# Tag Info

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According to the principle of Microscopic Reversibility all elementary reactions, i.e. those that proceed in a single step, are reversible. And you are right: For a reversible reaction $\Delta G$ is very small while for an irreversible reaction $\Delta G$ is large. This can be most easily described using a simple schematic energy curve diagram for your ...

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At the transition state, you have exactly one vibration mode with imaginary frequency (which corresponds locally to the reaction coordinate), and $3N-1$ other “normal” vibration modes. In the transition state theory, you separate out the reaction coordinate (consider it infinitely slow compared to the other movements). Thus, when you consider the energy of ...

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As you have figured out, this is a non-trivial problem. It's okay to find a boat conformation that might be a bit higher in energy and see if it can map to a TS. Keep in mind that for 6-membered rings, there are many conformations beyond chair and boat. A half-chair, for example, might be a useful test. In general, for 6-membered rings, using ring-...

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The picture below shows a simple potential energy surface with a transition state at the maximum point on the reaction path, which is at the 'saddle point'. The reaction path is the solid black line. (Oscillations in this show that the reactant and product are vibrating). The contours are at different energy with large negative being lowest. This picture is ...

7

Your question is really about what you consider to be a reasonable rate constant and so is somewhat subjective. The fastest a bimolecular reaction can be in solution is given by how fast the reactants can collide and then react on first collision. i.e reaction is diffusion controlled. In this case as reaction is at first meeting the activation energy must be ...

6

I have looked at the original article of Henry Eyring (J. Chem. Phys. 1935, 3, 107). In that article, Eyring explains the derivation of his famous equation. It basically boils down to the following: You assume an equilibrium between the initial and transition state and consider that once a species crosses the barrier, it goes to the product state: $$R \... 5 There are many sources that you can check out, in particular the reviews of Truhlar: Current Status of Transition-State Theory by Truhlar et al. This article was used in my undergraduate course of chemical kinetics, so I think that it could be a good source. Also we used: Variational transition-state theory by Truhlar and Garrett Maybe they are a bit ... 5 There is nothing special in having an activation barrier between reactants and products although, obviously, very many reactions do. But some reactions, such as electron transfer, are observed experimentally to have no barrier which happens when the product potential crosses that of the reactant at its minimum (i.e. between the 'normal' and 'inverted' ... 4 Intrinsic reaction coordinate (IRC) is not some internal coordinate of a molecular system (such as, bond length, bond angle, etc.), rather it is a curvilinear coordinate that describes the intrinsic reaction path (IRP), which is the reaction path along the direction of the gradient. Only for some simple reactions one can visually (and approximately) identify ... 4 A transition state does not ‘represent the energies of […] reactants and […] products’. A transition state is a certain arrangement of atoms somewhere between the most stable atom arrangement the reactants can have and the most stable atom arrangement the products can have. The key characteristic of a transition state is that it is directly en route from ... 3 I copied your input into a new file, added some empty lines at the end and changed the input line to be "#p opt=qst3 hf/6-31..." instead of redundant (and maybe erroneous(?) "opt ... opt=qst3"). I dropped the memory and processor line, as I defined some defaults in the Default.Route file. Using the following input, the calculation runs smoothly towards a ... 3 I already tested the TS search calculation of this reaction based on your proposed structure coordinates. The calculation has been terminated quickly at link9999.exe sub-routine program in which its error message being implies that your initial geometry coordinate is not good enough for TS search by using QST3 method. When you deal with QST3 you have to ... 3 The statement is essentially correct. Unimolecular reactions refer to the Lindemann scheme and its improvements called RRKM theory. The Lindemann model is \ce{A + M <=> A^* +M ;\; A^*\rightarrow P} where \ce{A^*} is an energised molecule and P product. M is an inert gas or another A type molecule that gives up some of its energy in a collision to ... 3 I'm guessing this quote is given in the context of classical transition rate theory. Although the statement may seem very strong at first, it is generally true but has some important caveats. Energy Landscapes Lets consider a diatomic molecule. It's geometry can entirely be described by one degree of freedom - the bond length. As we move along that ... 3 You are wrong when you say the potential energy of particles decrease when they move apart. The forces between the particles are attractive not repulsive. Recall the definition of potential energy. It is the work done by external force on a charged particle under influence of electric field. Therefore, the potential energy for particles, which attract ... 3 A chemist would say that the reaction proceeds through an "aromatic transition state". The bridged, chloronium ion transition state drawn in your question involves 2 electrons. No electrons from the electrophile were used, just 2 electrons from the pi-bond that it added to (draw the resonance structures for the bridged intermediate, each has 2 electrons in ... 3 First you are correct, there is no fundamental difference in reactions being described as reversible or irreversible, unlike in thermodynamics. A reaction will be called irreversible (a)if the product is removed from reaction i.e. by precipitation or physical removal and (b) if the rate of returning from product to reactants is so slow that it cannot be ... 2 Energy itself is usually not measured, but rather change in energy (\Delta E) is the quantity of practical importance. One may be interested in change in free energy as well, \Delta G, changes in enthaly \Delta H, and even change in entropy \Delta S. A common way to evaluate the increased energy of transition state (\Delta E^{\ddagger}) above the ... 2 First question: The answer is true from a reaction view. You are right. No matter how the reaction is favored in terms of Gibbs free energy change. The reverse process will happen in small or tiny amount. From a thermodynamics view, they are different. Similar to the reversible and irreversible process in physics, reversible reaction has no energy loss but ... 2 The Eyring equation is numerically correct, despite the apparent units problem. To understand the origin of the problem, one must go all the way back to the underlying statistical and quantum mechanics, since Eyring treated the motion across the transition state as being effectively a translation (J Chem Phys 3: 107, 1935, p109, emphasis added): The ... 2 First, the question may pose itself why 3N-6 at all? And for that, we should take a step back and ask ourselves: Why 3N? These are the degrees of freedom of a molecule. If you break a molecule down into its atoms, each atom would have three degrees of freedom: x, y and z. In a molecule, these degrees of freedom are generally retained; however, you ... 2 A reaction intermediate is nothing more than a plain old intermediate, which you probably understand. There is no stipulation as to how stable the intermediate must be. The IUPAC Gold Book defines intermediate as: A molecular entity with a lifetime appreciably longer than a molecular vibration (corresponding to a local potential energy minimum of depth ... 2 Only some of the statements you quote are true, certainly a transition state cannot be isolated since it lasts for less than a picosecond. In fact there is hardly any direct measurement of transition states, such as a spectrum or time profile or any structure. It is not that they are too fleeting to measure but that they occur very infrequently in any ... 2 If you don't have any idea what k value would be appropriate, arguably the best option is to identify a catalyst that is experimentally verified to do the reaction of interest and is assumed to involve the same mechanistic step you computed. Identify the lowest temperature the experiments are successfully run at. Determine the rate constant for the ... 2 The Arrhenius equation$$k=A\exp\left(-\frac{E_a}{RT}\right)$$places all of the T-dependence in the exponential factor. The pre-exponential factor is not assumed to be temperature-dependent. By contrast, the Eyring equation$$k=\frac{\kappa k_\mathrm{B}T}{h}\exp\left(-\frac{\Delta ^\ddagger G^⦵}{RT}\right) which is justified on the basis of ...

