# Tag Info

15

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 ...

10

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 ...

8

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 ...

8

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-...

8

You could convert the rate constant($k$) to half-life($t_{1/2}$) which would give you an idea of the time scale required for the reaction to finish at a certain temperature. The equation to obtain half-life from $k$ is different depending on the order of the reaction. For an unimolecular reaction($\ce{A->P}$), it is: $$t_{1/2}=\frac{\text{ln 2}}{k}$$ For ...

8

From the Eyring equation, we can simply calculate the $k$ value for it. \begin{align} k &= \frac{k_\mathrm{b} T}{h}\exp\left(\frac{-\Delta G^\ddagger}{RT}\right)\\ k_\mathrm{b} &= \pu{1.38E-9 J K^-1}\\ T &= \pu{355 K}\\ h &= \pu{6.626E-34 J s}\\ R &= \pu{8.3145 J K^-1 mol^-1} \end{align} Here, I am assuming the average of $\pu{29 kcal/mol}... 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 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 ... 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 \... 6 Why do these ambimodal transition states occur? I don't know that there is a general reason for that. Potential energy surfaces are complicated. Very often, there is an abundance of local minima, separated by potential energy barriers. It is similar to what happens with mountain chains and valleys, but with many dimensions. One could argue that the opposite ... 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 To study surface reactions I recommend the growing string method (GSM) for surfaces. Here is a nice quote from the paper which developed it (Ref. 1) GSM’s efficacy was confirmed by comparison with CI-NEB on an extensive set of reactions characteristic of modern surface chemistry studies. In these cases, GSM reduces the computational cost (in terms of ... 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 ... 4 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 ... 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 degree ... 3 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 ... 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 The equation$$\ln\left(\frac{k}{T}\right) = \frac{-\Delta H^{\ddagger}}{RT} + \frac{\Delta S^{\ddagger}}{R} + \ln\left(\frac{k_\mathrm{B}}{h}\right)$$does not assume that \Delta S^{\ddagger} is temperature independent. To evaluate its T dependence you might proceed as follows:$$\left(\frac{\partial \Delta G^{\ddagger} }{\partial T}\right)_p = -\Delta S^{... 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 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 ... 3 Using vibrational partition functions to define the reaction rate constant produces an equation of the form$\displaystyle k=aT^be^{-\Delta U_0^\mathrm{O}/(RT)}$where$a,b$are constants independent of$T$and$\Delta U_0^\mathrm{O}$is the difference in zero point energies at the transition state compared to reactant, this is related to what is commonly ... 3 This is deeper discussed in Rate Constant Units and Eyring Equation, but I am going to post a very short take-home message here. First, the three assumptions you have cited are not complete. They completely gloss over the most important one, i.e. nuclear and electron motion can be separated, energies (translation, vibration, rotation) can be treated ... 2 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 ... 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 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 ...

Only top voted, non community-wiki answers of a minimum length are eligible