There are reactions with large Delta G negative values. Why these reactions are irreversible? As in: out of 10 steps of Glycolysis, 3 are irreversible steps. I need an explanation for why they are irreversible.
$\begingroup$
$\endgroup$
2
asked Apr 7, 2014 at 23:28
-
$\begingroup$ About which reactions do you think? $\endgroup$– Chris ♦Apr 8, 2014 at 6:44
-
$\begingroup$ A short answer is that they are irreversible because in the conditions of the cell they happen far from equilibrium (which is actually what having a "large \Delta G" means). $\endgroup$– The QuarkMay 8, 2022 at 7:52
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
7
The term "irreversible" means that the reverse reaction occurs so rarely that it is considered negligible. This means that you do not have to consider equilibrium, as you have to for reversible reactions. Instead, you can assume that all of the reactants will eventually become product.
As you stated, this is true for reactions that have a very negative Gibbs free energy. Remember the formulation of the Gibbs free energy in terms of the equilibrium constant:
$$
\Delta G^o = -RTln(K)
$$
This can be rewritten as follows: $$ K = e^{-\Delta G^o/RT} $$
As $\Delta G^o$ gets very negative, you can see that $K$ gets very positive.
The equlibrium constant has two definitions that are both useful. First, it is the ratio of the products to the reactants. Therefore as $K$ gets very positive, the equilibrium ratio of products to reactants approaches infinity, implying that all of the reactants will be consumed.
More useful is the definition of the equilibrium constant as the ratio of the rates of the forward and reverse reaction. From this definition we can see that as $K$ gets very large, the ratio of the rates of the forward and reverse reaction approaches infinity. This means that the rate of the reverse reaction becomes negligible compared to the rate of the forward reaction. This is exactly what it means to be irreversible.
Note that it is possible to make (what I think is) a more convincing argument using the concept of activation energy, using Hammond's Postulate to describe the character of the transition state. If you'd like, I can write this out. However, I think the argument given above is more easily understood with a basic chemistry background.
-
$\begingroup$ yes! all chemical reactions are reversible. its not said enough in frosh chem. $\endgroup$– shigetaApr 8, 2014 at 13:56
-
$\begingroup$ I understood it properly, its very helpful but if you are comfortable to explain by using the concept of activation energy then, Then i want to understand that concept too.. :) $\endgroup$ Apr 8, 2014 at 23:22
-
2$\begingroup$ @katherinebridges Have a look into the Wikipedia on activation energy. It contains an image which explains the concept pretty nicely. Basically most chemical reactions (even those which set free energy) doesn't start on its own. To do so, they need an initial amount of energy to start. Think of a sparkler. It contains chemicals which are capable of setting free a lot of energy when reacting, but under normal conditionss nothing happens. If you add the activation energy through a match, the reaction will go off and set off a lot of energy. $\endgroup$– Chris ♦Apr 9, 2014 at 18:07
-
$\begingroup$ @Chris Ok, I will consult wikipedia to understand that concept,Thanx for the help.. $\endgroup$ Apr 10, 2014 at 14:06
-
$\begingroup$ "[...] you can assume that all of the reactants will eventually become product" this characterises a total reaction. A total reaction is irreversible by definition, but the opposite is not necessarily true. $\endgroup$ May 8, 2022 at 7:48
$\begingroup$
$\endgroup$
Whether a reaction is reversible or irreversible, in the presence or absence of an enzyme, is not related to activation energy. An enzyme lowers activation energy in both directions.
A reaction is reversible or irreversible depending on the conditions in which this reaction occurs. If there's a sustained high concentration of substrates, most of the reaction flux goes from substrates to products, and the reaction is practically irreversible. The same happens the other way. Note that whether the concentrations are high enough or not depends on the kinetic constants involved.
