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A rough calculation for the pH of 0.2-ᴍ formic acid gives: $$ \mathrm{pH} = 1/2 (\mathrm{p}K_\mathrm{a} - \log(0.2)) = 1/2 (3.74 + 0.70) = 2.22 $$ You can check by calculating the equilibrium constant from the concentrations of all the species, and the estimate is pretty good. The pH of the buffer is 3.44 using the Henderson Hasselbalch expression (see other ...


2

The flaw is in the graph if you take its labels in their literal sense. So technically, the label "at equilibrium" is incorrect for the left panel. You can either weaken the label language (center) or change the graph (right) to have a watertight graph and labels. Source of original graph: https://chem.libretexts.org/Courses/University_of_Kentucky/...


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If a catalyst is not supposed to affect the reaction's final equilibrium position how do we explain the catalyst selectivity seen here? If you wait long enough so that all three reactions attain equilibrium, the presence or absences of catalysts have no effect on the product mixtures. In the examples, however, the reactions without catalysts are all slow. ...


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Catalysts very much do affect end products because they may act differently on competing reactions. For instance, given ethylene and oxygen a suitable catalyst may promote formation of ethylene oxide and not as strongly promote oxidizing the ethylene to carbon dioxide and water. (In this particular case, a silver catalyst with carefully controlled ...


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See, I will explain my concept to you, it might be helpful In a reversible reaction in equilibrium, N2 + 3 H2 —— 2 NH3 For example, if you apple pressure in this system, then reactant had more than products, so reactant molecules will aggregate and increase in amount at a particular place (saying just to explain), then according to Le-Chateliar Principle, if ...


1

The gas would become solid, as the gas to solid transition does not have critical temperature, at least for usual orders of temperature. The needed pressure is just a question of the gas temperature. If we can achieve such a pressure is another question.


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