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Question: In the Arrhenius \(K = A{\exp ^{( - {E_a}/RT)}}\). A may be termed as rate constant at infinite temp...

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

Explanation

Solution

In the Arrhenius equation, K=Aexp⁡(−Ea/RT)K = A{\exp ^{( - {E_a}/RT)}}, you must know each term like A is the Arrhenius constant, K is known as the rate constant, Ea{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=Aexp⁡(−Ea/RT)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.
Ea{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=Aexp⁡(−Ea/R(∞))=Aexp⁡(−Ea/∞)=Aexp⁡(0)K = A{\exp ^{( - {E_a}/R(\infty ))}} = A{\exp ^{( - {E_a}/\infty )}} = A{\exp ^{(0)}}
When temperature is infinite, the value of the term (−Ea/RT {- {E_a}/RT}) becomes zero.
And, exp⁡0=1\exp ^{0} = 1
Thus,
K=Aexp⁡(0)=AK = 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=Aexp⁡(−Ea/RT)K = A{\exp ^{( - {E_a}/RT)}}
ln⁡K=ln⁡(Aexp⁡(−Ea/RT)) ln⁡K=ln⁡A+ln⁡exp⁡(−Ea/RT) ln⁡K=ln⁡A+−EaRT ln⁡K=ln⁡A−EaRT \begin{aligned} & \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{aligned}