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(A) ${E_{{a_1}}}$>${E_{{a_2}}}$

(B) ${E_{{a_1}}}$=${E_{{a_2}}}$

(C) ${E_{{a_1}}}$<${E_{{a_2}}}$

(D) ${E_{{a_1}}}$≥ ${E_{{a_2}}}$

Answer

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$k = A{e^{^{ - \dfrac{{{E_a}}}{{RT}}}}}$

$k$ is the rate constant

$T$ is the temperature in Kelvin

$R$ is the Gas Constant=\[8.314\]$J{K^{ - 1}}mo{l^{ - 1}}$

\[{E_a}\] is the activation energy expressed in \[J/mol\]

$e$ is a mathematical quantity

$A$ is called the pre-exponential factor. $A$ in the Arrhenius equation, has been ignored because it is not directly involved in relating temperature and activation energy, which is the main practical use of the equation.

It is given in the question that for two reactions, activation energies are ${E_{{a_1}}}$ and ${E_{{a_2}}}$.Also, rate constants are ${k_1}$ and ${k_2}$ at the same temperature. If ${k_1}$ > ${k_2}$ then from Arrhenius equation we have

${k_1} = A{e^{^{ - \dfrac{{{E_{{a_1}}}}}{{RT}}}}}$ and ${k_2} = A{e^{^{ - \dfrac{{{E_{{a_2}}}}}{{RT}}}}}$

Hence it is clear from the Arrhenius equation that rate of a reaction is inversely proportional to its activation energy. Therefore if ${k_1}$ > ${k_2}$ , then ${E_{{a_1}}}$ < ${E_{{a_2}}}$ .

We have to remember that the Arrhenius equation allows us to calculate activation energies if the rate constant is known, or vice versa. As well, it mathematically expresses the relationships we established earlier: as activation energy term Ea increases, the rate constant k decreases and therefore the rate of the reaction decreases. The activation energy can also be calculated algebraically if k is known at two different temperatures.