Equilibrium Constant and Rate Constant

[Dr Gerrard]

So what is the relation between these two. Is there any?

Rxn is aA + bB -> cC + dD

I think the equilibrium constant is always [C]^c * [D]^d / [A]^a * ^b

Correct? According to the Law of Mass Action, right?

However, EK makes the distinction between elementary and complex reactions.

The forward rate is only equal to k[A]^a^b if the reaction is elementary. If not, the exponents cannot be found. Correct?

The same can be said with the reverse rate, but with the products instead of the reactants.

So, does that mean Keq = Kfor/Krev ONLY if the reaction is elementary?

EK, on Page 57 of Lecture 3 in Chemistry, on the little red thing on the side, says that a catalyst changes rate constants by the same proportion so it also does not change the equilibrium constant. I know that this is true, but how can they make the assumption that the rate constants and equilibrium constants are related in this way?

Is this because an enzyme only catalyzes an elementary reaction at a time? Thus, when an enzyme catalyzes a complex reaction, it does both of the elementary reactions separately?

---

Members don't see this ad.

 

[sleepy425]

Well, let's try to figure this out:

Let's take this reaction (we'll leave out stoichiometry/exponents to make it simpler):

A+B+C<--->P+Q

(As always, <----> is going to represent equilibrium arrows, even though this is actually the resonance arrow)

So Keq for this, the way it's written, is [P][Q]/([A][C]).

But let's say the reaction is really made of the following elementary reactions:


A+B+C<--1-->ABC<--2-->PQ<--3-->P+Q

So basically, A, B, and C bind to form the ABC complex, then ABC reacts chemically to form the PQ complex, then the PQ complex dissociates to P and Q. I put numbers (1,2, and 3) to denote each equilibrium arrow.

Now, the rate limiting step of this reaction is step 2 (I'm just making this up, it could be any of the steps, although in most cases, the chemical step is the rate limiting step). Since step 2 is rate limiting, the forward rate can be written as RateF=k2[ABC] where k2 is the forward rate constant. The reverse rate can be written as RateR=k-2[PQ] where k-2 is the reverse rate constant.

Now, equilibrium occurs when the forward rate and the reverse rate are equal, so RateF=RateR

so: k2[ABC]=k-2[PQ]

rewriting: [PQ]/[ABC]=k2/k-2

Since [PQ]/[ABC] is the equilibrium constant for the reaction ABC<---->PQ, that means that k2/k-2 is also that equilibrium constant, so we can call k2/k-2 K2 (note the capital letter K which denotes equilibrium constant, while lowercase letters denote rate constants).

Ok, so K2=[PQ]/[ABC].

But PQ and ABC are intermediates, we don't add them to make the reaction go, we add A, B, and C separately, and it makes P and Q, so we've got to rewrite all of this in terms of A, B, C, P, and Q. How can we do that? Well, we're going to take what's called the rapid equilibrium assumption (it basically says that the rate limiting step is slow enough that all of the other steps come to equilibrium quickly and then just chill until the rate limiting step occurs. It can't always be taken, but for many cases, including many enzymatic reactions, it is valid, and is the simplest way to do it).

Ok, so, if we have rapid equilibrium, then:

for equilibrium 1:

K1=[ABC]/([A][C])

(remember, K1 is an equilibrium constant)
rewriting:

[ABC]=K1[A][C]

for equilibrium 3:

K3=[P][Q]/[PQ]

rewriting:

[PQ]=[P][Q]/K3

Ok, now we said that K2=[PQ]/[ABC], so we can plug in what we just found for [PQ] and [ABC]:

K2=[P][Q]/(K1K3[A][C])

rewriting:

K1*K2*K3=[P][Q]/([A][C])

but remember, we defined [P][Q]/([A][C]) as the overall Keq, so that means:

Keq=K1*K2*K3.

So in this case, even though the forward rate constant for the reaction is just k2 and the reverse rate constant for the reaction is just k-2, the equilibrium constant is not k2/k-2, it's K1*K2*K3, which is:

k1*k2*k3/(k-1*k-2*k-3), so in other words, it's the product of the ratio of the rate constants for each step of the reaction.

It turns out that this result can be extended to all multi-step equlibrium reactions: The equlibrium constant is the product of all of the forward rate constants, divided by the product of all of the reverse rate constants.

For enzymes, each forward rate constant is changed by the same proportion as the corresponding reverse rate constant. So the product of all of these forward and reverse constants will be the same as if the enzyme weren't present

 

[sokol8]

In the enzyme there are two redox sites. The first site accepts electron and passes it to the second site, where the substrate is reduced to product using the electron. Both redox sites stay in thermodynamic equilibrium of G=0, no driving force, Keq = 0 (the redox potential of both sites is the same). According to the equilibrium constant (thermodynamics), the forward rate of electron transfer from the first site to the second site is equal to backward rate, k1 = k-1. Are these kinetic rates already determined by thermodynamics equilibrium as equal, or they can still be different, for example k1 is 10x faster than and k-1? If yes, is it possible when thermodynamics equilibrium stay constant during reaction, I mean there are no new equlibrium set between. Does the equilibrium constant always say about the ratio between the forward and backward reaction rates, but it does not give the actual numbers, which you tha get from kinetic measurements?

