To solve the problem step by step, we will use the first-order reaction kinetics formula.
### Step 1: Understand the Reaction
The reaction given is A to B, and it follows first-order kinetics. We know that for a first-order reaction, the rate constant (k) can be calculated using the formula:
\[
k = \frac{2.303}{T} \log \left( \frac{[A]_0}{[A]} \right)
\]
Where:
- \([A]_0\) = initial concentration of A
- \([A]\) = final concentration of A
- \(T\) = time taken for the reaction
### Step 2: Calculate the Rate Constant (k)
From the problem, we know that:
- Initial amount of A, \([A]_0 = 0.8 \, \text{moles}\)
- Amount of B produced = 0.6 moles, so the final amount of A, \([A] = 0.8 - 0.6 = 0.2 \, \text{moles}\)
- Time taken, \(T = 1 \, \text{hour}\)
Now, substituting these values into the formula:
\[
k = \frac{2.303}{1} \log \left( \frac{0.8}{0.2} \right)
\]
Calculating the logarithm:
\[
\frac{0.8}{0.2} = 4 \quad \Rightarrow \quad \log(4)
\]
Thus,
\[
k = 2.303 \times \log(4)
\]
### Step 3: Set Up for the Second Reaction
Now we need to find the time taken for the conversion of 0.9 moles of A to produce 0.675 moles of B.
- Initial amount of A, \([A]_0 = 0.9 \, \text{moles}\)
- Amount of B produced = 0.675 moles, so the final amount of A, \([A] = 0.9 - 0.675 = 0.225 \, \text{moles}\)
### Step 4: Use the Rate Constant (k) to Find Time (T)
Using the same formula for the second reaction:
\[
k = \frac{2.303}{T} \log \left( \frac{0.9}{0.225} \right)
\]
Calculating the logarithm:
\[
\frac{0.9}{0.225} = 4 \quad \Rightarrow \quad \log(4)
\]
Now we can equate the two expressions for \(k\):
\[
2.303 \log(4) = \frac{2.303}{T_1} \log(4) \quad \text{(for the first reaction)}
\]
\[
2.303 \log(4) = \frac{2.303}{T_2} \log(4) \quad \text{(for the second reaction)}
\]
### Step 5: Solve for Time (T)
Since \(\log(4)\) and \(2.303\) are common in both equations, they cancel out:
\[
T_1 = T_2
\]
Thus, \(T_2 = 1 \, \text{hour}\).
### Conclusion
The time taken for the conversion of 0.9 moles of A to produce 0.675 moles of B is **1 hour**.
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