Answer :
The corresponding half-life for the reaction is 1.7 s. Option A
It can be determined using the first-order rate constant formula:
t₁/₂ = ln(2) / k
Where t₁/₂ is the half-life, ln(2) is the natural logarithm of 2, and k is the rate constant.
Given that the rate constant k is 25 min⁻¹, we can plug this value into the formula and solve for the half-life:
t₁/₂ = ln(2) / 25 min⁻¹
t₁/₂ = 0.028 min
Converting minutes to seconds:
t₁/₂ = 0.028 min × (60 s / 1 min)
t₁/₂ = 1.68 s
Therefore, the corresponding half-life for the reaction is 1.68 s, or approximately 1.7 s.
The correct answer is option A) 1.7 s.
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The half-life of a first-order chemical reaction with a rate constant of 25 min⁻¹ is given by the equation t1/2 = 0.693 / k, which yields approximately 1.7 minutes.
The question resembles a classic problem in chemical kinetics, which is the study of reaction rates and their mechanisms. Specifically, we're asked to determine the half-life of a first-order chemical reaction with a given rate constant. The half-life (t1/2) of a first-order reaction is related to the rate constant
(k) through the equation: t1/2 = 0.693 / k. With a rate constant (k) of 25 min⁻¹, the half-life can be calculated as follows:
t1/2 = 0.693 / 25 min⁻¹ = 0.02772 min or approximately 1.7 min.
Therefore, the correct answer is option b: 1.7 min.
