[Ahmed Abdullah]

Question Details: We know that
del G=del H - T(del S)

if del H is positive and del S is negative then del G is always positive. There is no way of making del G negative. My question is whether such a reaction take place at all (to any extent)?
thx

[macman104]

Question Details: We know that
del G=del H - T(del S)

if del H is positive and del S is negative then del G is always positive. There is no way of making del G negative. My question is whether such a reaction take place at all (to any extent)?thx

Sure.  It just means the reaction won't be spontaneous.  It doesn't mean that if you put in enough energy into reaction, you can't make it happen.

[Ahmed Abdullah]

Does the form of energy matter?

I think heat energy cannot make the reaction happen. Because whatever high the temperature is the reaction is still unfavored by a positive change in free energy.
So it must involve some mechanical form of energy or some ordered form of energy (I don't know what it is).

[macman104]

Oh, I see what you mean.  Yes, you do bring up a good point.  I agree, maybe something similar to a way that some enzymes work by bringing reactants closer together, or perhaps a change in temperature brings something from a gas phase to a liquid phase.  Not sure though, interesting question!

[enahs]

Lower the transition state energy or make the products more stable and lower in energy.

[Ahmed Abdullah]

I've finally learnt that:
If delta(G) of a reaction is positive, the reaction can also happen, it only have a lower equilibrium constant:
when delta(G) < 0 then K >1, when delta(G) = 0 then k = 1, when delta(G) > 0 then k < 1:
delta(G) = -RTlnK
If del G is positive the reaction will only occur if energy is supplied to force it away from the equilibrium position (i.e when del G=0). If del G is negative the reaction will proceed spontaneously to equilibrium.

[phenyl_ninja]

Lower the transition state energy or make the products more stable and lower in energy.

Lowering the transition state energy won't effect the thermodynamics/spontaneity of the reaction, just the rate at which it takes place.  Same thing goes for using an enzyme. Which basically accomplishes the same thing.

I've finally learnt that:
If delta(G) of a reaction is positive, the reaction can also happen, it only have a lower equilibrium constant:
when delta(G) < 0 then K >1, when delta(G) = 0 then k = 1, when delta(G) > 0 then k < 1:
delta(G) = -RTlnK
If del G is positive the reaction will only occur if energy is supplied to force it away from the equilibrium position (i.e when del G=0). If del G is negative the reaction will proceed spontaneously to equilibrium.

Yep, if delta(G)(o) is positive, this just means K is less than 1, products will still be formed in small quantities. 

On the other hand if delta(G), as opposed to delta(G)(o), is positive, the reaction will go backwards until equilibrium is reestablished. 

For understanding this sort of thing it's important to understand the difference between delta(G)(o) at standard conditions and regular old delta(G).

[Ahmed Abdullah]

may you state the difference, please..