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Question
AnswersFor two reactions, activation energies are ${E_{{a_1}}}$ and ${E_{{a_2}}}$ ; rate constants are ${k_1}$ and ${k_2}$ at the same temperature. If ${k_1}$> ${k_2}$ then
(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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Hint:We need to know the relationship between activation energy and rate constant of a reaction. It is obvious that chemical reactions speed up when the temperature is increased. In other words, if the temperature changes, the rate constant also changes. The most important parameter for this change is the Activation energy ( \[{E_a}\] ).It is that energy which needs to be overcome before the reaction starts taking place. It is believed when the temperature is increased, reactions speed up by lowering the activation energy, i.e. lower the activation energy, faster the rate of the reaction. The same theory applies when the temperature is decreased ( \[{E_a}\] increases).This can be mathematically shown with the help of Arrhenius Equation as:
$k = A{e^{^{ - \dfrac{{{E_a}}}{{RT}}}}}$
$k$ 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.
Complete step by step answer:
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 ${E_{{a_1}}}$ < ${E_{{a_2}}}$ .
Therefore the correct option is option (C).
Note:
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.
$k = A{e^{^{ - \dfrac{{{E_a}}}{{RT}}}}}$
$k$ 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.
Complete step by step answer:
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 ${E_{{a_1}}}$ < ${E_{{a_2}}}$ .
Therefore the correct option is option (C).
Note:
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.