[anna123]

I know that if there are more moles of product than reactant there is likely to be an increase in entropy, and that it also depends on states, BUT, if there is a reaction where there are the same number of moles on each side of the equation, but of product they are the same molecule and of reactant they are different, is that an increase or decrease in entropy and whhhy?

For example, the reaction,
N2 (g) + O2 (g) --> 2NO (g)

I hope that makes some sense, and thanks very much for any help

[m1a1tank]

Im not really sure, but I think it might be a decrease in entropy. Because you have two separate moles of in the gaseous state forming one molecule in the gaseous state. So there would be more disorder on the reactants side?
I think thats what it is.

[DarkTitan]

i think it will be the same entropicially because the end "product" is the same in moles. What changes is the enthalpy. quote me if i'm wrong, i'd be interested to know if i am

[00raiser]

I doubt you need to know this. In the exam, for these questions you will be given a table of entropy values for each molecule, which allows you to work out change in entropy.


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[m1a1tank]

(Original post by DarkTitan)
i think it will be the same entropicially because the end "product" is the same in moles. What changes is the enthalpy. quote me if i'm wrong, i'd be interested to know if i am

Thats what i thought initially. That the entropy before and after will probably be the same and so there is no change in entropy. But surely there would have to be even just a small change in the entropy for the reaction to happen.

[TiTo20]

Entropy is also affected by temperature though, is it not? So surely it will depend on the conditions of the reaction that will cause an increase or decrease in entropy?

Forgive me if I'm wrong, I haven't done entropy in over a year =P


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[Mathlete 4 the win]

It depends on how much you're required to know.

As I'm sure a lot of the above have said, entropy isn't just the amount of ways that molecules can be arranged but it's the way in which the quanta of energy the molecules posses can be arranged.

The structure of NO will have different vibrational, rotational and translational energy levels so the difference between the energy levels (the quantised energy difference) will also be different. I'd guess that NO has closer vibrational energy levels because the bonds are weaker so therefore has more quanta of energy for the given temperature and this means that they can be arranged in more ways.

[Bradshaw]

Broadly speaking, the entropy change will be approximately zero. However, you can go into a bit more depth.

Entropy can be split into translational, rotational, vibrational and electronic parts.

The translational parts rigorously cancel.
The NO bonds will be weaker than those in N2, but stronger than those in O2. This will mean that the bond lengths will not be quite the same, and so there will be a small change in rotational entropy. Similarly, there will also be a small change in vibrational entropy due to the different number of accessible states. There will also be slight changes due to the degeneracies of the ground electronic states, and an increase due to the reduction in symmetry of the molecule.

None of these changes will be significant though.

[m1a1tank]

(Original post by Bradshaw)
Broadly speaking, the entropy change will be approximately zero. However, you can go into a bit more depth.

Entropy can be split into translational, rotational, vibrational and electronic parts.

The translational parts rigorously cancel.
The NO bonds will be weaker than those in N2, but stronger than those in O2. This will mean that the bond lengths will not be quite the same, and so there will be a small change in rotational entropy. Similarly, there will also be a small change in vibrational entropy due to the different number of accessible states. There will also be slight changes due to the degeneracies of the ground electronic states, and an increase due to the reduction in symmetry of the molecule.

None of these changes will be significant though.

Im hoping what you've just said i'm not supposed to know for a level chemistry?

[EierVonSatan]

(Original post by m1a1tank)
Im hoping what you've just said i'm not supposed to know for a level chemistry?

Nah, don't worry

[anna123]

Thank you! By the sounds of it, it isn't something I'll need to know for my course but its interesting nonetheless thank you all very much for you help, much appreciated