Catalysts and Rate Law: The Role of Catalysts in Determining Reaction Rates

[zmike]

are catalysts suppose to be in the rate law? because I was looking at a reaction of 2H2O +I -> H2O + I + O2 via a catalyst of I and was in the rate law and the justification was that

**the rate of the reaction is limited by the amount of I available?

I am really confused are catalysts suppose to be in the rate law?

thanks

 

[Ygggdrasil]

Yes. Because catalysts affect the rate of reaction, they should be included in the expression for the rate law. Catalysts do not, however, affect the overall equilibrium of the reaction and are therefore not included in the expression for the equilibrium constant.

 

[zmike]

thank you yggdrasil,

just to follow up, the question I was working on is about enzymes and my book has that

- Dependence on [E] : rate (or velocity) = k [catalyst]1
- When is low : rate (or velocity) = k 1
- When is high : rate (or velocity) = k 0 = Rmax = Vmax

Why is it that when is high, the rate no longer depends on E? and would you happen to know the difference between Vo and Ro? where Rateo= Ro=Vo

thanks

 

[Ygggdrasil]

According to the Michaelis-Menten equation, the kinetics of an enzyme catalyzed reaction obeys the following equation:

[tex]v_o = \frac {k_{cat}[E]}{K_m + }[/tex]

Where v_o represents the rate of the reaction (how many moles of substrate are converted into product per second), is the total concentration of substrate, [E] is the total concentration of enzyme, and k_t and K_m are constants describing how good the enzyme is at catalysis (k_cat is called the turnover number and K_m is the Michaelis constant).

In biochemistry, we basically assume that any product produced must have come from an enzyme catalyzed reaction (a good assumption because the rates of most uncatalyzed reactions in biochemistry are extremely slow compared to the catalyzed reaction). The first step of an enzyme catalyzed reaction is binding between the enzyme (E) and its substrate (S). Since the substrate can be converted into product only when bound by the enzyme, the rate of the reaction is proportional to the amount of enzyme-substrate complex (ES).

There is, of course, an equilibrium between free substrate (S) and enzyme-bound substrate (ES), E + S <--> ES, and this equilibrium is affected by both the concentration of substrate and the concentration of enzyme. As you increase [E], you drive the equilibrium towards producing more ES, increasing the rate of reaction. Similarly, increasing , drives the equilibrium towards producing more ES. As you increase more, however, your rate does not increase forever. At some point, all of your enzyme molecules will have a bound substrate molecule. If 100% of the enzymes have bound substrate, increasing the amount of substrate will not produce more ES (there's no more free enzyme for the substrate to bind), so the rate of reaction will not increase.

We can see this mathematically, by plugging in the condition where is high (specifically, when >> K_m) into the Michaelis-Menten equation. When is much greater than K_m, +K_m is approximately equal to . Therefore, we can simplify the equation to:

[tex]v_o = \frac {k_{cat}[E]}{} = k_{cat}[E][/tex]

Notice that at high , the rate of reaction still does depend on [E]. We call the quantity k_cat[E] as V_max because, for a fixed amount of enzyme, you cannot achieve a higher reaction rate than V_max no matter how much substrate you add to the reaction.

I don't know the specific contexts where v_o and r_o are used in your book, but they usually mean the same thing, the rate of reaction.

 

[LudusRex]

for your question. Yes, catalysts play a crucial role in determining reaction rates, and they are often included in the rate law. The rate law is an equation that describes the relationship between the concentration of reactants and the rate of a chemical reaction. Catalysts are substances that increase the rate of a reaction without being consumed in the process.

In the reaction you mentioned, the catalyst is I, and its concentration, , is included in the rate law because it affects the rate of the reaction. The justification for including in the rate law is that the rate of the reaction is limited by the amount of I available. In other words, the more I present, the faster the reaction will proceed.

Including catalysts in the rate law allows us to understand the specific role they play in increasing the rate of a reaction. Without including them, the rate law would not accurately represent the effect of the catalyst on the reaction rate.

I hope this helps clarify the role of catalysts in determining reaction rates and their inclusion in the rate law. If you have any further questions, please don't hesitate to ask.