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For the reaction 2NO(2)+F(2) rarr 2NO(2)...

For the reaction `2NO_(2)+F_(2) rarr 2NO_(2)F`, the experimental rate law is `r=K[NO_(2)][F_(2)]`. Propose the mechanism of reaction.

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For the reaction : 2NO_(2)+F_(2) to 2NO_(2)F , the experimental rate law is given as : Rate =K[NO_(2)][F_(2)] , propose the mechanism.

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Knowledge Check

  • For the reaction NO_(2)+CO rarr CO_(2)+NO the experimental rate expression is (dc)/(dt)=k[NO_(2)]^(2) the number of molecules of CO involved in the slowest step will be:

    A
    0
    B
    1
    C
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    D
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  • For the reaction H_(2)(g) +Br_(2)(g)rarr 2HBr(g) the experimental data suggestion that r=k[H_(2)][Br_(2)]^(1//2). The molecularity and order of the reaction are respectively:

    A
    `2,(3)/(2)`
    B
    `(3)/(2),(3)/(2)`
    C
    not defined , `(3)/(2)`
    D
    `1,(1)/(2)`

  • The forward reaction rate for the nitric oxide-oxygen reaction 2NO+O_(2) rarr 2NO_(2) has the rate law as: Rate = k[NO]^(2)[O_(2)] . If the mechanism is assumed to be: 2NO+O overset(k_(eq))hArr , (rapid equilibration) N_(2)O_(2) + O_(2) overset(k_(2))rarr 2NO_(2) (slow step), then which of the following is (are) correct? (I) Rate constant = k_(eq)k_(2) , (II) [N_(2)O_(2)] = k_(eq)[NO]^(2) (III) [N_(2)O_(2)] = k_(eq)[NO] , (IV) Rate constant = k_(2) The correct option is

    A
    I, II
    B
    III, IV
    C
    I, III
    D
    None of these

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