# 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 ($$3$$) $$\ce{2HBr + 2 H2SO4 -> Br2 + SO2 + H3O+ + HSO4- \tag{3}}$$

• "This reaction occurs at all conditions of temperature and concentrations, even in dilute solutions" ||| What about chlorides and iodides? Why would the halides be protonated? Is this related to displacement of weak acid by strong acid? But aren't HCl, HBr, HI stronger than H2SO4? By the way, isn't NaHSO4 soluble? If we're in an aqueous environment, why would hydrogen halide be given off? Won't it just dissolve? Sorry ffor this barrage! Jul 11 at 11:55
• Why must H2O be removed? Jul 11 at 12:51
• @Cheng: I think it's better if you include this comment in your question. Jul 11 at 14:07
• 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
• I mean how is removing H2O supposed to improve yield? Jul 15 at 8:20