[leesoo05]

Does spontaneity of reaction depend on negative delta G value or positive delta S(universe) value?
According to Gibbs free energy equation, delta G = delta H - T X delta S(universe)
So it seems like even though we have negative delta S(universe) value we still can have negative delta G value repect to delta H value if it is much smaller than
T X delta S(universe), so I'm not sure increasing entropy can always make spontaneous reaction.
If I'm thinking wrong plz correct me.

Thanks

[banjo]

You are right.  Whether or not the reaction is spontaneous will depend on the enthalpy term as well as the T*deltaS term.

Just look at how the deltaH and the product of T*(deltaS) add together algebraically.

Also, rethink your idea of spontaneous. Delta G is negative for a spontaneous process.  Delta S needs to become more positive rather than negative to favor a spontaneous reaction. 

If delta H is positive and delta S is negative, then the reaction will not be spontaneous at any temperature.

If delta H is negative and delta S is positive, then the reaction will be spontaneous at all temperatures. 

[djiang87]

you should note that even if delta G is positive the reaction can still become spontanous due to coupling reactions

[Yggdrasil]

Does spontaneity of reaction depend on negative delta G value or positive delta S(universe) value?
According to Gibbs free energy equation, delta G = delta H - T X delta S(universe)
So it seems like even though we have negative delta S(universe) value we still can have negative delta G value repect to delta H value if it is much smaller than
T X delta S(universe), so I'm not sure increasing entropy can always make spontaneous reaction.
If I'm thinking wrong plz correct me.

Thanks

The correct equation for ΔG is ΔG = ΔH - TΔS_sys.  The change in entropy that you use to calculate the change in free energy is the change in entropy of the system only.  The total change in entropy does not figure into the equation for ΔG.

Ultimately, the condition for spontaneity is ΔS_universe > 0.  Under conditions of constant temperature and pressure, the statement that ΔS_universe > 0 is equivalent to the statement that ΔG < 0.

[leesoo05]

Thank you so much for all the replies.

Then if a reaction has negative delta G value, can we say this reaction is thermodynamically favorable? My book said so.
But although delta G is negative, delta H can be positive. Then it is wrong that the reaction is thermodynamically favorable.
I'm little bit confused.

[Yggdrasil]

There are two factors that contribute to thermodynamic favorability.  In general, systems will prefer to reduce their potential energy (enthalpy) and increase their disorder (entropy).  Often times, these two goals are in conflict.  Even if a process causes a system to raise its potential energy, this disfavored increase in enthalpy can be offset by an increase in the entropy of the system.

For example, when water melts, it moves toward a higher energy state (there are more hydrogen bonding interactions in ice than in liquid water and these interactions are stronger in ice) but it becomes much more disordered.  Therefore, at room temperature, the transition from ice to liquid water is thermodynamically favored despite the positive ΔH associated with the process.

ΔH is just one factor that determines thermodynamic favorability, but it is not the only factor.  Ultimately, it is ΔG that determines thermodynamic favorability (at least, under conditions of constant temperature and pressure).