## 9.33 (c)

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

Hannah Guo 3D
Posts: 56
Joined: Fri Sep 29, 2017 7:06 am

### 9.33 (c)

9.33 (c): Without performing any calculations, predict whether there is an increase or a decrease in entropy for each of
the following processes: SO2(g) + Br2(g) + 2 H2O(l) --> H2SO4(aq) + 2 HBr(aq).

I thought entropy decrease because there are fewer moles of gas on the product side, but why the answer from the SSM says that "entropy should decrease as the total number of moles decreases".

Thank you!

jillian1k
Posts: 54
Joined: Sat Jul 22, 2017 3:00 am

### Re: 9.33 (c)

The number of moles on the reactants side is 4. The number of moles on the products side is 3. Therefore the total number of moles decreases by 1 mole somewhere in the reaction. Entropy is based partly on mass or moles that are in a substance. Generally, the more mass or moles present, the higher the entropy. Therefore in this reaction the reactants have a higher entropy and entropy decreases as the products are formed.

You could also look at it in terms of moles of gas decreasing from 2 in the reactants to 0 in the products. This also shows a decrease in entropy. I think either logic works. Since they used the moles of gas example for part a, they probably just wanted to show the different ways to arrive at the same answer and show the different factors that determine entropy.