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The activation energy for a reaction is 9.0kcal/mol. The increase in the rate constant when its temperature is increased from 298K to 308K is:
Option
(A) 63 %
(B) 50 %
(C) 100 %
(D) 10 %

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Last updated date: 03rd Mar 2024
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IVSAT 2024
Answer
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Hint :
In chemistry, reaction rate refers to the rate at which a chemical reaction occurs. It is sometimes expressed in terms of the concentration (amount per unit volume) of a substance produced in a unit of time or the concentration (amount per unit volume) of a reactant absorbed in a unit of time.

Complete Step By Step Answer:
In chemistry and physics, activation energy is the minimal amount of energy required to cause a chemical reaction to occur. In joules per mole (J/mol), kilojoules per mole (kJ/mol), or kilocalories per mole (kcal/mol), the activation energy of a reaction is calculated. The amplitude of the potential barrier (also known as the energy barrier) dividing minima of the potential energy surface pertaining to the original and final thermodynamic states can be thought of as activation energy.
The Arrhenius equation establishes a quantitative foundation for the relationship between activation energy and reaction rate. The activation energy can be calculated using the equation and the relationship.
\[k = A{e^{ - {E_{\text{a}}}/(RT)}}\]
where A denotes the reaction's pre-exponential component, R denotes the universal gas constant, T denotes the absolute temperature (usually in kelvins), and k denotes the reaction rate coefficient. And if you don't know A, the difference in reaction rate coefficients as a function of temperature can be used to calculate \[{{\mathbf{E}}_{\text{a}}}\] (within the validity of the Arrhenius equation).
We know that
\[{\mathbf{2}}.{\mathbf{303}}\log \frac{{{{\mathbf{K}}_2}}}{{{{\mathbf{K}}_1}}} = \frac{{{{\mathbf{E}}_{\text{a}}}}}{{\mathbf{R}}}\left[ {\frac{{{{\mathbf{T}}_2} - {{\mathbf{T}}_1}}}{{{{\mathbf{T}}_1}{{\mathbf{T}}_2}}}} \right]\]
Using given values
\[{{\mathbf{E}}_{\text{a}}}\]= 9 kcal/mol = 9 x 10³ cal/mol
= 308 K
 = 298 K
\[\log \frac{{{{\mathbf{K}}_2}}}{{{{\mathbf{K}}_1}}} = \frac{{9.0 \times {{10}^3}}}{{{\mathbf{2}}.{\mathbf{303}} \times 2}}\left[ {\frac{{308 - {\mathbf{298}}}}{{308 \times {\mathbf{298}}}}} \right]\]
\[\frac{{{{\mathbf{K}}_2}}}{{{{\mathbf{K}}_1}}} = {\mathbf{1}}.{\mathbf{63}}\]
\[{{\mathbf{K}}_{\mathbf{2}}} = {\mathbf{1}}.{\mathbf{63}}{{\mathbf{K}}_{\mathbf{1}}}\]
\[\frac{{{\mathbf{1}}.{\mathbf{63}}{{\mathbf{K}}_1} - {{\mathbf{K}}_1}}}{{{{\mathbf{K}}_1}}} \times {\mathbf{100}} = {\mathbf{63}}.{\mathbf{0}}\% \]
Hence option A is correct.

Note :
The minimum amount of energy needed to induce a chemical reaction to occur is known as activation energy in chemistry and physics. The activation energy (Ea) of a reaction is measured in joules per mole (J/mol), kilojoules per mole (kJ/mol), or kilocalories per mole (kcal/mol). The amplitude of the potential barrier (also known as the energy barrier) that separates the initial and final thermodynamic states' potential energy surfaces can be thought of as activation energy.
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