[Schrödinger]

Hey guys. I have a problem understanding the thermodynamics behind the transition state theory.

I know that we assume that there exists an equilibrium between the reactants and the transition state(T.S)
I also understand that we also have the following relation:

$$\Delta G = \Delta H - T\Delta S /$$

where $$\Delta H /$$ is the enthalpy/heat of activation.
and $$\Delta S /$$ is the entropy of activation.

$$\Delta H > 0 /$$ or $$\Delta H < 0 /$$ depending on the reaction.

I also know that $$\Delta S /$$ is a measure of randomness/freedom in a system.

I read that $$\Delta S > 0 /$$ or $$\Delta S < 0 /$$ depending on the transition state.
This is the part I couldn't compehend.
Since the transition state more complex than the reactants, shouldn't it be more strained with its structure? Hence, shouldn't $$\Delta S < 0 /$$ always?

For example, should it not be easier for free rotation to happen in a methane molecule rather than in a methane-chlorine T.S during halogenation of methane, because there are too many species involved in the T.S?

[Grundalizer]

Only deals with products and reactants, not transition states. You are correct, the number of possible transition states can be high, and their individual entropy is basically uncalculatable.  I've never seen much literature about any machines or experiments that allowed the scientists to actually capture or directly observe a transition state complex and take measurements on it.  We can predict and guess at what they are, to help us understand how a reaction proceeds, but they are usually so unstable they turn into the product within micro/nano/femto? seconds.

I read that or depending on the transition state.
This is the part I couldn't compehend.

"depending on the transition state" is saying, does the transition state decay into a more stable product molecule(exothermic most likely, entropy decreases), or a less stable product (endothermic most likely, entropy increases)

So  :delta:S implies a change (products minus reactants) not just one species (transition state).

You have to be able to compare entropies of two systems, otherwise you have no frame of reference.  It's like saying, this soup is hot.  Ok...compared to what?  The surface of the Sun or this icecube

We can measure energies of rotations in stable molecules, but we can't measure them in a transition state complex because simply measuring them might push them to decay to the product/back to the reactant.

I think you are correct in saying that transition state complexes are more strained than the reactants though.

[Yggdrasil]

The relation ΔG = ΔH - TΔS also applies to the transition state complex.  We can measure these values by measuring the rate constant (k) of the reaction as a function of temperature.  You can then plot ln(k/T) vs 1/T, fit the plot to a line, and determine the enthalpy and entropy from the Eyring equation.

In general, the free energy of the transition state is much higher than the free energy of the reactants, so ΔG >> 0.  In general, this will imply that ΔH > 0 and ΔS < 0.  However, there is no physical law that I know preventing ΔS > 0.  For example, in a decomposition reaction (e.g. AB --> A+B), it is possible that the transition state may have more entropy than the reactants.

[Schrödinger]

Only deals with products and reactants, not transition states.

Please refer Morrison and Boyd.

You are correct, the number of possible transition states can be high, and their individual entropy is basically uncalculatable.  I've never seen much literature about any machines or experiments that allowed the scientists to actually capture or directly observe a transition state complex and take measurements on it.  We can predict and guess at what they are, to help us understand how a reaction proceeds, but they are usually so unstable they turn into the product within micro/nano/femto? seconds.

I don't think you understood my question. I'd like you to read my post again.

[Schrödinger]

In general, the free energy of the transition state is much higher than the free energy of the reactants, so ΔG >> 0.  In general, this will imply that ΔH > 0 and ΔS < 0.  However, there is no physical law that I know preventing ΔS > 0.  For example, in a decomposition reaction (e.g. AB --> A+B), it is possible that the transition state may have more entropy than the reactants.

If ΔG>>0 , then there will be no reaction.

[Yggdrasil]

By ΔG I mean , the difference in free energy between the transition state and the reactants.  But, yes, if  the reaction will be very slow.  Perhaps it's better to just state that for any reaction, .

[Schrödinger]

By ΔG I mean , the difference in free energy between the transition state and the reactants.  But, yes, if  the reaction will be very slow.  Perhaps it's better to just state that for any reaction, .

If  , then no reaction can take place right?
Isn't a condition for spontaneity of a reaction?






sorry, i didn't know how to put that superscript on G, but you know what i mean.

[Oxy]

If  , then no reaction can take place right?
Isn't a condition for spontaneity of a reaction?
sorry, i didn't know how to put that superscript on G, but you know what i mean.

When you talk about of a reaction, it's of standard condition or , which fugacities or activities of any reactants are 1.
But condition for spontaneity isn't . It's :

Q_r is reaction quotient.
If , reaction will occur. The same for transition state .

[Yggdrasil]

If  , then no reaction can take place right?
Isn't a condition for spontaneity of a reaction?

sorry, i didn't know how to put that superscript on G, but you know what i mean.

