# NaBr reacting with H2SO4

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Hey guys, im a bit stuck on this quesion:

Sodium bromide reacts with concentrated sulfuric acid in a different way from sodium chloride. Write an equation for this reaction of sodium bromide and explain why bromide ions react differently from chloride ions (3)

I thought the answer would be

NaBr + H2SO4 -> NaHSO4 + HBr + SO2

Br- -> Br2

(1)2Br- -> Br2 +2e-

H2SO4 -> SO4

(2) 2H+ + H2SO4 + 2e- -> SO2 + 2H2O

Combining the 2 equations (1) + (2):

2Br- + 2H+ +H2SO4 -> Br2 + SO2 + 2H2O

But the ans was 2NaBr + 2H2SO4 -> Na2SO4 + Br2 + SO2 + 2H2O

Im not quite sure where they got the 2NaBr Na2SO4 Br2 and H2O from.

(Sorry it looks a bit messy)

Sodium bromide reacts with concentrated sulfuric acid in a different way from sodium chloride. Write an equation for this reaction of sodium bromide and explain why bromide ions react differently from chloride ions (3)

I thought the answer would be

NaBr + H2SO4 -> NaHSO4 + HBr + SO2

Br- -> Br2

(1)2Br- -> Br2 +2e-

H2SO4 -> SO4

(2) 2H+ + H2SO4 + 2e- -> SO2 + 2H2O

Combining the 2 equations (1) + (2):

2Br- + 2H+ +H2SO4 -> Br2 + SO2 + 2H2O

But the ans was 2NaBr + 2H2SO4 -> Na2SO4 + Br2 + SO2 + 2H2O

Im not quite sure where they got the 2NaBr Na2SO4 Br2 and H2O from.

(Sorry it looks a bit messy)

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Does anyone know why the general equation for NaBr and H2SO4 is

2NaBr + 2H2SO4-> Na2SO4 +Br2 + SO2 +2H2O

The closest I could get was

NaBr +H2SO4 -> 2NaHSO4 + 2HBr

2HBr + H2SO4 -> Br2 + SO2 + 2H2O

combined them to get

NaBr + 3H2SO4 -> 2NaHSO4 + Br2 + SO2 + 2H2O

I am not sure how they got 2H2SO4 and Na2SO4 , especially where the H has gone in Na2SO4

I looked at some solutions online but they only give the ionic equations, I get how they got the ionic equations but I don’t know how to write the full equations

Thanks for any help

2NaBr + 2H2SO4-> Na2SO4 +Br2 + SO2 +2H2O

The closest I could get was

NaBr +H2SO4 -> 2NaHSO4 + 2HBr

2HBr + H2SO4 -> Br2 + SO2 + 2H2O

combined them to get

NaBr + 3H2SO4 -> 2NaHSO4 + Br2 + SO2 + 2H2O

I am not sure how they got 2H2SO4 and Na2SO4 , especially where the H has gone in Na2SO4

I looked at some solutions online but they only give the ionic equations, I get how they got the ionic equations but I don’t know how to write the full equations

Thanks for any help

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#3

(Original post by

Does anyone know why the general equation for NaBr and H2SO4 is

2NaBr + 2H2SO4-> Na2SO4 +Br2 + SO2 +2H2O

The closest I could get was

NaBr +H2SO4 -> 2NaHSO4 + 2HBr

2HBr + H2SO4 -> Br2 + SO2 + 2H2O

combined them to get

NaBr + 3H2SO4 -> 2NaHSO4 + Br2 + SO2 + 2H2O

I am not sure how they got 2H2SO4 and Na2SO4 , especially where the H has gone in Na2SO4

I looked at some solutions online but they only give the ionic equations, I get how they got the ionic equations but I don’t know how to write the full equations

Thanks for any help

**asa_1224**)Does anyone know why the general equation for NaBr and H2SO4 is

2NaBr + 2H2SO4-> Na2SO4 +Br2 + SO2 +2H2O

The closest I could get was

NaBr +H2SO4 -> 2NaHSO4 + 2HBr

2HBr + H2SO4 -> Br2 + SO2 + 2H2O

combined them to get

NaBr + 3H2SO4 -> 2NaHSO4 + Br2 + SO2 + 2H2O

I am not sure how they got 2H2SO4 and Na2SO4 , especially where the H has gone in Na2SO4

I looked at some solutions online but they only give the ionic equations, I get how they got the ionic equations but I don’t know how to write the full equations

Thanks for any help

2Br

^{-}==> Br

_{2}+ 2e

SO

_{4}

^{2-}+ 4H

^{+}+ 2e ==> SO

_{2}+ 2H

_{2}O

---------------------------------------------------------------------------------------------------------- add together

2Br

^{-}+ SO

_{4}

^{2-}+ 4H

^{+}==> Br

_{2}+ SO

_{2}+ 2H

_{2}O

If you need the balanced equation (not the ionic equation) you have to put in the balancing (spectator ions)

For example, the bromide arrives in the form of sodium bromide, so you must also add two sodium ions, the sulfate ions come from the sulfuric acid and the hydrogen ions from more sulfuric acid, etc.

