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For the Ellingham diagram: (1) We write down the reaction/reactions that we are interested in. The $\Delta G_f$ (formation) per mole of product is calculated using the individual free energies of any intermediate reaction, and using Hess' law to sum them up. (2) This energy is then normalised per mole of $O_2$, using stoichiometry. (3) The normalised ...

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The problem begins with your first statement: $$\left(\frac{\partial S}{\partial V}\right)_U=0$$ Because in fact: $$\left(\frac{\partial S}{\partial V}\right)_U=\frac{p}{T},$$ which is clearly non-zero (except in limit of zero pressure or infinite temperature). Further, if we are considering a system (rather than system+surroundings), it's not even ...

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We may interpret the energy variable as free energy of formation per mole of $\ce{O2}$. Thus, for instance, silicon would be favored to react with a limited amount of $\ce{O2}$ versus iron, because silica has a more negative free energy of formation per mole of $\ce{O2}$ than iron oxides; even though $\ce{Fe3O4}$ might be more negative per mole of compound ...

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The Ellingham diagram doesn't actually use molar Gibbs energies of formation $\Delta G_\mathrm{f}^\circ$ per se; it is more accurate to say that it uses molar Gibbs energies of reaction $\Delta G_\mathrm{r}^\circ$. The difference is that the formation energy is only relevant to one specific chemical equation, for example: $$\ce{Ca + 1/2O2 -> CaO} \qquad \... 1 \require{begingroup} \begingroup \newcommand{\pd}{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}} If you write$$dS=\pd SVT dV + \pd STV dT$$then$$\pd SVU=\pd SVT + \pd STV \pd TVU \tag{1}$$Next use$$dT = \pd TUV dU + \pd TVU dV$$to obtain$$\pd TUS = \pd TUV + \pd TVU \pd VUS$$or$$\pd TVU = \pd UVS \left[ \pd TUS - \pd TUV \...

