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In the Arrhenius $K = A{\exp ^{( - {E_a}/RT)}}$. A may be termed as rate constant at infinite temperature. If true enter 1, if false enter 0.

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Last updated date: 20th Jun 2024
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Answer
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Hint: In the Arrhenius equation, $K = A{\exp ^{( - {E_a}/RT)}}$, you must know each term like A is the Arrhenius constant, K is known as the rate constant, ${E_a}$ is the activation energy, R is the gas constant, T is the temperature in Kelvin. Put the value of temperature infinity, and then solve the Arrhenius equation.

Complete step by step solution:
In 1899, Swedish chemist combined the concepts of activation energy and the Boltzmann distribution law into one of the most common important relationships which are known as the Arrhenius equation:
$K = A{\exp ^{( - {E_a}/RT)}}$
Here, in the Arrhenius equation, each term has its specific meaning.
A is known as the Arrhenius constant or pre-exponential factor.
K is the rate constant of the reaction.
${E_a}$ is the activation energy required for a reaction.
R is the gas constant and T represents the temperature in Kelvin.
Now, let us put the value of temperature equals to infinity in the Arrhenius equation. The equation will be then as follows:
\[K = A{\exp ^{( - {E_a}/R(\infty ))}} = A{\exp ^{( - {E_a}/\infty )}} = A{\exp ^{(0)}}\]
When temperature is infinite, the value of the term ($ {- {E_a}/RT}$) becomes zero.
 And, \[\exp ^{0} = 1\]
Thus,
\[K = A{\exp ^{(0)}} = A\]
Hence, the value of the Arrhenius constant (A) becomes equal to the rate constant. Thus, we can say A may be termed as the rate constant at infinite temperature.

Hence, the given statement in the question is true.

Note: Arrhenius equation can also be written in a non-exponential form and this form is more convenient to use and interpret. Taking the natural log on both sides and separating the exponential and Arrhenius factor, the Arrhenius equation is:
$K = A{\exp ^{( - {E_a}/RT)}}$
$\begin{align}
& \ln K = \ln (A{\exp ^{( - {E_a}/RT)}}) \\
& \ln K = \ln A + \ln {\exp ^{( - {E_a}/RT)}} \\
& \ln K = \ln A + \dfrac{{ - {E_a}}}{{RT}} \\
& \ln K = \ln A - \dfrac{{{E_a}}}{{RT}} \\
\end{align} $