[WilliamYWT]

I was preparing for exam when I came across this question~ None of my friends know why the answer is C. May somebody help me?
Which one of the equations below represents a reaction that is feasible at all temperatures?
A. P_(s)  Q_(s) + R_(g) endothermic
B. 2L_(g) + M_(g)  2N_(g) exothermic
C. S_(g)  2T_(g) exothermic
D. A_(g) +B _(g)  C_(g) endothermic

[Dan]

What property must a reaction have to be spontaneous?

Hint: Gibbs

[WilliamYWT]

Free molecules ready for successive collisions? Or the chemical is reactive enough to make something less reactive? Sorry if the answers look stupid, and Gibbs' theories never appeared on my textbooks, I've searched wiki and I've found symbols spinning around my head...

[Corribus]

Here's another hint.  The title of your thread is wrong.

[WilliamYWT]

Maybe the title should be something like "entropy question" (actually I don't know what entropy is) or simply "exothermic/endothermic reaction question"? Some of my friends thought this may be related to equilibrium but I think otherwise.

[Dan]

Hint: It is a question about Gibbs free energy and spontaneous reactions

[WilliamYWT]

I've found this:
ΔG=ΔH-TΔS
When ΔS is positive and ΔH is negative, a process is always spontaneous
but I still don't know how to calculate entropy value and what does it mean..

[Corribus]

You don't need to calculate anything.  You need only consider concepts.

When is ΔH negative?
When is ΔS negative?

Think about the information that is given to you in the problem.  We're not dealing with real chemicals here.  There's nothing to look up.  Everything you need is written in the problem.  We have the exo/endotermicity of each reaction, we have states of matter (solids, gas) and we have stoichiometry.  How do those concepts relate to the thermodynamical quantities?

[WilliamYWT]

ΔH is negative when the reaction is exothermic, ΔS is negative when phase is changing from gas to liquid, gas to solid and liquid to solid (which I've just found out), so in
A. ΔH is +ve and ΔS is +ve
B. ΔH is -ve and ΔS is 0 (I'm not sure about this)
C. ΔH is -ve and ΔS is 0 (Still not sure about this)
D. ΔH is +ve and ΔS is 0 (Still not sure about this)
P.S.
 I've found this statement on yahoo answer: When entropy is positive, entropy is increasing and the products occupy more microstates than do the reactants
So in:
B. ΔS is -ve
C. ΔS is +ve
D. ΔS is -ve
So when  ΔS is positive and ΔH is negative, a process is always spontaneous, meaning only C is feasible at all temperatures.
So only C is the answer.
And this question is out of syllabus..
Thank you very much! You are very inspiring, I'm sure I won't get lost in such questions again.

[Borek]

Solids have always lower entropy than gases do, and the more gas there is, the larger its entropy is. Try to apply these simple rules to ΔS.