# Activation energy and entropy

First assertion

If a system is already in a high temperature, adding energy, will increment the entropy in a low amount (compared with a system in a lower temperature).

Question (if assertion is right)

What if the heat is enough that let molecules breaks (activation energy), this would lead to new multiplicity (more freedom) so more entropy. It is a higher entropy grow than if the temperature were lower! that seems to be in contradiction with first assertion.

I see there is something wrong here, but I don't know where.

## 2 Answers

The assertion is based on the assumption that you either have only ‘small’ increases in temperature (and hence small increases in entropy, think of all the $dS$ and $dT$ you encounter in standard thermodynamics) or that your system is sufficiently homogenous that the change in entropy is a continous function of the change in temperature. This obviously breaks down if your molecules start to break up.

• +1 I think that's the key, system is out of equilibrium, those are not small increases, then, it's a mistake to maintain temperature constantly "low" while increasing energy, that won't happen – HDE Nov 16 '12 at 18:52

By definition of temperature

$$\frac{1}{T} = \left( \frac{\partial S}{\partial U} \right)_{N_j}$$

If temperature is higher adding the same amount of energy $\delta U$ at constant composition $N_j$ results in a lower change $\delta S$ in the entropy. But if composition is changing due to chemical reaction $\mathrm{AB} \rightarrow \mathrm{A} + \mathrm{B}$ then there is an extra variation in the entropy due to change in composition

$$\frac{\mu_j}{T} = - \left( \frac{\partial S}{\partial N_j} \right)_U$$

The total change in the entropy is given by the variation of energy plus the variation on composition

$$\mathrm{d}S = \frac{1}{T}\mathrm{d}U - \sum_j \frac{\mu_j}{T}\mathrm{d}N_j$$