# Why is entropy a driving force for a reaction

1. May 9, 2013

### arp001

I understand the Gibbs Free Energy equation that was drilled into us in Chemistry class but not even my teacher could explain what entropy had to do with free energy, and furthermore, why entropy is a driving force for a reaction. From what I understand, entropy is the distribution of energy across a system. It would be great if someone could explain why entropy is a driving force for a reaction in simple terms as my online research has come to no avail. Thanks!

2. May 9, 2013

### Staff: Mentor

How do you mean that?

States with large entropy just have more possible "arrangements" (micro-states), therefore they are more likely to occur.

3. May 9, 2013

### Jano L.

No, that is not correct. Entropy has nothing to do with energy.

Entropy is an abstract quantity S defined for equilibrium states. The law of entropy (2nd law of thermodynamics) says that

In an isolated system, the entropy cannot decrease.

Usually when some processes go on, like chemical reaction, the entropy is not even defined. However, when you compare the entropy $S_1$ at the beginning and entropy $S_2$ at the end, when the system is again in equilibrium, if the system was isolated during the whole time then $S_2 \geq S_1$.

However, for the case of chemical reaction, the system is presumably not isolated because usually the beginning and the end of the chemical reaction occur at the same temperature and the same pressure. The restoration of the original temperature can occur because the reacting mixture is in thermal contact with the surroundings.

In such case the entropy law does not apply for the system, since the mixture $m$ is not isolated. The entropy of the mixture can decrease.

However, the entropy law does apply again if we include the surroundings (laboratory) to the system. Quantities $S_1$, $S_2$ then refer to entropy of such large system.

However, working with such entropies of the enlarged system is not practicable, as we do not know entropy of room etc. For this reason the entropy law is reformulated using Gibbs energy G = U - TS + PV (here S and other quantities are those of the mixture only).

The idea is as follows:

From the law of entropy, the heat absorbed by the mixture

$$\Delta Q \geq T\Delta S,$$

since the process is most probably not reversible.

Since $\Delta U = \Delta Q - P\Delta V$, we obtain the equation

$$\Delta U + p\Delta V - T\Delta S \leq 0.$$

This is the same as

$$\Delta (U - TS + PV) \leq 0,$$

because the temperature and pressure do not change - they are given by the atmosphere. The quantity in the brackets is the Gibbs energy $G$.

So, the entropy law leads to another law:

In a process which has the same T and P at the beginning and at the end, the Gibbs energy cannot increase.

If the reaction is spontaneous, G will decrease. Now from the above formula you can see that if final state has greater entropy, G decreases. Also, if the final state has lower volume, G decreases.

If the reaction occurs, $\Delta G$ is negative.

If $\Delta G < 0$ because the enthalpy $U + PV$ decreased enough, we say the reaction is driven enthalpically - it is an exothermic reaction; the system gives off heat (e.g. hydrogen + oxygen => boom + water + heat).

If $\Delta G < 0$ because of increase in entropy, one says the reaction is driven entropically; the system can also absorb some heat. For example, dissolving salt in some alcohol solutions makes them cool down. As it gets to the original temperature, it receives heat and in the end its entropy increases.

4. May 10, 2013

### DrDu

Let me try to give you an alternative explanation which also holds if there are other kinds of work than pV work as e.g. in a battery.
The total change of entropy in a chemical reaction has two components:
1. The change of entropy of the system (e.g. the reaction vessel) $\Delta S$ and the change of entropy of the surrounding $\Delta S_\mathrm{surrounding}=Q/T$. Now the heat can be expressed as $Q=-\Delta H$, as long as the pressure of the system and the surrounding is equal, at least at the beginning and end of the reaction.
For $\Delta G$ to be a thermodynamic potential, we also have to assume that the initial and final temperature of the system T is equal to that of the surrounding $T_\mathrm{surrounding}$.
So $0\le\Delta S_\mathrm{total}=(\Delta S+\Delta S_\mathrm{surrounding})= \Delta S -\Delta H/T=\Delta S -\Delta H/T =-\Delta G/T$ or
$0\ge\Delta G$.
So we have derived $\Delta G\le 0$ from the principle of the increase of entropy.

