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By principle, all catalysts accelerate both forward and backward reactions in the same extent, so the equilibrium with and without the catalyst is the same. So the catalyst catalysing the "reversed combustion" would catalyze the normal combustion as well. The equilibrium would be far at the combustion products side, as if the catalyst was not used. If ...


4

I don't know if this qualifies as an intuition answer or not, but it's very easy to show mathematically. The key is that the rate constant is not linearly proportional to the activation energy. By the Arrhenius equation, $k = Ae^{-E_\mathrm{A}/(RT)}$, so the rate constant is dependent on an exponential of the activation energy. The equilibrium constant $K$ ...


4

Yes Not only is this possible it can be done while extracting electricity from the reaction. This is the key reaction in a hydrogen/oxygen fuel cell. Fuel cells have been known for well over a century and Hydrogen oxygen fuel cells have been used commercially since the 1930s. Platinum is often used as a catalyst for the key reaction in the cells.


4

Despite the absurd lack of data, this problem attracted my attention, probably because it closer resembles a real-life challenge rather than a textbook problem. It looks like it's an adaptation of the problem from Schaum’s Outline of Theory and Problems of College Chemistry [1, p. 39]: 3.37 What is the empirical formula of a catalyst that can be used in ...


4

As far as I am aware, the selection of the best ligands for transition-metal catalysed reactions is still trial and error. In fact, not only is the selection of ligand trial and error but so is the choice of starting materials to result in that metal–ligand combination. What the original discoverer of a reaction method typically do is run a number of ...


4

All of the paths given by the OP can lead to discovery of catalysts. Usually, it is a combination of strategies and an accumulation of knowledge from different sources. I will give two examples, ammonia synthesis and enzymatic reactions. Ammonia synthesis $$\ce{N2(g) + 3H2(g) <=> 3NH3(g)}$$ The synthesis of ammonia from the elements is one of the ...


3

Alumina absorbs water to give aluminium oxide hydroxide, a solid which removes the water from the reaction environment. According to the paper by Pethrick et al. [1]: Aluminium oxide exists in several forms which on addition of water give a complex range of oxide-hydroxide; boehmite, bayerite and gibbsite. References Pethrick, R. A.; Hayward, D.; ...


3

What is the crystal structure of bismuth oxyhydroxyphosphate (BOHP)? Not exactly sure what an answer looks like so here is everything. First off looking at the diffraction pattern from the paper, bismuth oxyhydroxyphosphate matches that of petitjeanite so that is definately the prototype structure. Using the information about diffraction, from the ICDD ...


2

In those preparations, the foam arises from the reaction between water and isocyanate, which leads to the production of CO2 via formation of a carbamic acid, which upon decarboxylation generates the blowing agent, CO2. The trick would be that of making water more reactive towards the isocyanate, without (as expected by a catalyst) partecipating actively in ...


2

The activation energy of a single reaction does not change with time but there might be more than one reaction happening An activation energy is always constant for a given single reaction. But that doesn't mean the reaction speed will be constant (it might depend on concentration or temperature). But what about autocatalytic reactions? There are two ...


2

The conceptually easiest case is that of a positively charged active site with a negatively charged substrate. The substrate (i.e. reactant) enters the active site with kinetics that are faster than diffusion because there is a long-range electrostatic interaction. Catalase works near the diffusion limit. The document the OP cites is a hypothesis, without ...


2

As Curt F. has correctly showed that, in general, if the rate of reaction is plotted against time for an autocatalytic reaction, it would be the shape of curve B (in your question). Since Curt F. has already worked with the mathematical part, only I can do is show you some literature evidence. One of research on an epoxy resin composite hardening (Ref.1) ...


2

Catalysts generally decrease the electrode overpotential, which leads to decreasing the needed voltage to perform electrolysis at given current densities toward its minimal theoretical value. If higher current is possible at the same voltage due the catalyst, the lower voltage is needed for the same current, both leading to better efficiency. As the energy ...


2

Heating hydrate copper sulfate at 105 degrees does nothing. The chemically bound water takes a higher temperature to remove all water molecules (water is lost in steps as a function of temperature). This low temperature removes only two water molecules out of five. I think the major role of copper sulfate, besides being a gentle dessicant, is to oxidize ...


2

I suspect that you could be forming an aqueous solution of $\ce{(NH4)2PtCl6}$. I would like to make an observation as a person who has done platinum group metal chemistry, the anionic chloro complexes of PGMs like $\ce{PdCl4^2-}$ and $\ce{PtCl4^2-}$ are very toxic. They can induce a nasty allergy to PGMs. I would suggest that you do not work at home with ...


2

If the substrate still binds to the protein, other parts of the mechanism are still in place. Specifically, the main chain amide groups that help to stabilize the tetrahedral intermediate are still present. In the absence of serine, water will attack the carbonyl. In the presence of serine and the absence of the histidine, serine will still be the ...


1

Below is one way to write it that shows what the enzyme does and avoids the uncertainty of the details: $$\ce{E + H2O2 -> E-O + H2O}\tag{1}$$ $$\ce{E-O + H2O2 -> E-O2 + H2O}\tag{2}$$ $$\ce{E-O2 -> E + O2}\tag{3}$$ In words, the enzyme binds to the first molecule of hydrogen peroxide, release one molecule of water and holds on to the other oxygen ...


1

The decomposition of H2O2 is complex and has been the source of several studies. On iron oxide surface, here is a cited work `Catalytic Decomposition of Hydrogen Peroxide on Iron Oxide: Kinetics, Mechanism, and Implications’. Some selected quotes: "As depicted in Figure 2, the decomposition rate of H2O2 appears to be independent of the goethite particle ...


1

The rate of the reaction between a solute and a solid catalyst cannot be defined in the usual way, namely with an expression which is proportional to the concentration of the catalyst. Such a rate is proportional to the surface of the catalyst. And the surface of a powder is a parameter difficult to know with precision


1

Manganese(IV) oxide decomposes around 500 °C (German Wikipedia lists $\pu{450 °C},$ [1, p. 396] lists $\pu{530–585 °C}):$ $$\ce{4 MnO2 ->[\sim\pu{500 °C}] 2 Mn2O3 + O2}\label{rxn:1}\tag{R1}$$ On the other hand, according to [2, p. 12], $\ce{Mn3O4}$ reacts with oxygen only above $\pu{500 °C}$ and forms manganese(III) oxide: $$\ce{4 Mn3O4 + O2 ->[>\...


1

Platinum is a good catalyst for this reaction. This technique is well known. It has been developed in the 19th century by Döbereiner.


1

Let's suppose the mechanism is: $$\ce{A + B <=> (AB) -> B + B}$$ Here, A is decomposed catalytically by B into B through the formation of a transient (AB) complex. We can decompose two-step process into two reactions: $$\ce{A + B <=>[k_1][k_2] (AB)} $$ $$\ce{(AB) ->[k_3] B + B}$$ Supposing these are "elementary" reactions, then, $$\...


1

You may follow the procedure in this reference. I included the abstract for your convenience: Abstract: The hydroamination of styrene with aniline catalyzed by phosphine-ligated palladium triflates exhibits a substantial $\ce{^{13}C}$ isotope effect at the benzylic carbon. This supports rate-determining nucleophilic attack of amine on a $\eta^3$-phenethyl ...


1

You are asking a quantitative question, but you are hoping for an answer besides the "mathematical explanations" already on the site. That might be impossible, but here is an attempt to talk about it without explicitly using the Arrhenius relationship. It seems that if you add a catalyst it should decrease the activation energy by a constant for both ...


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