For a first-order reaction, the time required for a specific fractional conversion (or percentage decomposition) is independent of the initial concentration .

In the first case: Initial amount of A, $[A]_0 = 0.8 \text{ mol}$ Amount of A reacted to form B = $0.6 \text{ mol}$ Amount of A remaining, $[A]_t = 0.8 - 0.6 = 0.2 \text{ mol}$ Fraction of A remaining = $\frac{0.2}{0.8} = \frac{1}{4}$ Time taken = 1 hour.

In the second case: Initial amount of A, $[A]'_0 = 0.9 \text{ mol}$ Amount of A reacted to form B = $0.675 \text{ mol}$ Amount of A remaining, $[A]'_t = 0.9 - 0.675 = 0.225 \text{ mol}$ Fraction of A remaining = $\frac{0.225}{0.9} = \frac{1}{4}$

Since the fraction of reactant remaining ($\frac{1}{4}$) is the same in both cases, the extent of the reaction is identical (both require exactly 2 half-lives to complete). Therefore, the time taken will be exactly the same, i.e., $1 \text{ hour}$.