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