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Question: A reactant (1) forms two products : ![](https://cdn.pureessence.tech/canvas_237.png?top_left_x=217&...

A reactant (1) forms two products :

Ea2E_{a_{2}}= 2Ea1E_{a_{1}}

Frequency factors for both the reaction are equal.

Therefore-

A

k₂ = k₁ e^{- Eₐ₁/RT}

B

k₂ = k₁²/A

C

k₁ = A k₂ e^{Eₐ₁/RT}

D

Both (1)&(2) are correct

Answer

Both (1)&(2) are correct

Explanation

Solution

k2 = A e−Ea2/RTe^{- E_{a2}/RT} ; k1 = A e−Ea1/RTe^{- E_{a_{1}}/RT}

ln k2 = ln A – Ea2RT\frac{Ea_{2}}{RT} & ln k1 = ln A –Ea1RT\frac{Ea_{1}}{RT}

or Ea2= RTln A/k2 & Ea1E_{a_{1}} = RT ln Ak1\frac{A}{k_{1}}

\ 2RT ln Ak1\frac{A}{k_{1}} = RT lnAk2\frac{A}{k_{2}}

\ ln (Ak1)2\left( \frac{A}{k_{1}} \right)^{2} = ln Ak2\frac{A}{k_{2}}

\ (Ak1)2\left( \frac{A}{k_{1}} \right)^{2} = Ak2\frac{A}{k_{2}}

or A2k12\frac{A^{2}}{k_{1}^{2}} = Ak2\frac{A}{k_{2}}

\ k2 = k12A\frac{k_{1}^{2}}{A}

And k2 = Ae−Ea2/RTe^{- E_{a2}/RT}

k1 = Ae−Ea1/RTe^{- E_{a_{1}}/RT}

k2k1\frac{k_{2}}{k_{1}} = eEa1−Ea2/RTe^{E_{a_{1}} - E_{a_{2}}/RT}= e−Ea1/RTe^{- E_{a_{1}}/RT}

or k2 = k1 e−Ea1/RTe^{- E_{a_{1}}/RT}