How does a catalyst affect equilibrium?

In summary: Oxidation states go from 0 (no oxidation) to 8 (fully oxidized).In summary, the catalyst speeds up the reaction by increasing the rate of conversion back and forth between the reactants. The product is determined by the atom's electronegativity, with more electronegative atoms breaking weaker bonds.
  • #1
Callan Madden
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OK I clicked a bunch of similar discussions, but it still makes no sense, I barely know what a catalyst is, besides the fact that they do affect the rate of reaction but not the equilibrium expression or pressure, which just got me hella fudged up. And i don't want any over complicated explanations lmoa, but any help would be greatly appreciated.

---Also, when writing a reaction, I get really confused on how the products are set up, after that I can't easily calculate PH and such, and I do know it involves how strong or weak the acids and bases are, and if it lends protons or not, but I don't really understand how any of this works together to tell you how the products are set up
Ex; HOCl + H20 <--> H30+ + OCL- obviously this is correct but I don't understand how they got H30, and not OH, and H2OCL, any descriptions on this would help. I hope I am not restricting any guidelines cause this isn't really a problem in my hw, its just helpful in many parts to it.

Sorry if my questions sound stupid, I honestly don't know sometimes how I manage AP Chemistry, But I am always looking for more ways to succeed!
 
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  • #2
So, what your question is? Especially as you already know the answer to the one posted in the thread subject is "it doesn't"?

As to the latter question - do you ask about "how did they originally found out these things" or "how you predict what to expect"?
 
  • #3
Catalyst speeds up a reaction. If it is an equilibrium, it speeds up both the forward and backward reactions by same factor. The molecules will be converting back and forth faster but their total numbers will remain same at any time.

Acidity is based on how acids dissociate. In this case, the O-H bond is weaker than O-CL and so, the O-H bond is broken easily. This is common with most oxoacids with electronegative atoms, including sulfuric acid, nitric acid, perchloric acid and even carboxylic acids. Only when metal is involved, the M-O bond breaks (or it may take a proton from water). So, in a H-O-X system, the electronegativity of the atom X decides which bond will break.
 
  • #4
Kanesan said:
Only when metal is involved, the M-O bond breaks (or it may take a proton from water).

Explain that to permanganic acid (and many others).
 
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  • #5
Sorry Borek, I missed those! Apparently it looks like high oxidation state metal is similar to highly electronegative atom. I admit, I did not research a lot on this, simply realized this connection after your question.
This even seems to apply to electronegative atoms! Perchloric acid is stronger than hydrochloric acid...
 
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FAQ: How does a catalyst affect equilibrium?

How does a catalyst affect equilibrium?

A catalyst does not affect the equilibrium constant (K) of a reaction. However, it does decrease the time needed for the reaction to reach equilibrium, by lowering the activation energy required for the reaction to occur. This results in an increase in the rate of the forward and reverse reactions, but the ratio of products to reactants at equilibrium remains the same.

Can a catalyst shift the equilibrium position of a reaction?

No, a catalyst does not shift the equilibrium position of a reaction. It only speeds up the rate at which equilibrium is reached, but does not change the final ratio of products to reactants. The equilibrium position is determined by the relative stabilities of the reactants and products.

How does the concentration of a catalyst affect equilibrium?

The concentration of a catalyst does not have an effect on the equilibrium constant (K) of a reaction. However, increasing the concentration of a catalyst can increase the rate of the reaction, as more catalyst molecules are available to lower the activation energy. This does not change the equilibrium position, but only speeds up the process of reaching equilibrium.

Can a catalyst be used to create more products in an equilibrium reaction?

No, a catalyst cannot be used to create more products in an equilibrium reaction. As stated before, a catalyst only speeds up the rate at which equilibrium is reached, but does not change the final ratio of products to reactants. The amount of products formed is still determined by the equilibrium constant and the relative stabilities of the reactants and products.

What is the role of a catalyst in a reversible reaction?

The role of a catalyst in a reversible reaction is to speed up the rate at which equilibrium is reached, by lowering the activation energy. This allows the reaction to reach equilibrium faster and can also increase the overall yield of the reaction. However, a catalyst does not affect the equilibrium position or the equilibrium constant of a reversible reaction.

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