t_1/4 is the time taken for the concentration of a reactant to drop to 3/4 of its initial value. If the rate constant is K , t_1/4 is: - Chemistry MCQ AIEEE 2026 | ExamTest.live

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t_{1/4} is the time taken for the concentration of a reactant to drop to 3/4 of its initial value. If the rate constant is K, t_{1/4} is:

Solution & Step-by-step Explanation

For a first order reaction: t = \frac{2.303}{K} lo g \frac{[ A ] _{0}}{[ A ]} Given [A] = \frac{3}{4} [A]_{0} : t_{1/4} = \frac{2.303}{K} lo g (4/3) = \frac{2.303}{K} \times (0.602 - 0.477) t_{1/4} = \frac{2.303 \times 0.125}{K} \approx \frac{0.29}{K} .

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t_{1/4} is the time taken for the concentration of a reactant to drop to 3/4 of its initial value. If the rate constant is K, t_{1/4} is:

0.10/ K

0.29/ K

0.69/ K

0.75/ K

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