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21 February, 02:33

He second-order decomposition of hi has a rate constant of 1.80 x 10-3 m-1s-1. how much hi remains after 27.3 s if the initial concentration of hi is 4.78 m?

a. 4.55 m

b. 0.258 m

c. 3.87 m

d. 2.20 m

e. 2.39 m

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Answers (1)
  1. 21 February, 02:51
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    c. 3.87 M. After 27.3 s, the concentration of the remaining HI is 3.87 M

    Whenever a question asks you, "How long to reach a certain concentration?", you must use the appropriate integrated rate law expression.

    The integrated rate law for a second-order reaction is

    1/[A]t = 1/[A]0 + kt

    where

    • [A]t = the concentration of A at time t

    • [A]0 = the concentration of A at time 0 (i. e., at the beginning of the reaction)

    • k = the rate constant for the reaction

    1/[A]t = 1 / (4.78 M) + 1.80 * 10^ (-3) M·s^ (-1) x 27.3 s

    = 0.2092 M^ (-1) + 0.04914 M^ (-1) = 0.2583 M^ (-1)

    [A]t = 1/[0.2583 M^ (-1) ] = 3.87 M
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