2NaBr + 2H

_{2}SO

_{4}==> Br

_{2}+ SO

_{2}+ 2H

_{2}O + Na

_{2}SO

_{4}

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(Original post by

Sulfuric acid is an oxidising agent, not a very strong one, but an oxidising agent nevertheless. With the larger halide ions, bromide and iodide, which are better reducing agents, there is a redox reaction:

2Br

SO

---------------------------------------------------------------------------------------------------------- add together

2Br

If you need the balanced equation (not the ionic equation) you have to put in the balancing (spectator ions)

For example, the bromide arrives in the form of sodium bromide, so you must also add two sodium ions, the sulfate ions come from the sulfuric acid and the hydrogen ions from more sulfuric acid, etc.

2NaBr + 2H

**charco**)Sulfuric acid is an oxidising agent, not a very strong one, but an oxidising agent nevertheless. With the larger halide ions, bromide and iodide, which are better reducing agents, there is a redox reaction:

2Br

^{-}==> Br_{2}+ 2eSO

_{4}^{2-}+ 4H^{+}+ 2e ==> SO_{2}+ 2H_{2}O---------------------------------------------------------------------------------------------------------- add together

2Br

^{-}+ SO_{4}^{2-}+ 4H^{+}==> Br_{2}+ SO_{2}+ 2H_{2}OIf you need the balanced equation (not the ionic equation) you have to put in the balancing (spectator ions)

For example, the bromide arrives in the form of sodium bromide, so you must also add two sodium ions, the sulfate ions come from the sulfuric acid and the hydrogen ions from more sulfuric acid, etc.

2NaBr + 2H

_{2}SO_{4}==> Br_{2}+ SO_{2}+ 2H_{2}O + Na_{2}SO_{4}(NaBr + H2SO4 -> NaHSO4 + HBr

2HBr + H2SO4 -> Br2 + SO2 + 2H2O)?

Thanks

Last edited by asa_1224; 1 year ago

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#5

(Original post by

Thank you so much for answering, so all I needed to do was write the half equation and add Na back to the equation to get the balanced equation? Also, how do I know when to answer with half equation and when to use the full equation? The question only says write an equation

(NaBr + H2SO4 -> NaHSO4 + HBr

2HBr + H2SO4 -> Br2 + SO2 + 2H2O)?

Thanks

**asa_1224**)Thank you so much for answering, so all I needed to do was write the half equation and add Na back to the equation to get the balanced equation? Also, how do I know when to answer with half equation and when to use the full equation? The question only says write an equation

(NaBr + H2SO4 -> NaHSO4 + HBr

2HBr + H2SO4 -> Br2 + SO2 + 2H2O)?

Thanks

NaBr + H

_{2}SO

_{4}==> NaHSO

_{4}+ HBr

is NOT a half-equation

Br

^{-}+ H

_{2}SO

_{4}==> HSO

_{4}

^{-}+ HBr

is NOT a half-equation - it is an ionic equation

An equation shows the reactants and products of a chemical reaction and their stoichiometries

An ionic equation shows only the species that actually take part in the reaction, omitting the spectator ions

A half-equation shows only the reduction or oxidation of a redox reaction.

You add the two half-equations (one oxidation and one reduction) together to get the ionic redox equation.

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(Original post by

Be careful with the term "half-equation", it means either the reduction or the oxidation part (half) of a redox reaction.

NaBr + H

is NOT a half-equation

Br

is NOT a half-equation - it is an ionic equation

An equation shows the reactants and products of a chemical reaction and their stoichiometries

An ionic equation shows only the species that actually take part in the reaction, omitting the spectator ions

A half-equation shows only the reduction or oxidation of a redox reaction.

You add the two half-equations (one oxidation and one reduction) together to get the ionic redox equation.

**charco**)Be careful with the term "half-equation", it means either the reduction or the oxidation part (half) of a redox reaction.

NaBr + H

_{2}SO_{4}==> NaHSO_{4}+ HBris NOT a half-equation

Br

^{-}+ H_{2}SO_{4}==> HSO_{4}^{-}+ HBris NOT a half-equation - it is an ionic equation

An equation shows the reactants and products of a chemical reaction and their stoichiometries

An ionic equation shows only the species that actually take part in the reaction, omitting the spectator ions

A half-equation shows only the reduction or oxidation of a redox reaction.

You add the two half-equations (one oxidation and one reduction) together to get the ionic redox equation.

Thank you

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#7

(Original post by

Hi, so what should I write if the question only says write the equation or the general equation? Do I combine the ionic equations (Do I write the ionic redox equation)?

Thank you

**asa_1224**)Hi, so what should I write if the question only says write the equation or the general equation? Do I combine the ionic equations (Do I write the ionic redox equation)?

Thank you

If you were to write the ionic equation I doubt if you would lose a mark, unless it specifically asked for a full chemical equation showing spectator ions.

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(Original post by

You write a balanced equation for the reaction showning reactants and products.

If you were to write the ionic equation I doubt if you would lose a mark, unless it specifically asked for a full chemical equation showing spectator ions.

**charco**)You write a balanced equation for the reaction showning reactants and products.

If you were to write the ionic equation I doubt if you would lose a mark, unless it specifically asked for a full chemical equation showing spectator ions.

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