A rate is equal to 0.0200 m/s. if [a] = 0.100 m and rate = k[a]2[b]2, what is the new rate if the concentration of [a] is increased to 0.200 m?

Answers (1)
  1. Noelle
    The rate equation is r = k[a]2[b]2.

    Take logarithms of both sides.

    ln r = ln k + 2 ln [a] + 2 ln [b]

    When [a] = 0.0.100 m, r1 = 0.0200 m/s.

    Therefore

    ln (0.0200) = ln k + 2 ln (0.0100) + 2 ln [b] (1)

    When a = 0.200 m, obtain the new rate, r2, as

    ln r2 = ln k + 2 ln (0.0200) + 2 ln [b] (2)

    Subtract (1) from (2) to obtain

    ln r2 - ln (0.0200) = 2{ln (0.0200 - ln (0.0100) }

    ln (r2/0.0200) = 2 ln (0.0200/0.0100)

    = 2 ln (2)

    = ln (22) = ln (4)

    Therefore

    r2/0.0200 = 4

    r2 = 0.0800 m/s

    Answer: 0.0800 m/s
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