Last edited: May 10, 2013
5. May 10, 2013

### morrobay

Also there is the statistical mechanics term for entropy: S = k ln W
that can be equated with thermodynamic entropy
http://www.eoht.info/page/S+=+k+ln+W

6. May 10, 2013

### Jano L.

DrDu, your derivation seems to require too special assumptions. You use $T$ and $T_{surrounding}$ as different quantities, so I infer that you consider infinitesimal changes during the process. However, the process may not allow thermodynamic description at the intermediate times. For this reason it is better to refer to only the state before, and the state after the reaction ceases.

Also, you assume that

$$\Delta S_{surrounding} = \frac{Q}{T_{surrounding}}$$
but this may not be true, as the process is irreversible (spontaneous reaction).

7. May 10, 2013

### DrDu

Jano you are partly right here. I simplified the expression. T now is the temperature of the surrounding, only. One of the conditions for Delta G to be relevant to the determination of spontaneity of a reaction is that the surrounding has a fixed temperature T and pressure p. Then the change of the entropy of the surrounding can always be calculated using the equilibrium expression Q/T even if the temperature of the system is different from that of the surrounding during the reaction.

8. May 10, 2013

### Jano L.

This is interesting. In my first post, I derived the equation $\Delta G \leq 0$ from the assumption $\Delta S_{sys} \geq \frac{\Delta Q}{T}$. Nothing is supposed on the increase of entropy of the environment $S_{env}$; it may be equal to or higher than $-\frac{\Delta Q}{T}$.

However, in your derivation, you assume $\Delta S_{env} = -\frac{\Delta Q}{T}$, and from the law of entropy increase we can infer that $\Delta S_{sys} \geq \frac{\Delta Q}{T}$.

I wonder, is your assumption more strong (restrictive), as it seems, or is it equivalent to mine?

It seems more restrictive, since it treats the environment as if it underwent a reversible process.

On the other hand, I am not sure that my assumption $\Delta S_{sys} \geq \frac{\Delta Q}{T}$ for the overall result of the process is valid if your assumption is not...

9. May 10, 2013

### DrDu

I just had a look at the book "Vorlesungen ueber Thermodynamik", by Max Planck, 9th edition, Walter de Gruyter, Berlin, Leipzig, 1930. Planck wrote his thesis about thermodynamics before finding his quanta and getting awarded the Nobel prize and certainly is one who got things right. His derivation is very similar to mine. He also starts out from $dS+dS_0 \ge 0$ where S is the entropy of the system and S_0 that of the surrounding medium. Especially he writes that:"Let's assume that any volume changes of the surrounding medium occur in a reversible way" in §140.
Then he sets $dS_0=-Q/T$.

10. May 10, 2013

### Jano L.

I think that one also needs to assume that the reservoir passes through states which have the same temperature, in order to be sure that $\Delta S_{res} = Q/T$. This assumption bothers me - it seems quite restrictive. My question is: when we relax this assumption, what will happen? Will the equation $\Delta S_{res} = Q/T$ still hold?

11. May 11, 2013

### morrobay

ΔS = dQ/T , Entropy change equals the change in heat/Temperature.
For a simple example why an entropy increase is a driving force: Consider two objects at T1 and T2 and T1 is hotter than T2
Place them next to each other and in a very short period of time, a quantity of heat, dQ , will transfer from the hotter object to the cooler object. So the change in entropy for this irreversible process is: dQ/T2 - dQ/T1 And since T1 > T2 there has been an increase in entropy.

12. May 11, 2013

### morrobay

I just did the above numerically with two Aluminum 100g cubes , specific heat .22cal/g/°C
T1 = 80 C before dQ and 75 C after
T2 = 20 C before dQ and 25 C after
With S2 - S1 > ∫12 dQ/T
Total entropy before dQ was 22.3 J/K and after 27.56 J/K
With dQ = 460 J then what value of T in ∫12 dQ/T is correct ?
For evaluation : ln 460J/T - ln 460J/T

Last edited: May 11, 2013