[LogicX]

for the same reaction?

For example, if you have the reaction 6A₄--> 4A₆ and you calculate that free energy is higher (more positive) than enthalpy for this reaction, how would you reason why free energy is larger?  Is it because free energy takes into account entropy and thus is more unfavorable than enthalpy?

[XGen]

I am not sure what you are asking.

Do you know the relationship between enthalpy, entropy, and Gibb's free energy?

[LogicX]

Yeah, G= H-TS.

The question is to explain why dG (which was determined computationally) is larger than dH (also determined computationally) based on the above reaction.  Entropy was also determined computationally.  I mean, I can explain it mathematically I guess, but not conceptually.

[juanrga]

for the same reaction?

For example, if you have the reaction 6A₄--> 4A₆ and you calculate that free energy is higher (more positive) than enthalpy for this reaction, how would you reason why free energy is larger?  Is it because free energy takes into account entropy and thus is more unfavorable than enthalpy?

Assuming constant temperature :delta: G = :delta: H - T :delta: S

If :delta: S < 0 for a given reaction, then :delta: G > :delta: H

At the left of your reaction you have 6 units of A₄, at the right you have 4 units of A₆. Entropy must be low at the right because (i) you have less units (entropy is extensive) and (ii) each unit of A₆ is more constrained (has more bonds) that each unit of A₄ and has less accessible states.