Recent content by timntimn

Dear colleagues! I've asked the same question on the Russian forum at http://chemport.ru After intense discussions, we've almost came to a solution. Its key is that chemical reaction is non-equilibrium process (otherwise, affinity is zero and my question "vanishes") and hence, an entropy...

But it seems to me that you obtain dG=0 only if you add a small amount of reactants and products as well. The following makes me think so: ## \eqalign{ & G\left( {T,p,N_i + \delta N_i } \right) = G\left( {T,p,N_i } \right) + \sum\limits_i^{} {\left( {\frac{{\partial G}} {{\partial N_i }}}...

Let us consider two systems, initially separated by the wall, first system of particles of type A, and of particles of type B the second. Let the interparticle potential U(1,2) be zero for ## \left| {\vec r_1 - \vec r_2 } \right| > \sigma ##, but for ## \left| {\vec r_1 - \vec r_2 } \right| <...

I'm afraid that this is not obvious for me. ## G = \sum\limits_i^{} {G_i } ## since ## G = \sum\limits_i^{} {\mu _i N_i } ## This means that if number of particles in subsystems change, this may affect total Gibbs free energy of the system. (unless affinity is non-zero) Furthermore, if there...

Ok. In canonical ensemble statistical sum for the mixture will be Z_{mix} = \frac{1} {{N_A !}}\frac{1} {{N_B !}}\left( {z_A } \right)^{N_A } \left( {z_B } \right)^{N_B } where z_i = V/\lambda _i ^3 with \lambda _i = \frac{h} {{\sqrt {2\pi m_i kT} }} The free energy is then...

So, as I right that your suggestion is that in each moment of the reaction (if we consider it as being a very-very slow one) its affinity equals zero, i.e., reactants and products are in chemical equilibrium ?

It is difficult for me to accept this, since we can imagine a box with impenetrable wall which splits the total volume into two parts, each containing a gas of particles A and B respectively. Such system as a whole is clearly closed one: no "external" particle can enter or leave it. However, it...

Hi! In most textbooks on chemical physics/thermodymanics it is said that under fixed pressure the heat of reaction equals change of enthalpy of the system since dU = \delta Q - p\cdot dV, and hence d(U+pV) = \delta Q. But my question is: why they do not write a term +\mu\cdot dN which describes...

Thanks for your reply! Isn't it a somewhat puzzling way to define exchange 'part' of the energy via the value of the exchange DFT functional ? Moreover, it seems to me that your answer implies that so many fundamentally important papers aimed at the development of purely exchange functionals...

Dear Forum members! I'm wondering, what is an exact definition of the exchange energy in atomic physics and/or quantum chemistry ? For the best of my knowledge, the case is quite simple for correlation energy, namely E_{corr} = E_{exact} - E_{HF}, where E_{exact} is the exact solution of...