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As The Graph Shows; $E_a$ (Activation Energy) = Energy Of transition State(Threshold Energy) $-$ Energy of Reactants.

So let This be the graph at Temperature $T_1$, Now say We Increase The Temperature to say $T_2$ and so the energy of reactants must increase (as kinetic energy of molecules increases). So at $T_2$, reactants would start with more energy. So Shouldn't the Activation Energy decrease as now less energy is needed to reach Transition State.

Or I can say that since $E_a$ (Activation Energy) = Energy Of transition State(Threshold Energy) $-$ Energy of Reactants, So as Energy of Reactants has changed so should Activation Energy.

I already saw answers on a similar question but couldn't figure out what's wrong with my approach.

source - https://www.saburchill.com/IBbiology/chapters01/images/011204001.jpg

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    $\begingroup$ Say you have to travel to some place. If you drive faster, you'll get there sooner, but that doesn't mean that the distance got shorter. Same thing here. $\endgroup$ – Ivan Neretin Aug 21 at 17:27
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    $\begingroup$ The problem is that the activation energy is not derived the way you think. It is obtained by measuring a reaction rate constant at different temperatures. Afterwards, when reporting the logarithm of this rate constant versus the inverse of the absolute temperature, you obtain a line. And the slope of the line is equal to the energy of activation divided by the gas constant R. $\endgroup$ – Maurice Aug 21 at 17:29
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You are on the right track. The problem is with the diagram you show, which is just a cartoon, and describes something that is actually not the activation energy, although it is generally interpreted as this and as such is a good 'rule of thumb'.

The activation energy is experimentally derived as $\displaystyle E_A=-R\frac{\ln(k(T)}{d(1/T)}=\frac{RT^2}{k(T)}\frac{d k(T)}{dT}$.

where $k(T)$ is the rate constant at temperature $T$. However, this mathematical method imparts little or no physical information about what the activation energy is.

It is not just the height of the barrier above the bottom of the potential well, but has to reflect the fact that, in any experiment, the molecules are at a given temperature and so have a Boltzmann distribution of energies and conform to a Maxwell–Boltzmann distribution of collision speeds. Tolman interpreted this as shown in the figure.

Tolman

Figure 1. Tolman interpretation of the activation energy. Shown are the rate constant with energy $k(E)\equiv \sigma P(E)$, the Boltzmann factor $\exp(-E / k_BT)$ and the product $f(E)=\sigma P(E)\exp(-E/k_BT)$ which is the probability of having energy $E$ at the transition state. Note that $E_T\ne E_0$ where $E_0$ is the threshold energy for reaction. ($f(E)$ is multiplied by a large factor to make it visible on the graph).


In the Tolman interpretation, the activation energy is the average total energy $\langle E^*\rangle$ (relative translational plus internal) of pairs of species that have reacted, minus the average energy of all $\langle E_R\rangle$ pairs of species whether or not they react, or $\langle E_T\rangle =\langle E^*\rangle -\langle E_R\rangle $.

Direct calculation via the Equipartition theorem gives the average energy of all pairs as $\langle E_R\rangle =3RT/2$ but the average of the reactive pairs is more complicated and depends on knowing how the rate constant $k(E)\equiv \sigma P(E)$ changes with energy and then $\langle E^*\rangle$ can be calculated.

The essential point in this idea is that the activation energy is the difference in the average energy in the reactants and the average energy in the transition state which is the average of $f(E)$ as shown in the figure.

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  • $\begingroup$ So finally does it depend on temprature or not?...Thanks for the answer. $\endgroup$ – Yashasv Prajapati Aug 22 at 14:34
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    $\begingroup$ why yes, both $\langle E^*\rangle$ and $\langle E_R\rangle$ depend on temperature. The former because there is a Boltzmann distribution in the transition state and the latter is 3RT/2 as mentioned in the text. $\endgroup$ – porphyrin Aug 22 at 15:41
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You have conceptual misunderstanding. In the activation energy context, energy of reactants, products and transition state are all potential energies, independent on the kinetic energy of molecules.

The kinetic energy of molecules is used to reach the peak of the potential energy, similarly as e.g. a acrobatic aircraft uses its kinetic energy to climb to the apex of an acrobatic loop.

Higher temperature means higher kinetic energy, but it does not mean the potential energy barrier is lower. It means there is higher probability to overcome it, leading to a faster reaction.

There is a famour Arrhenius' equation for the reaction rate:

$$k = A \exp{\left(\frac{-E_\mathrm{A}}{kT}\right)}$$

where $A$ is the frequency factor, describing the rate of molecular collisions, potentially able to take part of the reaction and $\exp{\left(\frac{-E_\mathrm{A}}{kT}\right)}$ is the term from the Boltzmann distribution, telling us the probability a molecule at temperature $T$ has sufficient energy, if the activation energy is $E_\mathrm{A}$, and $k$ is the Boltzmann constant.

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  • $\begingroup$ I completely understood your answer and I got why I was wrong, but then I went on to thinking that Due to the vibrational degree of freedom Wouldn't the vibrational energy also increase with with temperature, hence leading to a higher potential energy, so the at higher temperature reactants would start with higher potential energy. And thanks a lot for your help. $\endgroup$ – Yashasv Prajapati Aug 22 at 5:31
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    $\begingroup$ Well, I have been simplifying a little. The activation energy is potential energy of another kind, compared to the potential energy of band oscillators. So more properly would be the thermal energy, not just kinetic energy takes part in overcoming the potential wall of the activation energy. $\endgroup$ – Poutnik Aug 22 at 5:49
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    $\begingroup$ By other words, if a car vibrates, the hill is not smaller . $\endgroup$ – Poutnik Aug 22 at 5:51
  • $\begingroup$ Sir please take a look at this source especially the last equation socratic.org/questions/… . This seemingly contradicts your explanation. $\endgroup$ – Tony Stark Aug 31 at 10:55
  • $\begingroup$ I have noticed they incorrently use the total translational kinetic energy, following the Maxwell-Boltzman distribution, while only the KE of 1 degree of freedom ( motion along the line of collision ) is to be considered, following the Boltzmann distribution. the rest may be matter of the terminology confusion. $\endgroup$ – Poutnik Aug 31 at 12:57

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