[Winga]

Is it a exothermic process or endothermic process?

H⁺ + Ph-O^- ---> Phenol

[dexangeles]

which one is more stable?

[Winga]

which one is more stable?

This is what I want to ask next.

I want to know which one is more stable thermodynamically, phenol or phenoxide?

Protonation of phenoxide should be exothermic, therefore, phenol is more stable than phenoxide, right?

[Donaldson Tan]

given phenol dissociates exothermically in aq system, thus I would expect the above reaction to be endothermic actually, although at first glance this seems to be a simple bond-forming case.

(exception is that both ions are in gas state)

[Winga]

given phenol dissociates exothermically in aq system, thus I would expect the above reaction to be endothermic actually, although at first glance this seems to be a simple bond-forming case.

(exception is that both ions are in gas state)

Although the dissociation of phenol is exothermic, it doesn't mean the protonation of phenoxide is endothermic.

Just consider that there are both bond formation and bond breaking in the dissociation of phenol. If breaking the O-H bond of phenol requires less energy and more energy release from forming the O-H bond with water, the reaction is exothermic.

In other case, if I want to convert the phenol to phenoxide to cyclohexanol, phenoxide release more energy as it has one O-H bond formation more than phenol, am I correct?  

[movies]

Just consider that there are both bond formation and bond breaking in the dissociation of phenol. If breaking the O-H bond of phenol requires less energy and more energy release from forming the O-H bond with water, the reaction is exothermic.  

Yes, this is absolutely correct.

Phenol should be lower in energy because ot has an additional bond.

[GCT]

Depends on which process you're referring to, are you talking about dissolving phenol in water?

Note that H+ + -OH--->H20 is exothermic, however, you'll probably need to consider the net standard enthalpy of the process if you wish to be exact.  

there's probably a better answer to this question, I just am not able to see it at the moment.

[Donaldson Tan]

but protonation of phenoxide means destroying the extended pi electron system in the benzene ring, doesnt it?

[Winga]

but protonation of phenoxide means destroying the extended pi electron system in the benzene ring, doesnt it?

What a good question!

Let me see...
e.g. hydrogenation of benzene, the final product is cyclohexane.
If 1,3,5-cyclohexatriene is present, it releases more energy than that of benzene.
The main key is that the activation energy of benzene is higher than that of 1,3,5-cyclohexatriene, this only affect the rate of reaction but not the thermodynamic position, am I right?

[Garneck]

Ok, my 2 cents:

Get some Ph-O-Na and HCl, pour them into a flask with a thermometer and there you have it

Ok, but seriously, it's exothermic, because a neutralisation reaction is always more or less exothermic. Besides Ka for phenol is 1.28 x 10^-10, so that means that phenol will favour to be in an undissociated form, which is energetically lower. Going from a higher energy state to a lower energy state requires emission of energy.

[GCT]

The only way to know with certainty is to perform a calculation, which I may do later. One can use the Gibbs equation; find G with the Ka value for phenol.  This case pertains to that of aqueous solutions.

geodome, you do have a nice point, some organic reactions require the deprotonated form of the final product to drive the reaction to appreciable yields.  The case that you're referring to is strictly related to between phenol and its conjugate.  However, it may be that the OP was referring to the case where phenol is in an aqueous solution.  One will need to find the overall favorability of the reaction, perhaps at standard temperatures.

[Garneck]

          Ka
Ph-OH-----> Ph-O^- + H⁺

T=298 K
R=8,314 J/mol*K

delta G = -R*T*logKa
d G = - 8,314*298*log(1.28 x 10^-10) = -8,314*298*(-9,9) = 24528 J/mol = 24,5 kJ/mol

So the reaction going the other way is endothermic.

And that means the answer to this thread is: exothermic

Right?

[Winga]

         Ka
Ph-OH-----> Ph-O^- + H⁺

T=298 K
R=8,314 J/mol*K

delta G = -R*T*logKa
d G = - 8,314*298*log(1.28 x 10^-10) = -8,314*298*(-9,9) = 24528 J/mol = 24,5 kJ/mol

So the reaction going the other way is endothermic.

And that means the answer to this thread is: exothermic

Right?

What is the difference between these two equations?
delta G = -R*T*logKa
delta G = -R*T*lnKa

[GCT]

It seems that your calculation is correct, if so, the reaction is spontaneous...in aqueous solutions.  In terms of enthalpy, it is endothermic.

This case is a bit confusing, because in terms of enthalpy, we are predisposed to say that all formation of bonds are exothermic, while the breakage is endothermic, as movies implied.  In constrast one might believe that because the bonds of diamonds are formed at higher temperatures, that the formation is endothermic.  However, I believe that this would be due to entropical factors, which increase as temperature is raised.  From a physical perspective, energy will always be required to break up nuclei attractions between atoms.  

It all shows how little I know about this issue.

[Garneck]

You're correct. I haven't counted stuff like this for two years. I forgot that you have to use the natural logarythm.

delta G = -R*T*lnKa = 56,4 kJ/mol

Even more. It's still exotermic though.