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4

Question is solved. Poutnik taught me to do it using steady state approximation :) Just adding the image for somebody who might find this in future.


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The expression $$v/(\mathrm{mol\ s^{-1}\ kg^{-1}})=30$$ means $$v=30\ \mathrm{mol\ s^{-1}\ kg^{-1}}$$ This notation is used in the table, so that the entries are all just numbers without any unit symbols. For an explanation, I copy the following section from my meta answer, which is mainly a collection of rules and examples taken from various standards and ...


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There is a description of Marcus theory here How does the inverted Marcus region explain chemiluminescence? . The quantum nature can be added into the theory as necessary, it leads to a asymmetrical plot of rate constant vs free energy with the inverted region (large -$\Delta G$) decaying away more slowly than the normal region. The equilibrium is between ...


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I think Bard's is more general than Newman's, Newman's assumes Cx=Cx* where Bard does not. Bard says the current is proportional to the forward rate, and if there are metal ions near the electrode, the forward reaction removes electrons from the electrode to reduce the ions. That means the current flows into the electrode. Bard then says the applied ...


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Yes, it is the process of forming a dimer, and so n is specifically 2. See for instance: https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Dimerization https://en.wikipedia.org/wiki/Dimer_(chemistry) https://www.merriam-webster.com/dictionary/dimer


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No, it is not necessary to have excess water for reaction to occur. Yet usually what books and sources mean by hydrolysis means that water as a solvent reacts with the compound. As you may know without large excess of water you wouldn't be able to go very far with the reaction (Low Yield due to equilibrium). This is a reason why ester hydrolysis is also a ...


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A rate law is never deduced from a theoretical equation. NEVER. It is always obtained form experimental measurements. It may happen that the order of the reaction is equal to the stoichiometric coefficient.


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At equilibrium, the forward and reverse rates have to be equal. For mass action kinetics, this tells you that the reverse rate constant must be equal to the forward rate constant divided by the equilibrium constant.


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In general I would say 'no', thermodynamics only tells us about starting and ending points, thus you know the $\Delta G^\text{o}$ for the reaction but nothing about its actual mechanism, (since time does not come into thermodynamics). As $K_\mathrm{eq}$ is a ratio of rate constants these can take on many different values and still have the same ratio. To ...


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Summarizing from the excellent comments: Isn't the order of the reaction supposed to be the sum of powers of the concentration of the reactants? Yes, that is how you get the overall order of reaction. If you compare the rate with a given set of concentrations to the rate with all concentrations doubled, the rate goes up by two to the power of the overall ...


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