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

I did the question like this:

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

Why is the answer

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

  • $\begingroup$ Useful links for text and formula formatting (not to be applied to titles): Notation basics , Formatting of math/chem expressions and upright vs italic $\endgroup$
    – Poutnik
    Feb 19, 2022 at 8:13
  • $\begingroup$ 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. $\endgroup$
    – Poutnik
    Feb 19, 2022 at 8:23
  • 1
    $\begingroup$ Related: chemistry.stackexchange.com/questions/4750/… $\endgroup$ Feb 19, 2022 at 8:38

1 Answer 1


$\ce{H2SO4}$ 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 ($2$), and so improves the yield in $\ce{Br2}$, according to $$\ce{2HBr + 2 H2SO4 -> Br2 + SO2 + H3O+ + HSO4- \tag{3}}$$


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