Positive Delta S: H2 + I2 --> 2HI, N2O4 from NO2

[enigmatic]

Question: Which of the following would probably have a positive delta S value?

He (g, 2atm) --> He (g, 10 atm)
H₂(g) + I₂(s) --> 2HI (g)
2Ag(s) + Br₂(l) --> 2AgBr(s)
O₂(g) --> O₂(aq)
2NO₂(g) --> N₂O₄(g)

My answer:
Well, I know that when a gas is pressurized, there are less states it can be in, so the first one's out.
I know that when a liquid becomes a solid, there are less possible states it can be in, so the third one's out.
I know that when a gas becomes an aqueous solution, there are less possible states it can be in, so the fourth one's out.

That leaves the second and fifth ones.
Can someone explain this to me?

 

[chemisttree]

The fifth one has two molecules reacting to form one molecule.

The second one has two molecules that are diatomic becoming two molecules that are composed of different species.

 

[Blueshift5]

Based on the definition of entropy (S), which is a measure of the disorder or randomness of a system, the reaction that would likely have a positive delta S value is the second one, H2(g) + I2(s) --> 2HI(g). This is because the reactants (H2 and I2) are in a more ordered state (gas and solid, respectively) than the products (2HI), which are in a more disordered state (gas). This increase in randomness or disorder leads to a positive delta S value.

As for the fifth reaction, 2NO2(g) --> N2O4(g), it is not possible to determine the delta S value based solely on the given information. This is because the delta S value also depends on the temperature at which the reaction takes place. At low temperatures, the reactants (NO2) are in a more ordered state (gas) compared to the products (N2O4), which are in a more disordered state (gas). In this case, the delta S value would be negative. However, at high temperatures, the opposite would be true and the delta S value would be positive.