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$\require{begingroup} \begingroup \newcommand{\pd}{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}}$ I would start from $$dH = \pd HTp dT + \pd HpT dp$$ This gives rise to $$\pd HTV = \pd HTp + \pd HpT \pd pTV$$ in which you can recognize some components of the solution. Since $$dH = dU +PdV + VdP$$ so that $$\pd HTV = \pd UTV + V \pd ... 2 Because life uses whatever it finds useful. And organic compounds with nitrogen can be easily synthesised, metabolized and may have many useful properties. And also, various stable organic compounds with nitrogen will just come into into existence in abiogenic conditions. Why though? Well, nitrogen-carbon bonds are fairly strong and unlikely to decompose ... 6 According to the RNA-hypothesis, life did not start with DNA and proteins (separate molecules for catalysis and storing sequence information) but RNA instead (one molecule to store sequence information, act as polymerase and catalyze other reactions). While that hypothesis is realistic in terms of the awesome properties of RNA, it requires a lot of ... 8 Dusting off my geochemist hat here - In addition to elemental abundance, the matter of temperature & pressure is also key. The "solid" planet that we call home is undergoing solution chemistry on a vast scale, with most minerals differentiated mainly by local T/P conditions, rather than chemical composition (\ce{SiO_4}-based minerals being most common ... 9 My take is a little bit different. Organisms evolved to adapt to the chemistry that is there. And what's there, primarily, are hydrogen, carbon, nitrogen and oxygen. Plus, as we shall see, sulfur, which cannot be ignored in proteins. Here we find a list of the most common elements in the Universe. Although everything else us made from fusion of hydrogen,... 17 Because chemistry built on carbon, hydrogen and oxygen would be very dull and building complex structures from modular components would be impossible The key point to remember is that life depends on complex chemistry. Living things are, by definition, things that can replicate and things than can build structures that can do complex chemistry. Metabolism ... 0 You are mixing two areas, electrode kinetics and electrode thermodynamics. If we have 3 apples and 2 oranges, this is not equal to 5 oranges or 5 apples but there are 5 fruits in total. Electrode thermodynamics tells you a "yes" or "no" answer in the sense that it is saying the reaction is spontaneous or not. Electrode kinetics tells you how fast the ... 0 If you are confused about the electrical double layer, it is completely fine because electrochemists spent several decades to understand it and yet the concept remains "grey" in the minds of most electrochemists. You will find different stories and versions on the web and of course different terminologies. Your understanding is correct but I will call the ... -2 Your second and third explanation are good and not basically different from one another. On the other end the first sentence is not as good, as there is no gain or loss of electrons in or through the double layer. 2 As pointed out in the comments the rigid rotating molecule only gains kinetic energy when it rotates not potential energy. The degeneracy describes the fact that some levels have exactly the same energy and this depends the value of the angular momentum rotational quantum number J. The number of degenerate levels is given by the multiplicity 2J+1. The ... 2 I have two questions... Why does a molecule "gain potential energy" when it rotates? Does it want to stop rotating for some reason? What do degenerate energy levels correspond to physically in terms of the molecule's rotation? Do you remember the Newton's law of motion? A body in motion will always remain in motion until and unless there is a ... 2 The Joint Entrance Examination board should update their outdated set of questions relying on rote memorization. Why do they assume the student has to memorize the charge of the ferric hydroxide particles and in what medium? Charge on the colloids is dependent on the medium and pH! They should also mention that the molar concentration of the anions is the ... 0 There is a saturation level for the adsorption of any adsorbate on the adsorbent due to limited surface area of the adsorbent. So once that saturation level is reached , more pressure won't increase the number of particles adsorbed . 0 You may simply say that energy is needed to break a bond. Breaking a bond is endothermic. So the inverse is exothermic. Energy is released when a bond is formed. Apparently you want to just discuss what is happening when an electron approaches a proton from far away (x1 in your drawing) to a shorter distance (x2). The electron is supposed to have no ... 0 The Gibbs energy has not been derived this way. It came from a thought process. About 100 years ago, Gibbs was puzzled by the very existence of spontaneous endothermic reactions. In mechanics, objects always fall down. They never "fall up" spontaneously. They never gets more energy spontaneously. In exothermic reaction, the matter looses spontaneously energy,... 2 There is no such thing as "a salt that makes ice unable to refreeze". Well, sort of depending of the conditions. What happens is that salts added to pure water ice form a solution (of the salt in the melt water) that has a much lower point of freezing that pure water ice. This is known as the freezing-point depression (Wiki for the basics). For example, ... 1 The way to think about microstates is as alternative configurations of a system which satisfy constraints consistent with the observed macroscopic properties of the system. The collection of such microstates forms an ensemble. The misconception lies in the idea of "concentrating" kinetic energy, and in thinking that reducing the number of available ... 1 The Hamiltonian for the collection of charged particles (electrons and nuclei) comprising a pair of molecules or ions has terms describing the kinetic energy of individual particles and pair-potential terms describing Coulombic interactions, that is, all interactions between particles are Coulombic. Where a Pauli repulsion term can crop up is when ... 2 AFAIK, such a reaction has no name. However, if one were so inclined, one could call it a replacement reaction, because the \ce{MgCO3} is being converted into \ce{Mg(OH)2}. To calculate the "equilibrium point," which I take to mean the concentration of all the ions at equilibrium, we would need to know the k_\text{sp} of both \ce{MgCO3} and \ce{Mg(... 3 I found a nice figure and the relevant statement in a paper by Frenking and Krapp (Unicorns in the world of chemical bonding models, 2006, https://doi.org/10.1002/jcc.20543): The crucial term which is responsible for repulsive interactions in chemical bonds except in two‐electron systems such as H2 is the Pauli repulsion. The three terms (a) ... 0 The magnetization is composed of the sum of dipole moments associated with individual spins:$$\vec{M}=\sum_i \vec{\mu}_i$$and the component along the magnetizing field (assumed to lie along the z axis) is$$\begin{align}M_z&= \left( \sum_i \vec{\mu}_i\right)_z \\ &= \sum_i \mu_{iz}\end{align}$$Since the individual dipole moment components ... 5 You are correct that a given value of m_I and I does correspond to one single value of the angle \theta. For the most commonly used nuclei in NMR spectroscopy, the "allowed" values are I = 1/2 and m_I = \pm 1/2. The two states are commonly labelled \alpha and \beta; these are the eigenstates of the angular momentum projection operator \hat{I}... 1 The reference given by Andselisk does not bring an answer to the question about the origin of the prefix "ortho" in inorganic nomenclature. The prefix "ortho" is added before the name of an acid H_aX_bO_c if first b = 1 and if it corresponds to the highest possible number of OH groups attached to the central X atom. For example H_4SiO_4 is the ... 