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Arrhenius Equation

The rate cons...

Question

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

E_{1} > E_{2}

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E_{1} < E_{2}

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E_{1} = E_{2}

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E_{1} = 4 E_{2}

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Solution

The correct option is B $E_{1} < E_{2}$
The expression for the Arrhenius equation is $k = A e^{- E a / R T}$ .
Thus, when the activation energy increases, the rate constant of the reaction decreases.
The rate constant $(K_{1})$ of one reaction is found to be double that of the rate constant of $(K_{2})$ of another reaction.
Hence, its activation energy is less than that of the second reaction $E_{1} < E_{2}$ .

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Consider figure and mark the correct option.

(a) Activation energy of forward reaction is $E_{1}$ + $E_{2}$ and product is less stable than reactant

(b) Activation energy of forward reaction is $E_{1}$ + $E_{2}$ and product is more stable than reactant

(c) Activation energy of both forward and backward reaction is $E_{1}$ + $E_{2}$ and reactant is more stable than product

(d) Activation energy of backward reaction is $E_{1}$ and product is more stable than reactant

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