Temperature at which net activation energy for dissociation of X will be 10.5 KC | TestHub

Passage / Comprehension

A substance ' X ' dissociates by two parallel reactions with different ' Arrhenius constants $(A_{1}$ and $A_{2})$ " and different "Activation energies" ( $E_{a_{1}}$ and $E_{a_{2}}$ ) as shown.

$X (g) \frac{A _{1} = 2 \times 1 0 ^{14} s ^{- 1}}{E _{a_{1}} = 12 K cal / mol} Y (g); X (g) \frac{A _{2} = 2 \times 1 0 ^{14} \times e ^{- 2} s ^{- 1}}{E _{a_{2}} = 9 K cal / mol} Z (g)$

[Given : $R = 2 Cal / mol - K$ ]

Temperature at which net activation energy for dissociation of X will be 10.5 KCal/mol?

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Answer:

Solution:

For parallel reactions,

$E_{n e t} = \frac{E _{a 1} k _{1} + E _{a 2} k _{2}}{k _{1} + k _{2}}$

Given:

$E_{a 1} = 12$ , $E_{a 2} = 9$ , $E_{n e t} = 10.5 k c a l m o l^{- 1}$

Substituting,

$10.5 = \frac{12 k _{1} + 9 k _{2}}{k _{1} + k _{2}} \Rightarrow k_{1} = k_{2}$

Using the Arrhenius equation,

$A_{1} e^{- E_{a 1} / RT} = A_{2} e^{- E_{a 2} / RT}$

$2 \times 1 0^{14} e^{- 12000/2 T} = 2 \times 1 0^{14} e^{- 2} e^{- 9000/2 T}$

$\frac{3000}{2 T} = 2$

$T = 750 K$

Chemistry - Chemical Kinetics Question with Solution | TestHub

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