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I think there is some confusion between removing a single degree of freedom (the vibration that changes the length of the bond that gets broken) and - in the text quoted in the question - freezing all degrees of vibrational freedom. I also don't understand the part about the transition state having no structure at all. According to this paper's introduction,...

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My answer is probably unhelpful, but such is life. === To put it simply, transition states do not exist. They are artificial constructs. Thus, any attempts to work with them outside transition state theory are technically incorrect. Molecules are quantum systems. We do not have a clear theory for interaction of quantum systems, only for isolated quantum ...

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Using this website, input the activation energy and temperature and it gives half life. A reasonable half-life is up to the user, but more than 1 day is going to be super slow. Usually assume the transmission coefficient is less than 1 (e.g. 0.5). For example, an upper bound at room temperature is usually less than 25 kcal/mol $\approx$ 105 kJ/mol. With a ...

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Let's assume your reactive complex (reactants interacting) are optimised, i.e. its gradient equals zero, and your potential energy surface is continuous. Then any direction is either uphill or constant (otherwise the optimisation hypothesis would not apply). There are three possibilities: (a) going from reactants to intermediate gives a transition state ...

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In transition state theory, it is assumed that the forward and the reverse reaction occur via the same transition state. This is implied by assuming a quasi-equilibrium between the reactants and the products, which adheres to a common Boltzman statistics. Therefore the labelling $\ce{AB^\ddagger_r}$, $\ce{AB^\ddagger_f}$ is artificial, because it is the same ...

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