The rate constant $K_{1}$ of a reaction is found to be double that of rate constant $K_{2}$ of another reaction. The relationship between corresponding activation energies of the two reactions $E_{1}$ and $E_{2}$ can be represented as

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$E_{1} = E_{2}$

$E_{1} = 2 E_{2}$

$E_{1} > E_{2}$

$E_{1} < E_{2}$

answer is B.

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Detailed Solution

As we are aware, the Arrhenius equation is expressed as the equation:

$k = {Ae}^{\frac{- E_{a}}{RT}}$

It is discovered that the reaction's rate constant drops as the activation energy rises. It is discovered that the rate constant of the first reaction is two times greater than the rate constant of the second reaction. As a result, the first reaction's activation energy is lower than the second reaction's, which is represented by the expression:

$E_{1} < E_{2}$

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