# Tag Info

13

I would like to start off by saying this is first and foremost a thermodynamic problem. Secondly, and as a result of thermodynamics, refer to Gibbs Phase Rule which says \begin{equation} F = C - P + 2 \end{equation} Where F = degrees of freedom, C = number of components, and P = number of phases. You seem to be after a pure liquid so $C=1$, and, you are ...

13

I concur with the answer by Alone programmer, but his answer only gives half the story. The coefficients or the model parameters come into the picture later and are actually obtained by optimization of both the experimentally obtained and theoretically calculated data. Here I will discuss the experimental data since the DFT part has already been discussed. ...

12

This is a tricky question and there is a whole area of research dedicated to solving it. The definition of the thermodynamically stable phase of a system is that occupying the global minimum of the potential energy surface, or more precisely, the global minimum of the potential free energy surface at the temperature of interest. Let me focus on solid phases, ...

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Basically you need to find $L_{ij}^{m}$ coefficients in the Redlich-Kister polynomials from DFT calculations, related to temperature as: $$L_{ij}^{m} = a_{ij}^{m} + b_{ij}^{m} T + c_{ij}^{m} T \ln(T) + d_{ij}^{m} T^{2} + e_{ij}^{m} T^{3} + \frac{f_{ij}^{m}}{T}$$ Typically you need just the values for $m \leq 2$. I think this is a good reference for the ...

10

High frequency in this case is 'particles move all independently, with different directions and speed', and low frequency means 'particles can be represented as gradually changing field of speed and direction of particles in the region'. Something like 'frequency in space', how often to change speed and direction as you observe more and more particles. If ...

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Because the hydrogenation release has a barrier (activation energy) that is not too low. The fact that the dehydrogenation energy of $\ce{H3N-BH3}$ is negative does prove that it will eventually decompose if left at room temperature, but it does so extremely slowly so that you don't notice, and it's more useful to treat it as stable. It's like, your ...

9

You are correct that this is due to not including quantum effects. Ref 1 in your figure is the paper cited below. In this paper, they explicitly mention that $C_v$ calculated using the cell-cluster method is in good agreement only for sufficiently high reduced-temperatures. From section IV of this paper: ...the calculation is in acceptable agreement with ...

8

It is straightforward to show that in a typical $NPT$ setting the Zwanzig equation still only depends on the energy difference and not on the volume (here I define $H$ to be the Hamiltonian of each system, respectively and $x$ to represent all variables over phase space): $$\frac{Z_{B,NPT}}{Z_{A,NPT}} = \int \int e^{-\beta \Delta H(x)} \frac{e^{-\beta H_A(x)... 8 Short introduction to ergodicity Ergodicity is when the time-average equals the ensemble-average. A process is ergodic if the time-average "converges in the square mean" to the ensemble average. A sequence X_t converges in the square mean to X if:$$ \tag{1} \lim_{t\rightarrow \infty}\langle \left|X_t - X\right|^2 \rangle = 0, $$where \... 7 Thermochemistry (for a single molecule, e.g., ideal gas) depends on the Temperature. At T=0K thermochemistry is kind to us. At T = 0K U = G = H. This is because H = U + PV, which for an ideal gas is H = U + RT, and T = 0K, and also, G = H-TS, and again, T = 0K so that cancels The only direct calculation a QM program is doing, that I am aware of, is ... 7 The flying ice cube effect is when the kinetic energy leaks into the translations and rotations. In a constant energy simulation (NVE) this must come at the expense of vibrations, which has the effect of "freezing" the bonds, angles and torsions. Obviously, this is a purely classical effect because even at 0 K a bond will vibrate at the zero-point ... 6 if you had access to the database equations for a given alloy system, could you analytically solve (and thus solve quickly) for the conditions (i.e., temperatures and compositions) that would produce the phase combination that you desire? Generally, the answer to this question is no. If we knew how to predict which phases were stable analytically without ... 6 Energy in an NPT simulation is not conserved, but (once equilibrated), it will fluctuate around an average value, and that average value has meaning. That is the ensemble average for your NPT and is a valuable and useful property. You are also correct that the internal energy is the summation of potential and kinetic contributions. To be thorough, pp. 60 of ... 5 I don't have access to TC-Python, so I can't offer advice on how to use the API specifically. This is more of a workaround than a direct way to get Thermo-Calc to tell you whether the phase is ordered or disordered. Are you able to get the site fractions from an equilibrium calculation? In console mode, you can use the Y(CR,BCC_B2#2) syntax. My understanding ... 5 I am not sure about TC-Python but in console mode, you can access the Gibbs energy equations directly using the Gibbs Energy System (GES) module. In case you need the details about a particular phase then you have to first define the system as usual and then after retrieving the data go to the GES module and use the List_phase_data command. You will have to ... 5 I know it's a boring thing to say, but: It depends on what you want to do. The way I use Langevin thermostats, is to ensure good equipartitioning in my setup, so that I don't have any local hotspots in the system or something like that. If heat transfer is slow in the system, "global" thermostats like Nosé-Hoover and the like will equilibrate ... 5 High accuracy energy methods, are they worth it? They are absolutely worth it for many types of studies, but not worth it for other types of studies. It depends on precisely what the aim of your project is. As far as I understand they are used for high-end kinetics calculations of small molecules. They are also used for spectroscopy, benchmarking and for ... 4 Just throwing a couple of ideas out there: (i) Perform a free energy perturbation or potential of mean force calculation as available in typical classical MD setups. For known salt systems in water, this is well studied and should result in decent results (e.g., Joung and Chaetham's paper). (ii) Do a classical MD in explicit solvent for the solute of ... 4 I have faced a similar issue and it seems that something happens while calling calculate(). I think it may be because when you use the same single equilibrium calculation for determining the Gibbs energy it does without suspending the other phase and this might be creating the problems, but I am not sure. I would suggest trying two different single ... 4 As a retired professor in Computational Thermodynamics I am happy that young students try to use the Calphad methods to understand materials. However, to model a phase using Calphad there are some background needed. A Calphad database is built from a set of data for the pure elements in different phases describing how the Gibbs energy of the element vary ... 4 "This works for organic molecules, but what happens when the excited states are closer in energy to the ground state, for example in open-shell molecules or in atoms?" If there's excited states close to the ground state, the approximation you said Gaussian uses, where excited state contributions are neglected, seems not to be such a great idea ... 4 Thermo-Calc's DGM is the normalized parallel tangent driving force per mole of components. To be able to compute the driving force of a phase, it must be set to "dormant" status so that it cannot become stable, but the driving force can still be computed. Here's an example macro of how this could be done in Thermo-Calc: @@ These commands can be ... 3 So what happens when the contribution of excited determinants are included in the vibrational and rotational terms? Nothing, except that the Hessian and/or gradient are more difficult to evaluate, for example see coupled-perturbed Hartree-Fock. I think maybe your confusion is arising because the excited determinants are not exactly excited-states. A multi-... 2 This is not a complete answer, since I can't seem to work out the r_1=r_2 case, but I think this should get you most of the way to an answer. To start, its useful to give another definition for the grand potential in terms of the grand partition function and write the derivatives implicitly using this definition:$$\Omega=\frac{1}{\beta}\ln(\Xi)\tag{1}\...

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TLDR: The rate constant calculated by transition state theory depends on the standard-state (gas-phase: ideal gas at a partial pressure of 1 atm or any state concentration (1 cm$^3$ molecule$^{-1}$ or 1 mol L$^{-1}$), liquid-phase solutes as in an ideal solution of 1 mol L$^{-1}$) and all species (reactants and transition state) should be in that specific ...

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