spontaneous reaction

[joonkimdds]

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why is it thermodymatically favored?
and

the equilibrium mixture consist almost exclusively of C
in 2A+B==>3C with delta G = -1500KJ

 

[doc3232]

why is it thermodymatically favored?
and

the equilibrium mixture consist almost exclusively of C
in 2A+B==>3C with delta G = -1500KJ

So two things determine deltaG. One is the entropy and the other is the deltaH thing (which is determined by thermodynamic favorability), which I forgot the name for. Anyway, since the number of atoms does NOT change then we basically know that entropy does not change much. This is assuming that there is no phase change between reactants and products.
So what is making deltaG so negative: The reactions thermodynamic stability.
Hope I explained that right.

 

[joonkimdds]

so basically u r saying that
delta G = delta H - TdeltaS
but since delta S is 0(zero) we got
delta G = delta H
is that right?

but when we say thermodynamically favored, does that mean we want delta H to be positive or does that just mean it's depending on delta H becuz only -delta H would be -delta G which is spontaneous.

and what about the 2nd question about equilibrium mixture consistonly exclusively of C?

 

[tranv117]

From my understanding, since we are saying that delta S=0 assuming their are no phase changes and since both sides contains equal moles. This leaves deltaG=deltaH. deltaG from the given data is very negative meaning that it is a spontaneous rxn. It is also thermodynamically favored because delta H would also be negative (an exothermic rxn). In this case we are forming bonds, since A+B will form bonds to make C releasing energy via heat.

 

[water88]

why is it thermodymatically favored?
and

the equilibrium mixture consist almost exclusively of C
in 2A+B==>3C with delta G = -1500KJ

It's thermodynamically favored because deltaG < 0 i.e. spontaneous. The more negative deltaG is, the more the forward reaction is favored. Thus, the equilibrium mixture will consist of mostly C.

For spontaneity, the best possibilities are a negative deltaH (energy released, more stable product formed) and positive deltaS (entropy increases, 2nd law of thermodynamics).

 

[Andre3k]

The only thing in the problem that tells you the reaction is spontaneous is the negative deltaG.

 

[doc3232]

To the OP, yes that is right...because G is so negative then H will also so be so negative and hence thermodynamically stable.

 

[artanis]

delta S is negative, it goes from less ordered to more ordered.

But in order for delta G to be negative, delta H has to be big and low temperature (since delta H is negative) in order to get a Delta G of -1500kj.

 

[doc3232]

delta S is negative, it goes from less ordered to more ordered.

But in order for delta G to be negative, delta H has to be big and low temperature (since delta H is negative) in order to get a Delta G of -1500kj.

We don't know that it becomes more ordered. We can only say that the number of atoms didn't change so we really don't know about the change in entropy.