Activation energy and entropy - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-10-22T21:47:38Z https://physics.stackexchange.com/feeds/question/44362 https://creativecommons.org/licenses/by-sa/4.0/rdf https://physics.stackexchange.com/q/44362 1 Activation energy and entropy HDE https://physics.stackexchange.com/users/1916 2012-11-16T14:51:24Z 2012-11-17T12:24:28Z <p>First assertion</p> <p>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).</p> <p>Question (if assertion is right)</p> <p>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 <em>a higher entropy grow than if the temperature were lower!</em> that seems to be in contradiction with first assertion.</p> <p>I see there is something wrong here, but I don't know where.</p> https://physics.stackexchange.com/questions/44362/-/44369#44369 2 Answer by Claudius for Activation energy and entropy Claudius https://physics.stackexchange.com/users/12808 2012-11-16T17:55:29Z 2012-11-16T17:55:29Z <p>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.</p> https://physics.stackexchange.com/questions/44362/-/44421#44421 2 Answer by juanrga for Activation energy and entropy juanrga https://physics.stackexchange.com/users/12998 2012-11-17T12:19:15Z 2012-11-17T12:24:28Z <p>By definition of temperature</p> <p>$$\frac{1}{T} = \left( \frac{\partial S}{\partial U} \right)_{N_j}$$</p> <p>If temperature is higher adding the same amount of energy $\delta U$ <em>at constant composition</em> $N_j$ 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</p> <p>$$\frac{\mu_j}{T} = - \left( \frac{\partial S}{\partial N_j} \right)_U$$</p> <p>The total change in the entropy is given by the variation of energy plus the variation on composition</p> <p>$$\mathrm{d}S = \frac{1}{T}\mathrm{d}U - \sum_j \frac{\mu_j}{T}\mathrm{d}N_j$$</p>