# NaBr reacting with H2SO4

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#1
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)
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#2
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
0
1 year ago
#3
(Original post by 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
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- ==> Br2 + 2e
SO42- + 4H+ + 2e ==> SO2 + 2H2O
2Br- + SO42- + 4H+ ==> Br2 + SO2 + 2H2O

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 + 2H2SO4 ==> Br2 + SO2 + 2H2O + Na2SO4
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#4
(Original post by 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- ==> Br2 + 2e
SO42- + 4H+ + 2e ==> SO2 + 2H2O
2Br- + SO42- + 4H+ ==> Br2 + SO2 + 2H2O

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 + 2H2SO4 ==> Br2 + SO2 + 2H2O + Na2SO4
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
Last edited by asa_1224; 1 year ago
0
1 year ago
#5
(Original post by 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
Be careful with the term "half-equation", it means either the reduction or the oxidation part (half) of a redox reaction.

NaBr + H2SO4 ==> NaHSO4 + HBr
is NOT a half-equation

Br- + H2SO4 ==> HSO4- + 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.
0
#6
(Original post by charco)
Be careful with the term "half-equation", it means either the reduction or the oxidation part (half) of a redox reaction.

NaBr + H2SO4 ==> NaHSO4 + HBr
is NOT a half-equation

Br- + H2SO4 ==> HSO4- + 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.
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
0
1 year ago
#7
(Original post by 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
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.
0
#8
(Original post by 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.
Ok! thank you so much for your help 0
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