6 replies to this topic

[daraj]

Hi, I need more understanding on this topic.Pl help.

 

I have known fairly well about kinetically vs mass transfer controlled reactions in industrial reactors. Lot of reactors end up having some mass transfer limitations. But recently I came across a third angle to this , that of thermodynamically controlled reactions.

 That is reactions where equilibrium is quickly established, faster than typical reaction times, such that it is thermodynamics that controls the process and not kientics.

 

I want to know following:

 

1. How common is thermodynamic controlled reactions in industrial reactors/processes? any examples?

 

2. "every reaction a priori starts kinetically controlled" says a source. It makes sense.

If you have a reversible A=B, then FOR reverse reaction to take place you need some B to form first.

So logically, kinetics for forward reaction has to controlling first and only then equilibrium/thermo control can take over.

so how is a purely thermodynamic control regime ever possible?

 

Iam confused by this concept. Please help.

At a fundamental level Iam trying to understand when, where and how thermodynamic comtrol happens

 

 

 

 

 

[Lucian Gomoescu]

Hello,


Let's take a generic and simple example to illustrate the concept:

The following reversible reactions take please in liquid phase of a CSTR:

A (liq) <=1=> B(liq) <=2=> C(liq)

Feed is pure A.
Stability of components is C>>B>A, which is to say that chemical  equilibrium constants are Keq1<<Keq2. So thermodinamically, C is favoured vs B.

Kinetic constant of reaction 1 (forward)  is much higher than that of reaction 2 (forward). So kinetically, B is favoured vs C.

Now, it can be decided either
a) to produce more B than C by having a small residence time or/and low temperature such that reaction 2 is far from chemical equilibrium. In this case, it is said that kinetic reaction control is used.

 

,either
to produce more C than B by having a large residence time or/and high temperature, allowing the reaction 2 to get close to chemical equilibrium. In this case, it is said that thermodynamic reaction control is used.

Observation: B is called the kinetic product and C is called the thermodynamic product.

A good real example is represented by: addition of halogenic acids to alcadienes and Diels-Alder reactions.

 

Regards,
Lucian

 

 

[daraj]

Thanks Lucian. Thats a very good example.

 

For  a gas solid type heterogeneous catalytic reaction, say of type  A<->B which is reversible, I want to understand the following:

 

1. Is it possible just by looking at reported Keq numbers, to qualitatively judge whether the reaction is going to be limited by equilibrium or not? I mean, is there any thumb rule to know what values of K or what is the lower limit of K  for the forward reaction go further

I know the equation that connects delGa with Ln(K), but I do not have a qualitative feel for these numbers. Just by looking at the value of K I cannot say whether the reaction is severly equilibrium limited

 

2. Again for gas-solid catalytic reactions in industrial scale, I have heard people say that they are not commonly thermodynamically controlled and are most often kinetically controlled. Is that true?

 

3. what role does mass ransfer limitations(internal especially) play in changing the control regimes?

 

Thanks

[Lucian Gomoescu]

Hello Daraj,

 

You're welcome.

1. When a chemical reaction is at chemical equilibrium state, the forward and reverse reaction rates are equal, which is to say
k_f*P_A,eq=k_r*P_B,eq (let's assume ideal gas phase, so use pressures instead of fugacities for A and B concentration expression).

Further, the expression can be modified to: k_f/k_r=P_B,eq/P_A,eq. These ratios are the reaction chemical equilibrium constant, Keq.
So, if Keq>1 then, at equilibrium, the products pressures product (i.e. P_B,eq) will be higher than the one of reactants because forward reaction rate constant, k_f, is higher than the reverse reaction rate constant, k_r.

Keq gives no clue about how fast are the forward and reverse reactions, but shows how fast is the forward reaction compared to the reverse one.

 

Van't Hoff equation gives Keq dependence on temperature.

It can be told if a reaction is  limited in anyway, just by knowing Keq.

 

I want to point out that kinetic/thermodynamic control doesn't apply to a system with only one elementary reaction. Instead there should be at least two elementary reactions, from which one at equilibrium and the other, reversible or not, should have a rate constant much higher than the other forward reaction.

2. I don't know, but since there are catalyst involved (which don't affect Keq, instead increase both rate constants of some reactions) , i assume that kinetic control is the one used, if applicable (see 1).

3. Let's take an example. Assume that the kinetic product, X, is the one of interest and for high yield of X and low yield in Y (the thermodynamic product) a porous catalyst is used.
The internal structure of catalyst eases internal diffusion of reactants and products such that formation rate of X is high enough to avoid formation of too much Y (succesfull kinetic control of reaction system).
From various reasons, let's assume that pores get plug and internal active sites of catalyst are starved because of poor internal diffusion such that the formation of X is slowed and formation of too much Y is not avoided (failed kinetic control of reaction system).

So, when it comes to control regime, mass transfer has the same importance as kinetics.

 

[daraj]

Thanks again for a detailed response Lucian.

 

I welcome throughts of all others as well as I feel this is an interesting as well as important topic

[daraj]

Lucian, one more thought/question. Others, pl free to contribute. what you have mentioned are broad differences between the 2 control regimes giving extreme examples.

 

But, Is it possible that in reality a mix of Kinetic and thermo controls compete and exist at the same time?

For instance in the reaction

SO2+O2 = SO3, WHICH IS EXOTHERMIC, lower temp favors the product.

 

But lower temp also causes reaction to proceed too slowly due to kinetic reasons. Hence a :"compromise" temperature of 400-450C is generally used for faster throughputs, which is not very low, but at the same time not very high enough to cause serious equilibrium limitations. I got this info from open literature.

 

So if somebody gives me the above reaction , the catalyst used and actual reactor conditions like temp (450C) , how will I know which control regime it is being operated under? Is there a methodology?

 

1. Assume that I have evaluated the K(equilibrium constant) at that temperature as well as know the kinetic rate/rate constants.

2. In a second scenario I know the former but not the latter, i.e kientic rate constant is not precisely known but Keq is known from open literature.

 

for both above scenarios pl let me know the approach

[daraj]

Hi Lucian, also so far we have assumed different products being formed from different control regimes. But in the above scheme we have just one product SO3 being formed. So for this case how to know the ratios of the two regimes? if the products were different from different regimes you can measure the ratio of the 2 products at the outlet to see which regime is more controlling.

 

will a plot of kinetic rate constant and equilibium constant vs reaction temp reveal anything?