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Chemical Kinetics

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Question:

At 45^oC, the decomposition of hydrogen iodide into hydrogen and iodine is a second-order reaction. The rate constant for the reaction at 45^oC is 0.130 L*mol^-1*s^-1. How long does it take an initial concentration of 0.060 mol/L to reduce to one-quarter the initial concentration? Enter your answer as a time in seconds with one decimal place.

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Answers:

nglove

Novice

Karma Points: 31

Date Posted: 2/2/2008 9:31:01 PM Status: Live

Asker's Rating: Lifesaver

Response:

t_1/2 = 1/k[A]₀=1/1.130L/mol.s *0.060mol/L

= 0.8349/s

t_1/4= 2[1/k[A]₀]

=2[0.8349/s]

=1.6698/s

˜1.67/s

iLoVeScIeNcE's Comment:

Thank you for the explanation, but I don't understand how you got 1.130 as the rate constant when it is 0.130. Also, something is wrong in the calculation. I am getting 128 s and if you multiply that by 2, you get 256 s, but it is not the right answer.

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Chemical Kinetics

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