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24 June, 17:35

Consider the single‑step, bimolecular reaction. CH3Br+NaOH⟶CH3OH+NaBr When the concentrations of CH3Br and NaOH are both 0.120 M, the rate of the reaction is 0.0080 M/s. What is the rate of the reaction if the concentration of CH3Br is doubled? rate: M/s What is the rate of the reaction if the concentration of NaOH is halved? rate: M/s What is the rate of the reaction if the concentrations of CH3Br and NaOH are both increased by a factor of 5? rate: M/s

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  1. 24 June, 17:52
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    CH₃Br+NaOH⟶CH₃OH+NaBr

    It is a single step bimolecular reaction so order of reaction is 2, one for CH₃Br and one for NaOH.

    rate of reaction = k x [CH₃Br] [ NaOH]

    .008 = k x. 12 x. 12

    k =.55555

    when concentration of CH₃Br is doubled

    rate of reaction =.555555 x [.24] [.12 ]

    =.016 M/s

    when concentration of NaOH is halved

    rate of reaction =.555555 x [.12] [.06 ]

    =.004 M/s

    when concentration of both CH₃Br and Na OH is made 5 times

    rate of reaction =.555555 x. 6 x. 6

    = 0.2 M/s
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