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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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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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For those who don't know the Carius test with addition of barium chloride is for organic sulfur. Without the exact reaction conditions it is a bit difficult to be definitive however, it is clear that A & D will react with sodium thiophenoxide under mild conditions. I think this is the answer the question-setter expects. I think it possible with the right ...


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Both the forward and the reverse rate are proportional to the surface area of A. Steelwool rust faster than a block of steel. Crystals grow faster with microcrystals instead of a single crystal in the mother liquor. Once you update your rate expressions to include the surface area of A in contact with the solution, it should all make more sense.


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On the microscopic scale, the thermodynamic equilibrium is dynamic, i.e. there are constant back- and forward reaction. Here: dissociation of $\ce{BaSO4}$ to yield $\ce{Ba^{2+}}$ and $\ce{SO^{2-}_4}$, and association to yield again $\ce{BaSO4}$. Once the thermodynamic equilibrium is installed, these ongoing microscopic reactions do not change anymore the ...


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Recall the definition of chemical equilibrium. The chemical composition becomes "static" when equilibrium is reached i.e., it appears that neither the concentration of the reactants or the products is changing. At a microscopic level the system is dynamic. Apply the same idea to the partition of a substance in phase "A" and "B". ...


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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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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 ...


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