# Reaction between concentrated sulfuric acid and sodium bromide

What are the products of the reaction between sodium bromide and concentrated sulfuric acid?

I did the question like this:

\begin{align} \ce{2Br- &-> Br2 + 2e-} \tag{R1.1} \\ \ce{SO4^2- 4H+ + 2e- &-> SO2 + 2H2O} \tag{R1.2} \\ \hline \ce{SO4^2- + 2Br- + 4H+ &-> SO2 + Br2 + 2H2O} \tag{R1.3} \end{align}

$$\ce{NaBr + H2SO4 -> NaHSO4(s) + HBr(g)}? \tag{R2}$$

• Useful links for text and formula formatting (not to be applied to titles): Notation basics , Formatting of math/chem expressions and upright vs italic Feb 19, 2022 at 8:13
• E.g. type $\ce{2e- +4H+ + SO4^2- -> SO2 + 2H2O}$ to get $\ce{2e- +4H+ + SO4^2- -> SO2 + 2H2O}$. That is for inline formula or equation within the text. To display it in dedicated space, use double dollars: $$\ce{2e- +4H+ + SO4^2- -> SO2 + 2H2O}$$ // Use 4 trailing spaces for forced newline, or double enter for an empty line. Feb 19, 2022 at 8:23
• Feb 19, 2022 at 8:38

$$\ce{H2SO4}$$ reacts with $$\ce{NaBr}$$ in two steps. First $$\ce{H2SO4 + NaBr -> NaHSO4 + HBr \tag{1}}$$ This reaction occurs at all conditions of temperature and concentrations, even in dilute solutions, where the reagents and products are transformed into ions. Later on, if the sulfuric acid is hot and concentrated enough, $$\ce{HBr}$$ produced by the reaction ($$1$$) is partially oxidized by $$\ce{H2SO4}$$ according to a second equation $$\ce{2 HBr + H2SO4 <=> Br2 + SO2 + 2 H2O \tag{2}}$$ An excess of sulfuric acid helps getting rid of the $$2\ce{H2O}$$ molecules produced in ($$2$$) in a rection like $$\ce{H2SO4 + H2O -> HSO4^- + H3O+}$$ and so improves the yield in $$\ce{Br2}$$, as ($$2$$) in replaced by ($$3$$) $$\ce{2HBr + 2 H2SO4 -> Br2 + SO2 + H3O+ + HSO4- \tag{3}}$$
• Nobody can say that "$\ce{H2O}$ must be produced". I think that whatever the nature of the halogen $\ce{X}$, the reaction of $\ce{H2SO4}$ on $\ce{NaX}$ is the production of $\ce{HX}$. But this $\ce{HX}$ stays as it is, if $\ce{X}$ is fluorine and chlorine. But if $\ce{X}$ is bromine or iodine, the $\ce{HBr}$ or $\ce{HI}$ produced reacts with a new $\ce{H2SO4}$ molecule so that $\ce{Br}$ or $\ce{I}$ is oxidized into $\ce{Br2}$ and $\ce{I2}$ by this supplementary $\ce{H2SO4}$ molecule, which is reduced into $\ce{SO2}$ (and $\ce{H2O}$) Jul 11 at 16:23