## Entropy Change in reversible reaction

Volume: $\Delta S = nR\ln \frac{V_{2}}{V_{1}}$
Temperature: $\Delta S = nC\ln \frac{T_{2}}{T_{1}}$

Seohyun Park 1L
Posts: 73
Joined: Fri Sep 28, 2018 12:29 am

### Entropy Change in reversible reaction

Why is the total entropy change equal to 0 in a reversible, isothermal gas expansion?

makenzie2K
Posts: 31
Joined: Fri Sep 28, 2018 12:20 am

### Re: Entropy Change in reversible reaction

I’m not sure this is correctly interpreted. In reversible reactions entropy change does not = 0. The change in internal energy (deltaU) = 0 but this still constitutes a spontaneous process (in the gas phase) because there is still the factor of degeneracy, multiple potential states for the gas molecules to occupy, and this creates a spontaneous process that does in fact increase entropy.

Katelyn Phan 2A
Posts: 61
Joined: Fri Sep 28, 2018 12:23 am

### Re: Entropy Change in reversible reaction

The total entropy change does not equal 0 under such conditions. It is the change in internal energy that equals 0.

deepto_mizan1H
Posts: 65
Joined: Fri Sep 28, 2018 12:16 am

### Re: Entropy Change in reversible reaction

In addition to the responses above, we can see that entropy would not be 0 in any cases as it violates the laws of thermodynamics, where entropy only approaches 0 when it's at ~0 K. (Third Law). In an isothermal system our temperature is constant, so it doesn't approach 0. (Only delta T is 0). When calculating the change in entropy for an isothermal reversible reaction, we can use the equation: delta S = n (moles) * R (gas constant) * ln(Final volume/Initial volume).

(We can also substitute pressure for volume, and invert it to become Initial Pressure/Final Pressure).