[MilkyCar]

Stuck on this entropy problem:

Predict the sign of ΔSsurr for the following processes:

a. H2O (g) -> H2O (l)
- ΔSsurr is positive
- ΔSsurr is negative

b. I2 (s) -> I2 (g)
- ΔSsurr is positive
- ΔSsurr is negative

I think that if gas becomes liquid, the entropy would be negative whereas if a solid becomes a gas, the entropy would be positive, right?

[Aldebaran]

What formula are you familiar with which relates enthalpy of reaction with entropy of surroundings?

[MilkyCar]

ΔG = ΔH - TΔS?

[Aldebaran]

Google entropy of surroundings. You will find the formula; then consider your reactions and figure out whether delta H is positive or negative for these changes of state. The formula will then give you the sign for the entropy.

[MilkyCar]

The process above has the entropy of surroundings increasing, so ΔSsurr is positive, and the second process below has the entropy of surroundings decreasing, so ΔSsurr is negative. Is this right?

[Borek]

You are trying to cut corners.

How is the change in the entropy defined? The most general definition of dS (ΔS)?

[MilkyCar]

The change of disorder in a thermodynamic system. Does it have something to do with the change in disorder from reactants to products?

[Borek]

Nope, that's a handwavy definition, one that is common in the pop sci.

Compare https://en.wikipedia.org/wiki/Entropy_(classical_thermodynamics)#Definition

[MilkyCar]

The quantity of heat that is being transferred, divided by temperature? Does it also have to do with the difference between the initial and final states of entropies?

[Borek]

The quantity of heat that is being transferred, divided by temperature?

Yes. Processes that were mentioned - are they exothermic, or endothermic? Where does the heat go?

Does it also have to do with the difference between the initial and final states of entropies?

Not sure what you are asking - the definition we are talking about refers to the change, and the change is between initial and final state.

Note: one important thing here is that we can calculate by how much entropy changed, but we still don't know the absolute value.

[MilkyCar]

The first process is exothermic, and the heat goes from the system to its surroundings. The temperature of the surroundings rises.

The second process is endothermic, and the system absorbs heat from the surroundings. The temperature of the surroundings decreases.

In an exothermic reaction, the entropy of the surroundings increases (ΔSsurr > 0), and in an endothermic reaction, the entropy of the surroundings decreases (ΔSsurr < 0).

So, for the first process above ΔSsurr is positive, and for the second process below ΔSsurr is negative. Am I getting this right?

[Borek]

So, for the first process above ΔSsurr is positive, and for the second process below ΔSsurr is negative. Am I getting this right?

That's my understanding.

In general you can assume surroundings to be infinite and its temperature to be constant.

Also note the definition dS=dQ/T is exact only for reversible processes, in real world it is rather dS≥dQ/T - which doesn't change the qualitative predictions, it just makes calculating exact changes more difficult.

I can be missing something though, I never feel confident around thermodynamics.