# Relationship between Gibbs free energy, enthalpy, entropy and rate of reactions

I'm having a little trouble constructing the bigger picture of chemical reactions.

From the collision theory we know that molecules react if: 1) they collide 2) they collide bearing a sufficient amount of energy to surmount the activation energy and collide at a specific geometric orientation.

From the relationship between Gibbs free energy, enthalpy, entropy and temperature we know: $\Delta G = \Delta H - T\,\Delta S$, and that a reaction is only thermodynamically favored when $\Delta G < 0$. And even when $\Delta G < 0$ the reaction may still not occur at a measurable rate because it has too high an activation energy.

From the Arrhenius equation we know: $k = A ~\mathrm e^{-E_\mathrm a / (RT)}$

1. Reactions only occur when particles can collide with each other.
2. Colliding particles will only react if they bear sufficient energy (can be increased through temperature and/or adding other external energy sources) and have correct orientation (chances can't be increased?).
3. Only particles in thermodynamically favored reactions have enough energy to overcome $E_\mathrm a$ and so will spontaneously occur without external energy applied. But they may occur extremely slowly if they have a very high $E_\mathrm a$ (according to the Arrhenius equation).

Is this reasoning correct?

• You're forgetting about equilibrium ΔG=-RTlnK – Mithoron Apr 26 '16 at 0:13

I'm having a little trouble constructing the bigger picture of chemical reactions.

Let's go step by step.

From the collision theory we know that molecules react if: 1) they collide 2) they collide bearing a sufficient amount of energy to surmount the activation energy and collide at a specific geometric orientation.

Yes, in order to react, molecules taking part in a (bimolecular) reaction need to:

1. be in contact;
2. be correctly aligned to each other (a random alignment may not react);
3. have enough energy to overcome the activation barrier ($\Delta G^\neq$).

From the relationship between Gibbs free energy, enthalpy, entropy and temperature we know: $\Delta G = \Delta H - T\,\Delta S$, and that a reaction is only thermodynamically favored when $\Delta G < 0$. And even when $\Delta G < 0$ the reaction may still not occur at a measurable rate because it has too high an activation energy.

What you're describing is the Gibbs reaction energy: the Gibbs energy between products and reactants. This indeed tells us how thermodynamically favourably the reaction is ($\Delta G < 0$).

But the phrase in bold above is wrong. The reason is that the Gibbs activation energy ($\Delta G^\neq$) is the Gibbs energy difference between transition state and reactants.

As Mithoron mentioned, Gibbs reaction energy would give you the equilibrium constant for the reaction:

$$K = e^{-\tfrac{\Delta G}{R T}}$$

From the Arrhenius equation we know: $k = A ~\mathrm e^{-E_\mathrm a / (RT)}$

Generally, from Eyring theory,

$$k = \frac{k_B T}{h c_0} e^{-\tfrac{\Delta G^\neq}{R T}}$$

Comparing to Arrhenius equation, see:

$$k = \frac{k_B T}{h c_0} e^{-\tfrac{\Delta H^\neq - T \Delta S^\neq}{R T}} = \frac{k_B T}{h c_0} e^{\tfrac{\Delta S^\neq}{R}} e^{-\tfrac{\Delta H^\neq}{R T}}$$

Thus,

$$A = \frac{k_B T}{h c_0} e^{\tfrac{\Delta S^\neq}{R}} \\ E_a = \Delta H^\neq$$

That's what you would find from an Arrhenius's plot.

3. Only particles in thermodynamically favored reactions have enough energy to overcome $E_\mathrm a$ and so will spontaneously occur without external energy applied. But they may occur extremely slowly if they have a very high $E_\mathrm a$ (according to the Arrhenius equation).
The reasoning is correct, except for point three above (in bold), for which the same error is found as before: reactions may have a big $K$ (small $\Delta G$) but small $k$ (big $\Delta G^\neq$) or vice-versa.
1. In the second point, that happens because in Arrhenius equation, $e^{-\tfrac{E_a}{RT}}$ actually corresponds to the fraction of molecules that have an energy which more than $E_a$. (Part of Arrhenius' theory which explains his equation). So as $T$ increases, the negative exponent of $e$ decreases and hence the value of the number increases. And no, the chances can not be increased. All you can do is use a positive catalyst to increase the amount of colliding molecules and hence increase the rate of reaction.