## Entropy

$\Delta S = \frac{q_{rev}}{T}$

Vanessa Rojas 4F
Posts: 15
Joined: Wed Feb 20, 2019 12:17 am

### Entropy

When calculating the entropy change of a system at constant pressure why is q= delta H?

Posts: 97
Joined: Tue Feb 12, 2019 12:15 am

### Re: Entropy

This is what I found on the web. http://chemed.chem.purdue.edu/genchem/t ... emical.php

"The change in the enthalpy of the system in a reaction is equal to the change in internal energy plus the change in the product of the pressure times the volume of the system.
DeltaH= DeltaE + Delta(PV)

Because the reaction is run at constant pressure, the change in the enthalpy that occurs during the reaction is equal to the change in the internal energy of the system plus the product of the constant pressure times the change in the volume of the system.
deltaH = deltaE + PdeltaV (at constant pressure)

Substituting the first law of thermodynamics into this equation gives the following result.
deltaH = (qp + w) + PdeltaV

Assuming that the only work done by the reaction is work of expansion gives an equation in which the PdeltaV terms cancel.
deltaH = (qp - PdeltaV) + PdeltaV

Thus, the heat given off or absorbed during a chemical reaction at constant pressure is equal to the change in the enthalpy of the system.
deltaH = qp (at constant pressure)"

Julie Park 1G
Posts: 100
Joined: Thu Jul 25, 2019 12:15 am

### Re: Entropy

$P\Delta V$ becomes insignificant because there is no expansion work and the volume of the reactants = volume of the products. This means that $\Delta U$, which normally would be $q_{p} + w$, now only equals $q_{p}$, which in other words, is described as $\Delta H$