
Rate constant for a chemical reaction taking place at \[{\text{500K}}\] is expressed as \[{\text{k}} = {\text{A}}{{\text{e}}^{ - 1000}}\]. The activation energy of the reaction is:
A) $100{\text{kcal/mol}}$
B) $1000{\text{kcal/mol}}$
C) ${\text{1}}{{\text{0}}^{\text{4}}}{\text{kcal/mol}}$
D) ${\text{1}}{{\text{0}}^6}{\text{kcal/mol}}$
Answer
536.1k+ views
Hint:
To answer this question, you should recall the concept of the Arrhenius equation. We shall substitute the value of temperature and compare it with the given equation in the question.
The formula used:
The Arrhenius equation
\[{\text{k}} = {\text{A}}{{\text{e}}^{ - {{\text{E}}_{\text{a}}}{\text{/RT}}}}\]
where ${\text{A}}$: The frequency or pre-exponential factor, ${\text{T}}$: temperature, \[{{\text{E}}_{\text{a}}}\] : Activation energy and ${\text{R}}$: Universal gas constant $\left( {{\text{R}} \approx 2 \times {{10}^{ - 3}}{\text{ cal/mol}}} \right)$
Complete step by step solution:
We know that \[{{\text{e}}^{ - {{\text{E}}_{\text{a}}}{\text{/RT}}}}\] in Arrhenius equation is the fraction of collisions that have enough energy to react i.e., have energy greater than or equal to the activation energy \[{{\text{E}}_{\text{a}}}\] . This equation is used to study the dependence of a reaction on temperature.
The formula for the Arrhenius equation is given as follows:
\[{\text{k}} = {\text{A}}{{\text{e}}^{ - {{\text{E}}_{\text{a}}}{\text{/RT}}}}\]
But in the question, it is mentioned that:
\[{\text{k}} = {\text{A}}{{\text{e}}^{ - 1000}}\]
Now, comparing these two equations we can conclude that
\[\dfrac{{{{\text{E}}_{\text{a}}}}}{{{\text{RT}}}} = 1000\].
Substituting and solving the value of universal gas constant and temperature:
\[ \Rightarrow \dfrac{{{{\text{E}}_{\text{a}}}}}{{2 \times {{10}^{ - 3}} \times 500}} = 1000\].
$\therefore $ The value of activation energy will be${{\text{E}}_{\text{a}}} = 1000{\text{cal/mol}}$.
Hence, the correct answer to this question is option B.
Note:
The Arrhenius equation is used to study reaction rates and also to model the temperature variance of permeation, diffusion and solubility coefficients, and other chemical processes over moderate temperature ranges. You should know the importance of the Arrhenius equation. ${\text{RT}}$ is the average kinetic energy, and the exponent is just the ratio of the activation energy \[{{\text{E}}_{\text{a}}}\] to the average kinetic energy. Larger this ratio, the smaller is the rate constant. It can be concluded that high temperature and low activation energy favour larger rate constants, and thus speed up the reaction.
To answer this question, you should recall the concept of the Arrhenius equation. We shall substitute the value of temperature and compare it with the given equation in the question.
The formula used:
The Arrhenius equation
\[{\text{k}} = {\text{A}}{{\text{e}}^{ - {{\text{E}}_{\text{a}}}{\text{/RT}}}}\]
where ${\text{A}}$: The frequency or pre-exponential factor, ${\text{T}}$: temperature, \[{{\text{E}}_{\text{a}}}\] : Activation energy and ${\text{R}}$: Universal gas constant $\left( {{\text{R}} \approx 2 \times {{10}^{ - 3}}{\text{ cal/mol}}} \right)$
Complete step by step solution:
We know that \[{{\text{e}}^{ - {{\text{E}}_{\text{a}}}{\text{/RT}}}}\] in Arrhenius equation is the fraction of collisions that have enough energy to react i.e., have energy greater than or equal to the activation energy \[{{\text{E}}_{\text{a}}}\] . This equation is used to study the dependence of a reaction on temperature.
The formula for the Arrhenius equation is given as follows:
\[{\text{k}} = {\text{A}}{{\text{e}}^{ - {{\text{E}}_{\text{a}}}{\text{/RT}}}}\]
But in the question, it is mentioned that:
\[{\text{k}} = {\text{A}}{{\text{e}}^{ - 1000}}\]
Now, comparing these two equations we can conclude that
\[\dfrac{{{{\text{E}}_{\text{a}}}}}{{{\text{RT}}}} = 1000\].
Substituting and solving the value of universal gas constant and temperature:
\[ \Rightarrow \dfrac{{{{\text{E}}_{\text{a}}}}}{{2 \times {{10}^{ - 3}} \times 500}} = 1000\].
$\therefore $ The value of activation energy will be${{\text{E}}_{\text{a}}} = 1000{\text{cal/mol}}$.
Hence, the correct answer to this question is option B.
Note:
The Arrhenius equation is used to study reaction rates and also to model the temperature variance of permeation, diffusion and solubility coefficients, and other chemical processes over moderate temperature ranges. You should know the importance of the Arrhenius equation. ${\text{RT}}$ is the average kinetic energy, and the exponent is just the ratio of the activation energy \[{{\text{E}}_{\text{a}}}\] to the average kinetic energy. Larger this ratio, the smaller is the rate constant. It can be concluded that high temperature and low activation energy favour larger rate constants, and thus speed up the reaction.
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