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For CH(3)Br+OHrarrCH(3)OH+Br the rate ...

For `CH_(3)Br+OHrarrCH_(3)OH+Br`
the rate of reaction is given by the expression .

A

`rate =k[CH_(3)Br]^(0)`

B

`rate=k[OH-]

C

rate=k[CH_(3)Br]`

D

`rate=k[CH_(3)Br]^(0)[OH]^(0)`

Text Solution

Verified by Experts

The correct Answer is:
C
Rate of reaction` [r] =K [CH_(3) Br][OH]` .
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Knowledge Check

  • Consider the following reaction, CH_(3)Br+OH^(-)rarrCH_(3)OH+Br^(-) The true statement about the above process is

    A
    the rate does not change on doubling `[CH_(3)Br]` and making `[OH^(-)]` to half
    B
    the rate becomes half on doubling `[CH_(3)Br]`
    C
    the rate becomes half on doubling `[OH^(-)]`
    D
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  • In acidic medium the rate of reaction between (BrO_3^-) and Br^(-) ion is given by the expression (d[BrO_3^-])/(dt)=k[BrO_3^(-)] [Br^(-)] [H^+]^2 . The expression signifies that:

    A
    Rate constant of overall reaction is 4 `sec^–1`
    B
    Rate of reaction is independent of the concentration of acid
    C
    The change in pH of the solution will not affect the rate
    D
    Doubling the concentration of `H^+` ions will increase the reaction rate by 4 times

  • In acidic medium, the rate of reaction between BrO_(3)^(-) and Br^(-) is given by the expression -(d[BrO_(3)^(-)])/(dt)=k[BrO_(3)^(-)][Br^(-)][H^(+)]^(2) It means

    A
    Rate constant of the reaction depends upon the concentration of `H^(+)` ions.
    B
    Rate of reaction is independent of the concentration of the acid added.
    C
    The change in pH of the solution will affect the rate of reaction.
    D
    Doubling the concentration of `H^(+)` ions will increase the reaction rate by 4 times.
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