# Tag Info

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I feel Koushal has not seen the difference between $\Delta G$ and $\Delta G°$, because $G$ of all reactants and products change during the reaction. $G$ and $G°$ are very different concepts. $G°$ is the Gibbs energy of $1$ mole of any reactant and of any product in the pure state at 25°C and 1 atm. $G°$ is a constant of the substance, independent of the ...

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The Gibbs energy of reaction $\Delta_r G$ determines in which direction equilibrium lies, i.e. in which direction there has to be a net reaction (with a change in concentrations) to reach equilibrium. When equilibrium has been reached already, there is no net reaction (i.e. concentrations are constant). Nevertheless, at the molecular level, reactions in ...

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At the triple point the water system is assumed to be adiabatic. The system has 3 phases, ice from pure water, liquid water, and pure water vapor. The various equilibria are shown below. Note that the ice floats on the water. So the ice is in contact with both the liquid phase and the gas phase. Furthermore The ice can't cover the whole surface like a ...

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The triple point of water is an example of Thermodynamic Equilibrium, which is explained by Wikipedia as: simultaneously in mutual thermal, mechanical, chemical, and radiative equilibria Your equation only covers the chemical facet of this equilibrium and hence is not the correct way of representing total thermodynamic equilibrium. I am also not sure of ...

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I found this paper (it's in Japanese, but the relevant things are legible) https://www.jstage.jst.go.jp/article/yakushi1947/117/10-11/117_10-11_764/_pdf, which says that phenolphthalein has 2 pKas, pKa1 = 9.05 and pKa2 = 9.50. It also says that the pink form is the twice deporotonated; the 1- form exists, but is colourless.

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The Wikipedia article has a nice picture of the stages of deprotonation of phenolphthalein: https://en.wikipedia.org/wiki/Phenolphthalein However, there is a big jump here between H2In and In2-. In fact, smaller steps, showing just one deprotonation, are very reasonable. Whoever heard of two protons coming off at the very same time? http://www.ch.ic.ac.uk/...

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If you look at the graph of the reactivity with relation to temperature it increases proportionally until it reaches its activation energy and then it proceeds to decrease. For exothermic reactions since the activation energy is met early the reactivity decreases when more temperature is provided. While for endothermic reactions since they require a higher ...

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For a general idea, see Le Chatelier's principle When any system at equilibrium for a long period of time is subjected to change in concentration, temperature, volume, or pressure, (1) the system changes to a new equilibrium and (2) this change partly counteracts the applied change. For the changes of $p, V$, read about equilibrium constant $K$ and ...

1

Consider three compositions: A. 2NaF + CaCO3 B. CaF2 + Na2CO3, and C. NaF + 0.5 CaCO3 + 0.5 CaF2 + 0.5 Na2CO3. Using data from the CRC Handbook (62nd ed), the heats of formation of A and B are respectively 560.47 and 560.6 kcal, so there is little driving force to make a reaction go to completion. Note that A should be near neutral pH, but B ...

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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 $\ce{Mg(... 2 Yes, the calcium ion could lead to precipitation. The solubility of$\ce{CaCO3}$in distilled water is about 15 mg/L, which is about 0.15 mM calcium ion if there is no other source of carbonate. The solubility constant for$\ce{CaF2}$is about$4\times 10^{-11}$, which means that we can only have 0.5 mM fluoride ions before precipitation will start. That's ... 0 If you know what the final equilibrium products are or may be, and you know how much mass you start with, either as elements or molecules, you can calculate the Gibbs energy of the products as a function of (unknown) concentrations (assuming constant temperature and pressure) and then minimize the Gibbs energy with respect to those concentrations. You have ... 5 [Comment by Poutnik] Important is also en.wikipedia.org/wiki/Grotthuss_mechanism for the proton interchange. Mobility of H3O+ and OH- gives a hint it must be fast. If you compare the diffusion coefficient of hydroxide ($\pu{5.270e9 m^2/s}$) to that of fluoride ($\pu{1.460e9 m^2/s}$), you might be surprised to see such a difference despite their comparable ... 1 The key word is "mixing". If you take a glass cylinder and fill it halfway with D2O (the heavier water, so put it on the bottom) and fill the cylinder ever so gently with H2O, the mixing will be determined by diffusion (NO mechanical mixing). But the experimental techniques required to analyze and determine the rates would be expensive and time-consuming. ... 0 Why do you think so ? As I have missed the other answer, this is just a complement to it. In the equilibrium: $$\ce{ 2 NO2 <=> N2O4}$$ both are reactants and products at the same time. If you formally consider products on the right side, then you see by the naked eyes the red-brown reactant. What is or is not seen depends solely on the ... 1 Not really. It just depends on the reaction conditions. Consider the following reaction: $$\ce{N2O4 <=> 2NO2}$$ Here’s a picture showing this equilibrium in action: One of the reasons you can’t see the reactant here is because it is colourless, but this gives a pretty good example regarding visual equilibrium changes. this is a good answer giving ... 0 Calcium bicarbonate$\ce{Ca(HCO3)2}$simply does not exist. It is impossible to fill up any container with pure$\ce{Ca(HCO3)2}$. Nobody has ever been able to produce a powder containing$\ce{Ca(HCO3)2}$. It is possible and easy to produce a solution containing the ions$\ce{Ca^{2+}}$and the ions$\ce{HCO3^-}$, by bubbling$\ce{CO2}$into a solution of ... 1 The correct expressions are: $$\mathrm{pH} = \frac{1}{2}(\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2})\tag{1}$$ $$\ce{[H+]} = \sqrt{\frac{K_1K_2[\ce{HA-}] + K_1K_\mathrm{w}}{K_1 + [\ce{HA-}]}}\tag{2}$$ Equation 2 is an exact expression (neglecting activities vs. concentrations), but Expression 1 is an approximation. To dervice the approximate ... 5 Since the equilibrium constant for the formation of the complex ion is very large, I assume that$\ce{[Ni(CN)4^{2-}] >> [Ni^{2+}]}$From the comments: [comments:] That assumption is incorrect. The equilibrium constant seems large, but the exponents are high, so it is misleading. In fact, more of the cyanide is in the form of HCN than in complex ... -2 I had a different approach to the problem which I have highlighted below Notes : Initially, I have considered the entire reagents to have reacted and then considered the equilibrium to set in I have considered two equilibria simultaneously, one involving the dissociation of the Nickel complex and also the hydrolysis of the Cyanide ion because it is a ... 1 The former equation assumes $$[\ce{H2A}]\simeq [\ce{A^2-}]$$ due reaction $$\ce{ 2 HA- <=> H2A + A^2-}$$ The is possible with 2 simplifying conditions: The concentration of oxonium resp. hydroxide ions originated from water dissociation is much lower than concentration of the basic resp. acidic ampholyte form. $$[\ce{H2A}] \gg \sqrt{K_\mathrm{w}}$$... 5 I would guess$Ω^m$notation is a personal invention and I couldn't find any sources that would standardize it. The choice of these symbols for abbreviating "equilibrium" may be justified that the glyph "Libra" ♎ originates from the Greek letter Omega Ω. Further, from Merriam–Webster Online: Equilibrium contains a root from the Latin libra, meaning "... -1 My approach is the following. There is exactly enough$\ce{Ni^{2+}}$ion and CN- ions for making up the ion$\ce{Ni(CN)4^2-}\$. As the stability constant of this complex is huge, and HCN is an extremely weak acid. I would admit that the rare HCN molecules remaining in the solution have no effect on the pH of water, which must be 7.

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