0 To understand internal energy*, I prefer to start with this: Consider a thermodynamic system. There are only three ways we can increase its internal energy (which is the sum of its internal potential energy and internal kinetic energy): flow heat into it; do work on it; or add matter to it (and the converse for decreasing its internal energy). If the ... 0 The internal energy is defined in (classical) thermodynamics for a finite change as \Delta U=q+w where U is the energy contained in the 'system' called the internal energy, and q is the heat absorbed by the system and w the work done on the system. Often the word 'system' means some ideal gas (often imagined to be in a cylinder with a frictionless ... -2 The internal energy U is the sum of all energies stored in the chemical bonds of the examined sample of matter. It is a potential energy. The enthalpy H is the same, but with a correction due to the presence of the atmosphere. The difference between H and U is similar to the difference between weight and apparent weight. Let me go back to this measurement. ... 0 As others have answered, yes the potential of the cell will still change. It sounds like you're looking for a more "physics-y" answer than the ones provided so here goes. The Nernst equation (and equilibrium equations in general) work based on concentration and not total moles. To understand why, start with a homogeneous reaction, like \ce{HCl + NaOH <-&... 2 Firstly, I would replace the word "feasible" with "favorable". I'm also going to replace E_{cell} with E_{OCV}, where the open circuit voltage (OCV) is the potential of the cell without any applied electric potential. So you're right in that if the E_{OCV} is positive, the net reaction is thermodynamically favorable: it will occur spontaneously if ... 0 It is difficult to give a general answer. A glass is an amorphous solid that may change its composition in a continuous way from one point to another. When melting brown bottles and colorless bottles in the oven of a glass factory, you may obtain a non homogeneous mixture, with regions more or less colored. It is obviously one and the same phase all over the ... 1 The water which is distributed in the public is not polluted. It has been first filtrated through a layer of fine grains of activated wood charcoal that fixes most unwanted impurities. Then it is treated by gases like chlorine or ozone that eliminates unwanted substances or toxic bacteria that would have passed through the charcoal filter. Don't forget that ... 0 In an electrochemical cell, increasing the concentration of reactants will increase the voltage difference, as you have indicated. A higher concentration of reactant allows more reactions in the forward direction so it reacts faster, and the result is observed as a higher voltage. If you have adjusted the cell volume to keep the total amount of reactants ... 0 I'll address part of your confusion here: The following statement is not generally true: Because H and U are state functions, the amount of heat entering the surroundings is independent of the path; q is the same whether the transfer occurs reversibly or irreversibly. For instance, in a closed system \Delta U = q + w so what is independent of path isn'... 0 This has to do with the simplifications we make regarding the energy of molecules in an ideal gas (they've been largely discussed here if you want to have a look). But essentially, in an ideal monoatomic gas, we can state that all the energy is translational kinetic energy, and so we can write that the mean energy is:$$U=n \frac{3}{2}RT=n c_vT$$So, the ... 0 The characteristic property of all acids and bases is the pH. Their characteristic general reactions are the acid-base reactions. Let's concentrate on this kind of reactions. We don't want to consider here reactions of particular acids and bases (like e.g. oxidation, passivation, complexation). We have to distinguish between dissociation degree of an acid ... 2 The primary problem is too low permeability = too high resistence of the gloves, causing high voltage drop where it should be minimal one. The gloves are not hydrophilic enough for electrolyte soaking to get low resistance. The remaining voltage - after subtracting the glove voltage drop - is not high enough to cause electrolysis by significant current. ... 3 Entropy is indeed a state function, and thus depends only on the state of the system. Hence it doesn't matter how you get from state A to state B, the entropy change will be the same. The analogy would be that it doesn't matter which path you use to get from the base of a mountain to the summit, your elevation change will be the same. This is because ... 0 Assume: Water has a constant density of 1.000 g/ml from 20 ^\circ\pu{C} to 100 ^\circ\pu{C}. Given: 400\pu{g}\ \ce{Na2SO4} contains 400/142.04 = 2.8161 mole of \ce{Na2SO4}. 1000\ \pu{ml}\ \ce{H2O} = 1000\ \pu{g}\ \ce{H2O} = 1000/18.015 = 55.509\ \pu{moles}\ \ce{H2O} 19.4\ \pu{grams} of \ce{Na2SO4} per 100\pu{mL} of water = 19.4/142.04 ... 0 Ionic solids like NaCl, KOH, KBr, PbCl2 and KBrO3 have ratios of individual elements, but the "molecule", if you could use that term to describe it, would be the whole crystal, because all the ions are attracted to nearby ions, ad infinitum. The white solid could be written as "K23.4% Br47.8% O28.8%", and every time this experiment is run, the same product ... 2 While considering colligative properties we always assume solute to be non-volatile. So, it doesn't contribute to the vapor. The amount of vapor will only be due to solvent particles. So, in the problem we can use the following relation:$$ \Delta T = K_\mathrm b\cdot m $$Initially  \Delta T = 0.53\ \mathrm{^\circ C} so plugin the value of K_\mathrm b ... 4 The reasons for the change in internuclear separation and the imporance of the Franck-Condon factors, as has been clearly pointed out in answers and comments. The FC factors determine the strengths of transitions from M to M^{+.} and to clarify this figure below shows a simple calculation based on harmonic oscillator wavefunctions of the effect of ... 5 The reason why statement #2 doesn't make sense to you is because it is not generally true! For a closed system with P-V work only,$$dU = \left(\frac{\partial U}{\partial T}\right)_V dT + \left(\frac{\partial U}{\partial V}\right)_T dV = C_v dT + \left(\frac{\partial U}{\partial V}\right)_T dV The difference between that statement and your statement is ...

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This is related to the difference between how Cv is measured experimentally and how it is applied in solving practical problems. To measure Cv experimentally, we hold the volume constant and determine the change in internal energy U by measuring the amount of heat added. However, for practical purposes, we know that, for an ideal gas, U and Cv are ...

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I am a 12th grader, so I'll explain in the most colloquial way possible. Internal energy is a state function. A state function is a property whose value does not depend on the path taken to reach that specific function or value. ... Path functions are functions that depend on the path taken to reach that specific value. (1) This implies a derivation for ...

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You may have calculate dU and/or dH in any transformation, at constant volume or constant pressure, because U and H are state functions. These variations of U and H may not be easy to obtain, but they can and must always be calculated.

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Why is it unusual (as it seems to be implicitly implied) that the bond lengths of the molecular ion in its ground state somehow end up being larger than the bond lengths of the molecular ion in its vibrationally excited state? It's not so much about whether the bond length in $\ce{M^.+}$ is larger or smaller than that in $\ce{M}$; it's more about whether ...

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