# Chemistry

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For the reaction described by the chemical equation:
3C2H2(g) -> C6H6(l) .. Delta H rxn = -633.1 kJ/mol
a) Calculate the value of Delta S rxn at 25.0 C...
b) Calculate Delta G rxn...
c) In which direction is the reaction, as written, spontaneous at 25 C and 1 bar pressure?
- forward, reverse, neither, or both?

here's what i did:
Delta S = [1(173.4 J/mol-k)] - [3(200.9)]
Delta S = -429.3 J/mol-k

Delta G = Delta H - T(Delta S)
Delta G = (-633.1 kJ/mol) - (298 K)(-0.4293 kJ/mol-k) = -505.168 kJ/mol

if that's right so far.. yay! but can someone explain how a reaction can be spontaneous forwardly or reversely? right now i only know that a reaction is spontaneous if delta G is a negative number.. please help!

• Chemistry - ,

I didn't confirm the numbers you quote by looking in a table but other than that the process looks ok. The question may be worded a weird way but it makes sense if you know what it's asking. You know, as you pointed out, that the reaction is spontaneous if dG is <0. What if dG>0? That simply means the reaction is not spontaneous in the direction written BUT it is spontaneous in the REVERSE direction. That's what the problem means by forward or reverse. Any reaction that is spontaneous in one direction you can categorically say is not spontaneous in the reverse direction.

• Chemistry - ,

Oh I understand. If dG = 0 .. would that make the reaction direction neither? When would it ever be both?

• Chemistry - ,

Yes, dG = 0 is neither. When dG = 0 the system is at equilibrium and there is no net reaction in either direction.
If a reaction is spontaneous in one direction it is automatically non-spontaneous in the reverse direction. There never will be a situation in which it is spontaneous in both directions unless you want to argue the fine point that when dG = 0 it is proceeding both forward and reverse (however, there is no net reaction and I hardly call that spontaneous but I guess in the strict sense, and taken literally, you could argue that it is).

• Chemistry - ,

Oh okay trick question then thanks again dr bob!