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Yes. The universal condition for equilibrium is that, given the current constraints on the system, no further spontaneous change can take place. I.e., that the entropy of the universe (system + surroundings) is maximized. At constant temperature and pressure, a maximization of the entropy of the universe corresponds to a minimization of the Gibbs free ...


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The answer by A.K. is eminently reasonable, but I wondered if stability constants were available. I quote from the summary of Ref. 1 : The solubility of silver chloride in concentrated solutions of various chlorides was determined mainly at 25°. The solubility is nearly doubled in going from o° to 25°, the rate of increase above and below 25° being nearly ...


2

Two carbon dioxide scrubbers come to mind: lithium hydroxide and soda-lime. (https://en.wikipedia.org/wiki/Lithium_hydroxide)"Lithium hydroxide is used in breathing gas purification systems for spacecraft, submarines, and rebreathers to remove carbon dioxide from exhaled gas by producing lithium carbonate and water: 2 LiOH•H2O + CO2 → Li2CO3 + 3 H2O or ...


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Thermodynamics courses usually start by calculating amounts of heat and amounts of work enterring a container. And the work is $\pu{p\Delta V}$. No mention of concentration ! Just the pressure. Afterwards, enthalpy is introduced, then gas chemistry is developed, always using pressures. Equilibrium constants are then introduced, always with gases and ...


1

So at equilibrium, we can list the concentration of each species: $\ce{[CO_2] = 0.50 + 0.086 - x = 0.586 - x}$ $\ce{[H2] = 0.045 - x}$ $\ce{[CO] = 0.05 + x}$ $\ce{[H2O] = 0.040 + x}$ where x is the change in concentration that is given to all species. Therefore, we get the following expression at the new equilibrium: $\ce{ \frac{[H2O][CO]}{[CO2][H2]} = \frac{...


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Okay, so I managed to get a solution without too many equations by making the right approximations. First with your expressions: $\ce{K_{sp} = [Mg^2+][NH_4^+][PO_4^3-]}$ $\ce{K_{sp} = (S)(0.1)(S-x})$ Here we want to find S-x which is $\ce{[PO_4^3-]_{eq}}$. To do this, I did the following: $\ce{\frac{[PO_4^3-][H+]}{[HPO_4^2-]}=K_{a3} = 10^{-12.4}}$ Assuming $\...


1

Let's consider an example of your galvanic cell : the Daniell cell, made of a zinc anode ($\ce{Zn}$) and a copper cathode ($\ce{Cu}$). When zinc $\ce{Zn}$ is in contact with water, it "prefers" being transformed into the cation $\ce{Zn^{2+}}$ in order to be dissolved in water. In your language, you say that they are more "comfortable" in ...


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