In other words, when I assume thermodynamic equilibrium of G=0, no driving force, Keq = 0 between A and C (and I also assume there is no B), does it mean that k1 forward rate of 100s-1 has to be equal to k-1 backward rate of 100s-1, or maybe one of these rates can be 1000s-1 and the other one 100s-1 due to some kinetic effects, not related to thermodyanmics?

 

[fiedel]

The Law of Mass Action explains. Rate of forward reaction is Kf times concentarations of reactants in respective powers, rate of reverse reaction is Kr times concentrations of products in respective powers. In equilibrium both rates are equal, so you get Kc = Kf/Kr if you write it up.





_____________________________

Education Consultants in India http://www.imtpconsultants.com/

 

[sokol8]

The Law of Mass Action explains. Rate of forward reaction is Kf times concentarations of reactants in respective powers, rate of reverse reaction is Kr times concentrations of products in respective powers. In equilibrium both rates are equal, so you get Kc = Kf/Kr if you write it up.

OK, so if I only have A = B and K = 5 = kf/kr, so the equilibrium lies on right. I understand that this places you in situation where you have 5x more product than substrate. Once this situation is reached the equilibrium sets with kf=kr, so there is some small dynamic conversion of B to A and than back from A to B, but in general we stay at the ratio of products/substrates = 5 right? Why do we say than that at K=5, the forward rate is 5x faster than backward rate, e.g. kf = 10s-1 and kr = 2s-1? First this is not equilibrium, second if we continue with this rates we will quickly run out of A and the equilibrium will never set. Are we allowed to talk about thermodynamic equilibrium of K=5 between both anyway, or we need to say that the actual equilibrium sets/lies on right side?

 

[kalilza]

The Law of Mass Action explains. Rate of forward reaction is Kf times concentarations of reactants in respective powers, rate of reverse reaction is Kr times concentrations of products in respective powers. In equilibrium both rates are equal, so you get Kc = Kf/Kr if you write it up.

OK, so if I only have A = B and K = 5 = kf/kr, so the equilibrium lies on right. I understand that this places you in situation where you have 5x more product than substrate. Once this situation is reached the equilibrium sets with kf=kr, so there is some small dynamic conversion of B to A and than back from A to B, but in general we stay at the ratio of products/substrates = 5 right? Why do we say than that at K=5, the forward rate is 5x faster than backward rate, e.g. kf = 10s-1 and kr = 2s-1? First this is not equilibrium, second if we continue with this rates we will quickly run out of A and the equilibrium will never set. Are we allowed to talk about thermodynamic equilibrium of K=5 between both anyway, or we need to say that the actual equilibrium sets/lies on right side?

First of all, K=5=kf/kr is only true for single step reactions (or elementary ones), so be careful where you apply this. Second, I think where you are confused is in your interpretation of kf and kr. Remember, they are rate CONSTANTS, not the actual rates of the reactions. Keeping this in mind, and using your numbers:

in the reaction A<---->B,

The actual RATES are:
RateF (forward)=kf[A]
RateR(Reverse)=kr
(remember: kr and kf are constants, not the actual rates).

If K=5=kf/kr=/[A], then there is 5x more than [A] at equilibrium (or =5[A]), and kf is 5x bigger than kr (or kr=kf/5). Plugging the relative numbers in parenthesis into the rate laws above:
RateF=kf[A]
RateR=(kf/5)*(5[A])=kf[A]

This shows that at equilibrium, the forward and reverse reaction rates are equal, even though kf and kr are not.

 

[sokol8]

First of all, K=5=kf/kr is only true for single step reactions (or elementary ones), so be careful where you apply this. Second, I think where you are confused is in your interpretation of kf and kr. Remember, they are rate CONSTANTS, not the actual rates of the reactions. Keeping this in mind, and using your numbers:

in the reaction A<---->B,

The actual RATES are:
RateF (forward)=kf[A]
RateR(Reverse)=kr
(remember: kr and kf are constants, not the actual rates).

If K=5=kf/kr=/[A], then there is 5x more than [A] at equilibrium (or =5[A]), and kf is 5x bigger than kr (or kr=kf/5). Plugging the relative numbers in parenthesis into the rate laws above:
RateF=kf[A]
RateR=(kf/5)*(5[A])=kf[A]

This shows that at equilibrium, the forward and reverse reaction rates are equal, even though kf and kr are not.

 

[ajaxme]

In other words, when I assume thermodynamic equilibrium of G=0, no driving force, Keq = 0 between A and C (and I also assume there is no B)