mol} \) of \( A \) to produce \( 0.675 \, \text{mol} \) of \( B \) is approximately **1 hour**.
---
