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a certain first order reaction has a rate constant of 8.50x10^-3s^-1 how long will it take for the reaction concentration to drop to 1/8 of its initial value

User Terris
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Final answer:

To find out how long it takes for the concentration to reduce to 1/8 in a first-order reaction with a rate constant of 8.50x10^-3s^-1, apply the first-order kinetics formula and solve for time, yielding approximately 319.72 seconds.

Step-by-step explanation:

The question asks how long it will take for the concentration of a reactant in a first-order reaction with a rate constant of 8.50x10-3s-1 to drop to 1/8 of its initial value. To solve this, one can use the first-order reaction equation:

ln([A]_t/[A]_0) = -kt


Where [A]_t is the concentration at time t, [A]_0 is the initial concentration, k is the rate constant, and t is the time. For the concentration to drop to 1/8 of its initial value, we set [A]_t/[A]_0 to 1/8 and solve for t:

ln(1/8) = -8.50x10-3s-1 x t
t ≈ 319.72 s

User Zach Inglis
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