inorganic chemistry - Why do sodium halides react so differently with sulfuric acid?

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 $(\ref{NaF},\ref{NaCl})$?

Answer

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.