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The rate constant for this first-order reaction is 0.0720 s-1 at 400 °C.

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After how many seconds will 15.9% of the reactant remain?
t=___________S

User Buru
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1 Answer

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Final answer:

To find the time for 15.9% of a reactant to remain in a first-order reaction with a rate constant of 0.0720 s-1, we use the first-order kinetics integrated rate law. The calculation yields a time of approximately 24.671 seconds.

Step-by-step explanation:

The question relates to the first-order reaction and the determination of the time required for a certain percentage of reactant to remain. To calculate the time (t) when 15.9% of the reactant remains, we can use the first-order kinetics integrated rate law:



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



Where [A]_t is the concentration of reactant at time t, [A]_0 is the initial concentration, k is the rate constant, and t is the time.



Since 15.9% remains, 84.1% has reacted, and thus [A]_t/[A]_0 = 0.159. We can substitute the values into the equation and solve for t:



ln(0.159) = -0.0720 s-1 • t



t = ln(0.159) / (-0.0720 s-1)



t ≈ 24.671 seconds



The time after which 15.9% of the reactant will remain is approximately 24.671 seconds.

User Jichi
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