I want to resurrect this topic which has been discussed before:http://www.chemicalforums.com/index.php?topic=19788.0http://www.chemicalforums.com/index.php?topic=16293.0
I am aware of some of the past thoughts but I want to learn more. First of all, I do not wish to invoke electronegativity arguments as I believe this is a scale greatly biased by the redox reactions. Thus the electronegativity of fluorine is large because fluorine is very reactive and iodine is small because it is much less so. This is simply different than the electron withdrawing power. Thus HF is a weak acid and HI is strong. That simply restates the question, but it doesn't explain why.
It is easy to understand why HI is a strong acid. Its nuclear charge is large (+53) and thus it can pull its electrons inward. Compared to the other halogens, the repulsive field of an iodine nucleus grows faster than bond length. That is, if we set the repulsive force of HI to be equal to that if HF, then the bond length of HI would be 223 pm versus 161 pm actual, x^2=(53*92^2)/9. Granted, this is simplistic reasoning, but I think it is consistent with a general trend in bond lengths, e.g., CH>NH>OH>HF due to nuclear charge. For this reason, we might then expect that the electrons of fluoride to be more accessible, more basic, than those of iodide. This should be a general principle of physics that matches the acidity trends of simple acids, HI>HBr>HCl>>HF; HF>H2O>NH3>CH4.
You may see arguments that HF is a weak acid because of low entropy. While I think this is true, I don't agree with the interpretation of it. The entropy is lower because HF doesn't ionize, not that it doesn't ionize because its entropy is lower. I don't think of entropy as a normal force. Entropy can be small if a bonding force is present, otherwise molecules will behave as ideal substances, that is without interaction.
The issue I am trying to resolve is gas phase v solution phase reactions. I think they are simply different. McMurray states, “… bond dissociation energies refer to molecules in the gas phase and aren’t directly relevant to chemistry in solution.” The conservation of energy principle means the same energy difference would exist whether a reaction were gas or solution phase. I have used this argument for a solution phase Born-Haber cycle in a formation of NaCl(aq), -386 kJ/mol v -411 kJ/mol (s). In a solution phase reaction, no bond is formed between Na(+) and Cl(-). The energy is largely the energy of the redox reaction of Na->Na(+) and Cl->Cl(-). So, if you perform a reaction, it will be guided by Hess's Law.
I saw a post for the reaction of LiBr + MeI
LiI + MeBr, the net enthalpy of bonds formed versus broken would be +18 kJ/mol. However, I began to think the bonds of LiBr and LiI are actually not being broken and thus should not be part of the calculation. Now, the reaction enthalpy is a difference in the bond energy of methyl bromide and methyl iodide, -72 kJ/mol. However, as I understand this calculation, this is a measure of the homolytic bond energy. The homolytic bond energy is being used to calculate the thermodynamics of a heterolytic reaction.
Further, the commenter posted:
In another paper, they gave the following equilibrium constants for a (reversed Finkelstein) reaction LiBr + RI = LiI + RBr: MeI (K = 1.7), n-alkyl (K = 7), iPr (K = 30), t-Bu (K ca 100). They furthermore stated that for the reaction LiCl + RBr = LiBr + RCl, the equilibrium should even be further on the RCl side (K larger). (See: DOI: 10.1039/JR9550003173)
Now, I realized I could calculate a Gibbs free energy from the equilibrium. The methyl bromide reaction has a rather modest enthalpy of -1.32 kJ/mol and isopropyl -8.4 kJ/mol by my calculations. As an organic chemist, these values seem about right. I am finding these calculations daunting. The enthalpy values should include all of the factors in a reaction, not just the bond energy. I don't know how to adjust for an entropy change, though I would have thought it to be small and toward equilibrium in this substitution reaction.
Now I've seen the heat of neutralization of HF is -68 kJ/mol while HCl, HBr, and HI are -57 kJ/mol. The simplicity of HF makes this an ideal molecule to understand these energy differences. I can understand HCl, HBr, and HI as they simply form the same H3O(+). The enthalpy of neutralization is the neutralization of hydronium ion. This is analogous to my ignoring the bonds of LiI and LiBr as they are broken upon dissolving in acetone.
One analysis that seems to be in line as to how this problem may be solved has been contributed by Jim Clark
. This example illustrates the problem I have with understanding the data. It illustrates a Born-Haber cycle with enthalpies of vaporization, bond energy, ionization, electron affinity, and hydration. I find it difficult to verify any one of these data points and if I can, they may differ significantly from this source.
I don't know if I am trying to understand that which cannot be readily understood. I am dubious of gas phase acidity as it can often differ from solution phase acidity. For me, a clear warning to using values from one phase to predict reactions in another is the endothermic gas phase ionization of sodium. If sodium preferred to hold its electron, I would expect to find sodium in its ground state, yet metallic sodium reacts exothermally in solution.
I hope someone can clarify this problem.