5
$\begingroup$

Why do sodium halides react so differently with sulfuric acid? \begin{align} \ce{NaF + H2SO4 &-> NaHSO4 + HF} \tag{1a}\label{NaF}\\ \ce{NaCl + H2SO4 &-> NaHSO4 + HCl} \tag{1b}\label{NaCl}\\ \ce{2 NaBr + H2SO4 + 2H+ &-> Br2 + SO2 + 2H2O + 2Na+} \tag2\label{NaBr}\\ \ce{8 NaI + H2SO4 + 8H+ &-> 4 I2 + H2S + 4 H2O + 8Na+} \tag3\label{NaI} \end{align}

Conventional explanation: $\ce{NaI}$ is a strong enough reducing agent to reduce the sulfur, and $\ce{NaBr}$ is a little less strong, so sulfur is not reduced as much. Equivalently, $\ce{H2SO4}$ is not as strong an oxidising agent to reduce $\ce{NaF}$ and $\ce{NaCl}$.

My ensuing question: Why is reaction \eqref{NaI} more favourable than \eqref{NaBr} and $(\ref{NaF},\ref{NaCl})$ for sodium iodide, so that it reacts with sulfuric acid like it does. What factors are involved: what is more stable, for instance, (or there is lower Gibbs free energy or something else) about reducing sulfur as much as possible (as $\ce{NaI}$ 'chooses' \eqref{NaI} over $(\ref{NaF},\ref{NaCl},\ref{NaBr})$?

Ditto for sodium bromide. I understand why \eqref{NaI} is impossible, but why is \eqref{NaBr} chosen over $(\ref{NaF},\ref{NaCl})$?

$\endgroup$
10
$\begingroup$

A factor you didn't touch upon which may have some weight in determining the reactions is the acidities of the hydrogen halides formed by reactions of the first type (which is fundamentally a coarse qualitative analysis of the free energy of reaction). In aqueous solutions, the order of acidity is $\ce{HF} \ll \ce{HCl} < \ce{HBr} < \ce{HI}$. While concentrated sulfuric acid is obviously not the same as water, perhaps it is fair to say the relative acidities of the hydrogen halides are approximately the same, given they are both polar protic solvents with very high dielectric constants.

As you go down the family, the acid formed in reactions of the first type become stronger, and there is less thermodynamic drive to follow such a reaction pathway. $\ce{HI}$ is somewhat close in acidity to $\ce{H2SO4}$ since neither conjugate base is very coordinating. Meanwhile, as you say, the reducing power of the halides increases, and since sulfur is in a high oxidation number in $\ce{H2SO4}$, redox reaction pathways become more attractive. A detailed analysis of free energies of reactions should agree with the change in the mechanism since it is unlikely these reactions are kinetically controlled.

Edit: Another neat consequence of the acidity strength argument is that it explains why the reactions stop at the bisulfate anion instead of going all the way to the halide sulfates; $\ce{HSO4-}$ is a much, much weaker acid than $\ce{H2SO4}$, so there is no free energy to lose out of consuming it to produce a much stronger acid (in the case of $\ce{HCl}$, $\ce{HBr}$, and $\ce{HI}$ at least). $\ce{HF}$ is similar in acidity to $\ce{HSO4-}$ (in water), so it's not as readily explainable in such a qualitative analysis.

$\endgroup$
  • $\begingroup$ 'The acid formed in reactions of the first type become stronger, and there is less thermodynamic drive to follow such a reaction pathway'. Sorry, but I don't quite understand why stronger acids have lower entropy (that's what I assume makes it more thermodynamically favourable). $\endgroup$ – Meow Apr 20 '13 at 11:03
  • 1
    $\begingroup$ Free energy is what thermodynamically determines the paths of reaction, and it is a combination of enthalpy changes and entropy changes. Stronger acids have higher free energy content mostly due to enthalpy. You can think of strong acids as entities with a very weak tendency to hold onto $H^{+}$ ions, either because the $H^{+}$ is poorly bound or because the acid would be much stabler without it. Therefore, it becomes strongly exothermic for the $H^{+}$ to migrate to another substance that can better accept it, creating a weaker acid. The inverse reaction is highly endothermic, and disfavoured $\endgroup$ – Nicolau Saker Neto Apr 20 '13 at 12:46
  • $\begingroup$ "acidity" is something which happens in aqueous solutions only! Think of NaCl + SiO2 makes Na2SiO3 + HCl if temperature is sufficient. This kind of reaction is about volatility, not "acid strength"! $\endgroup$ – Georg Apr 23 '13 at 15:42
  • 3
    $\begingroup$ Not at all. The concepts of acidity and basicity can be extended to any ionizable liquid. As long as a solvent molecule can ionize into a "solvonium" cation and a "solvate" anion, one can define acidity, even if it seems strange at first. Check this wiki article. Your example doesn't fit an acidity analysis because neither $NaCl$ nor $SiO_2$ are ionizable compounds ($NaCl$ is dissociable, not the same thing! And $SiO_2$ is an extended solid. I'm actually not sure what its liquid molecular structure is like) $\endgroup$ – Nicolau Saker Neto Apr 25 '13 at 2:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.