Please refer to the following diagram from wikipedia:

http://en.wikipedia.org/wiki/File:Rxn_coordinate_diagram_5.PNG

When examining the spontaneity of a reaction, you need to look at the free energies of the reactants and products.  If the free energy of the reactants are higher than the free energy of the products, the reaction is thermodynamically favorable and can proceed spontaneously.  However, this does not necessarily mean that you will see the accumulation of product.  While the ΔG of the reaction can tell you whether it is possible to convert reactants to products, it does not tell you about the kinetics of the reaction (whether the reaction will occur quickly enough to observe).  For example, graphite has a lower free energy than diamond, yet you will never see diamond convert to graphite at normal temperatures because the reaction occurs much too slowly to observe.

In order to assess the kinetics of the reaction you need to look at the transition state.  What is this transition state?  For any chemical reaction, the reactants and products will be separated by an energy barrier (see image).  The molecular configuration corresponding to the highest point on this energy barrier is the transition state.  We refer to the energy difference between the transition state and the reactants as .  Note that is always greater than zero (i.e. the transition state always has a higher free energy than the reactants).  If this were not the case, the transition state would not be a barrier and the reactants would not be a chemical species.  If there were no barrier between the reactants and products, the reactants would instantaneously convert to product and there is no way you could actually isolate the reactants in the first place.

How does the molecule cross this energy barrier?  The answer is thermal energy.  Molecules that collide with enough thermal energy can use this energy to distort their molecular configuration into the configuration of the transition state.  Once the molecule crosses over this barrier, it can then relax to the lower energy of the products.  Molecules that do not have enough energy to cross the barrier will relax back to the configuration of the reactants.  Thus, for one to observe a reaction occurring on a reasonable timescale, the must be on the order of or less than the thermal energy of the system.

[Schrödinger]

Ok, let me see if I got what you were saying.

1. This $$\Delta G^{+}/$$ is not the same $$\Delta G /$$ that we deal with in thermodynamics.

2. This is a totally different theory. Here, this $$\Delta G^{+}/$$ is actually the activation energy of a reaction, just expressed in terms of other measurable thermodynamic properties.

3. As a result, the condition for spontaneity of a reaction would be $$\Delta G^{+}/$$. This is equivalent to saying E_act > 0.

4. Also, the smaller the $$\Delta G^{+}/$$, the faster the reaction. This is the same as saying smaller the E_act, the faster the reaction.

So, can we safely conclude that $$ E_{act} = \Delta H^{+} - T\Delta S^{+} ?/$$




Could you please tell me how to put that symbol on the 'G'

[Yggdrasil]

Ok, let me see if I got what you were saying.

1. This is not the same that we deal with in thermodynamics.

2. This is a totally different theory. Here, this is actually the activation energy of a reaction, just expressed in terms of other measurable thermodynamic properties.

3. As a result, the condition for spontaneity of a reaction would be . This is equivalent to saying E_act > 0.

4. Also, the smaller the , the faster the reaction. This is the same as saying smaller the E_act, the faster the reaction.

Yes, is the activation energy of the reaction.  Some of the times, people only consider the change in enthalpy as the activation energy, but it is just as valid to consider the Gibbs free energy of activation.  The value still represents a free energy, just that it represents the change from reactant to transition state not the change from reactant to product.

So, can we safely conclude that

Sometimes.  Certainly $$\Delta G^\ddag = \Delta H^\ddag -T\Delta S^\ddag /$$, but sometimes people use E_a to represent $$\Delta H^\ddag /$$.

Could you please tell me how to put that symbol on the 'G'

The latex code for the double dagger symbol is \ddag, so to write the expression for the Gibbs free energy of activation in latex, you would write \Delta G ^ \ddag

[Schrödinger]

Ok, let me see if I got what you were saying.

1. This is not the same that we deal with in thermodynamics.

2. This is a totally different theory. Here, this is actually the activation energy of a reaction, just expressed in terms of other measurable thermodynamic properties.

3. As a result, the condition for spontaneity of a reaction would be . This is equivalent to saying E_act > 0.

4. Also, the smaller the , the faster the reaction. This is the same as saying smaller the E_act, the faster the reaction.

Yes, is the activation energy of the reaction.  Some of the times, people only consider the change in enthalpy as the activation energy, but it is just as valid to consider the Gibbs free energy of activation.  The value still represents a free energy, just that it represents the change from reactant to transition state not the change from reactant to product.

So, can we safely conclude that

Sometimes.  Certainly , but sometimes people use E_a to represent .

If E_act can be represented as $$\Delta G^\ddag /$$ and also as $$\Delta H^\ddag /$$, then what is the difference between $$\Delta G^\ddag /$$ and $$\Delta H^\ddag /$$. I mean, they have got to represent two different quantities, right?

[Yggdrasil]

They do represent different quantities.  However, different books will sometimes have different conventions on how they define E_a.