Sciencemadness Discussion Board

how the inverse reaction could happen while the original reaction is happening from the view of thermodynamics?

ilwdx - 17-4-2004 at 06:01

for a equilibrium like

A + B = C DG=-33kj/mol
so the inverse reaction
C = A + B DG= 33kj/mol
which means the inverse reaction cannot happen spontaneously.

So HOW the inverse reaction in the equilibrium could happen? I'm really puzzled. Would someone tell?thx!~~~~~

[Edited on 17-4-2004 by ilwdx]

vulture - 17-4-2004 at 06:56

DG = - RT ln K at the equilibrium point.

If you calculate K from this you'll see this is an equilibrium reaction.

Another way to prove it is through kinetics, but this is somewhat more advanced.

Proteios - 17-4-2004 at 07:33

***** -state 1
|
|
|-difference in energy between 2 states DG
|
|
***** -state 2

the population of the two states at eq will just reflect the botlzman distribution for that Temp. These are state functions. It doesnt matter which way around you draw it.

'the reactions dont proceed unless the have a -ve DG' usually only refers to 'irreversible' reactions where the DG is far larger than the energy typically availabe at room Temp (KbT).

ilwdx - 17-4-2004 at 07:36

Quote:
Originally posted by vulture
DG = - RT ln K at the equilibrium point.

If you calculate K from this you'll see this is an equilibrium reaction.

Another way to prove it is through kinetics, but this is somewhat more advanced.


o...I still don't understand
my puzzle is why an reaction possessing a positive DG can happen as shown in my previous example.

Proteios - 17-4-2004 at 07:51

hmmm lets try again.

Reactions with +ve DGs dont happen is not true. They happen but you have to put energy in.

all reactions are microscopically reversible.

The equilibirium constant is just a numerical experssion for the populations of two states (defined by the difference in energy of the two states and energy available to the molecules (typically a boltzman distrubtion)).

If DG is very large compared to the typical energy avail. to the molecules, the reaction is typically termed irreversible (H2 + O2 for instance). However even here equilibrium constant can be defined, its just from a practical standpoint its kinda pointless.

Marvin - 17-4-2004 at 10:37

If A + B => C liberates energy, then if you react a large amount of it, this energy is around in the reaction mixture. So there is nothing to stop some of it being absorbed by some C that was produced earlier on, and converting it back into A + B.

If you take all this energy out, then eventually, C will be produced until either A or B has been depleted.

A slightly more formal explanation, is that macroscopically delta G depends on the temperature of the reaction mixture, if A + B => C liberates energy, then the temperature of the reaction mixture will rise as it goes forwards. Obviosly this cannot produce a higher temperature on avarage, becuase as the sign changes the reaction stops. But you can take this 'large' reaction and split it into a large number of smaller reactions.

If you do spit this macroscopic reaction into a large number of microscopic ones, say only a few molecules each, then there becomes a very good chance one or more of these 'reactions' will have a high enough 'temperature' to have a negative DG on the back reaction. So microscopically the reaction will go backwards to some extent so long as there are a few molecules with enough energy.

As Proteios is saying, its a question of scale.

ilwdx - 17-4-2004 at 10:42

Quote:
Originally posted by Proteios
hmmm lets try again.

Reactions with +ve DGs dont happen is not true. They happen but you have to put energy in.

all reactions are microscopically reversible.

The equilibirium constant is just a numerical experssion for the populations of two states (defined by the difference in energy of the two states and energy available to the molecules (typically a boltzman distrubtion)).

If DG is very large compared to the typical energy avail. to the molecules, the reaction is typically termed irreversible (H2 + O2 for instance). However even here equilibrium constant can be defined, its just from a practical standpoint its kinda pointless.


you clear up lots of things~~.

ilwdx - 24-4-2004 at 20:44

I've reached an understanding:

1. C absorbs energy from A + B-> (or from the environment) and reacts(however DG of decomposing C is +ev)

2. (for DG = DH - T*DS) A + B-> liberates energy and thus a small number of C have a higher "temperature" enough to made the DG of decomposing C negative. And C reacts.

right?