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Nutan Garg
Asked: 2022-11-05T17:13:08+05:30 In:

Some times a reacant undergoes chemical/radioactive changes following two or more different paths to yield two or more different produces respectively. Such reactions are called parallel path reactions. If `K_(1)` and `K_(2)` are rate constans for the reaction of `A` follwing two parallel paths, then
image
Then `K_(av) = K_(1) + K_(2)`
For image if `E_(1)` and `E_(2)` are energy fo activations, then
A. `E_(“Total”) = E_(1) + E_(2)`
B. `E_(“Total”) = E_(1) – E_(2)`
C. `E_(“Total”) = K_(1) E_(1) + K_(2)E_(2)`
D. `E_(“Total”) = (K_(1) E_(1) + K_(2)E_(2))/(K_(1) + K_(2))`

Some times a reacant undergoes chemical/radioactive changes following two or more different paths to yield two or more different produces respectively. Such reactions are called parallel path reactions. If `K_(1)` and `K_(2)` are rate constans for the reaction of `A` follwing two parallel paths, then
image
Then `K_(av) = K_(1) + K_(2)`
For image if `E_(1)` and `E_(2)` are energy fo activations, then
A. `E_(“Total”) = E_(1) + E_(2)`
B. `E_(“Total”) = E_(1) – E_(2)`
C. `E_(“Total”) = K_(1) E_(1) + K_(2)E_(2)`
D. `E_(“Total”) = (K_(1) E_(1) + K_(2)E_(2))/(K_(1) + K_(2))`
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1 Answer

  1. Correct Answer – `d`
    `K_(av) = K_(1) + K_(2)`
    `A_(T)e^(-E_(T)//RT^(2)) = A_(1) e^(-E_(1)//RT^(2)) + A_(2) e^(-E_(2)//RT^(2))` ….(1)
    Differentiating eqn. (1)
    `A_(T). (E_(T))/(RT).e^(-E_(T)//RT^(2)) = A_(1). (E_(1))/(RT).e^(-E_(1)//RT^(2)) + A_(2) (E_(2))/(RT) e^(-E_(2)//RT^(2))`
    `K_(av) (E_(T))/(RT) = (K_(1)E_(1))/(RT) + (K_(2) E_(2))/(RT)`
    `(K_(1) + K_(2)) (E_(T))/(RT) = (K_(1)E_(1))/(RT) + (K_(2) E_(2))/(RT)`
    `:. E_(T) = (K_(1) E_(1) + K_(2) E_(2))/(K_(1) + K_(2))`

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