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Step-by-Step Solution
Step 1: Identify the relevant first-order kinetics formula
For a first-order reaction, the rate constant $k$ is given by:
$$
k = \frac{2.303}{t} \log\left(\frac{[A]_0}{[A]_t}\right)
$$
where $[A]_0$ is the initial amount (or concentration) of A, $[A]_t$ is the amount (or concentration) of A at time $t$, and $k$ is the rate constant.
Step 2: Calculate the rate constant $k$ using the first data set
Given: Time taken ($t$) is 1 hour for A to reduce from 0.8 mole to 0.2 mole (since 0.6 mole of B is formed, the remaining 0.2 mole is A).
Substitute into the formula:
$$
k = \frac{2.303}{1} \log\left(\frac{0.8}{0.2}\right).
$$
Simplify inside the logarithm:
$$
\frac{0.8}{0.2} = 4,
$$
hence:
$$
k = 2.303 \log(4).
$$
Step 3: Express the unknown time $t_1$ for the second data set
In the second scenario, A decreases from 0.9 mole to 0.225 mole (since 0.675 mole of B is formed).
Using the same first-order equation but with time $t_1$:
$$
k = \frac{2.303}{t_1} \log\left(\frac{0.9}{0.225}\right).
$$
Step 4: Equate both expressions for $k$
From Step 2, we have $k = 2.303 \log(4).$
From Step 3, we have $k = \frac{2.303}{t_1} \log\left(\frac{0.9}{0.225}\right).$
Setting these equal:
$$
\frac{2.303}{t_1} \log\left(\frac{0.9}{0.225}\right) = 2.303 \log(4).
$$
The factor $2.303$ cancels on both sides:
$$
\frac{\log\left(\frac{0.9}{0.225}\right)}{t_1} = \log(4).
$$
Step 5: Solve for $t_1$
Notice $\frac{0.9}{0.225} = 4,$ so:
$$
\log\left(\frac{0.9}{0.225}\right) = \log(4).
$$
Hence,
$$
\frac{\log(4)}{t_1} = \log(4).
$$
This gives
$$
t_1 = 1 \text{ hour}.
$$
Answer
The time taken for 0.9 mole of A to produce 0.675 mole of B is 1 hour.
