Jump to content

Equalibrium v. completion

Featured Replies

So in referring to certain types of reactions as either reactions that reach chemical equilibrium or go to completion. The problem I have is that equilibrium reactions can have extremely high equilibrium constant values, like the reaction 2H2 +O2 <=> 2H2O which has a constant of something like 10^80 if I'm not wrong. So is it correct to say that all reactions are in reality equilibrium reactions, it's just that some have an extremely small concentration of reactants when the reaction reaches equilibrium?

So in referring to certain types of reactions as either reactions that reach chemical equilibrium or go to completion. The problem I have is that equilibrium reactions can have extremely high equilibrium constant values, like the reaction 2H2 +O2 <=> 2H2O which has a constant of something like 10^80 if I'm not wrong. So is it correct to say that all reactions are in reality equilibrium reactions, it's just that some have an extremely small concentration of reactants when the reaction reaches equilibrium?

Yes, that is a way to see it (and that's the way I approach it myself too).

In theory, every reaction possible is an equilibrium. The reaction can simply just go back the same way in came (which is why catalysts don't alter the position of equilibrium). However, some reactions lie so far to one side of the reaction that you might as well just say that they are complete.

 

Burning hexane for example is a reaction where the reverse basically doesn't happen. If you think about it, trying to get 6 carbon dioxide molecules are 12 water moelcuels to meet at exactly the right time is very unlikely.

Excellent question,

 

The law of microscopic reversibility guarantees that every process is at least theoretically reversible. That is true, however, it does not guarantee that every process will be reversible in practice. Most chemical reactions that are not, in practice, reversible suffer from a very high kinetic barrier for the reverse reaction, or extremely disfavorable entropy/enthalpy changes for the reverse product.

 

Some molecules will decompose backward from intermediate or transition state back to reactants. That's why physical chemists freak out when biochemists ignore the [math] k_{-1} [/math] (reverse rate constant for the first step) in Michaelis-Menton enzyme kinetics. You can evoke what is called the steady state assumption though: the concentration of the intermediate is roughly constant after the reaction gets going so that [math] \frac {d^{2}[X]}{dt^{2}} [/math] is zero; for any reaction with an intermediate and at least two elementary steps.

Edited by mississippichem

Archived

This topic is now archived and is closed to further replies.

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.

Configure browser push notifications

Chrome (Android)
  1. Tap the lock icon next to the address bar.
  2. Tap Permissions → Notifications.
  3. Adjust your preference.
Chrome (Desktop)
  1. Click the padlock icon in the address bar.
  2. Select Site settings.
  3. Find Notifications and adjust your preference.