Do we use sodium hydrogen carbonate to quench this reaction H2+I2->2HI?
The question asks to state a suitable technique to obtain rate data for the following reaction:
H2 (g) + I2 (g) -> 2HI (g)
At a first glance I know that this reaction is supposed to be reversible as we used to use it when demonstrating no change in equilibrium position on changing pressure as the two sides have the same number of moles etc.
Anyway, the equation given indicates the reaction goes to completion so that is what we will work with.
The answer is:
Withdraw samples at regular time intervals. Stop the reaction in the sample by adding sodium hydrogencarbonate. Add excess potassium iodide and titrate the liberated iodine with standard sodium thiosulfate(VI) solution
According to what I know, sodium hydrogen carbonate reacts with acids (which in this case is HI).
Can someone explain which species will react with sodium hydrogen carbonate?
Additionally, I would be grateful if you included a refresher on the reason we add iodate(V) ions before titrating with thiosulfate.
Thanks in advance!
H2 (g) + I2 (g) -> 2HI (g)
At a first glance I know that this reaction is supposed to be reversible as we used to use it when demonstrating no change in equilibrium position on changing pressure as the two sides have the same number of moles etc.
Anyway, the equation given indicates the reaction goes to completion so that is what we will work with.
The answer is:
Withdraw samples at regular time intervals. Stop the reaction in the sample by adding sodium hydrogencarbonate. Add excess potassium iodide and titrate the liberated iodine with standard sodium thiosulfate(VI) solution
According to what I know, sodium hydrogen carbonate reacts with acids (which in this case is HI).
Can someone explain which species will react with sodium hydrogen carbonate?
Additionally, I would be grateful if you included a refresher on the reason we add iodate(V) ions before titrating with thiosulfate.
Thanks in advance!
(edited 5 years ago)
Can you post the question? Not sure how you are removing samples when all the reactants and products are gaseous.
Original post by BobbJo
Can you post the question? Not sure how you are removing samples when all the reactants and products are gaseous.
The question is:
state a suitable technique to obtain rate data for each of the following reactions:
(a)Mg(s) + 2H+(aq) -> Mg2+(aq) + H2(g)
(b)CH3COOCH2CH3(l) + OH-(aq) -> CH3COO-(aq) + CH3CH2OH(aq)
(c)H2 (g) + I2 (g) -> 2HI (g)
The answer is on page 38 of https://www.pearsonschoolsandfecolleges.co.uk/AssetsLibrary/SECTORS/Secondary/SUBJECT/Science/Misc/Edexcel-AS-and-A-level-answers/Edexcel-Chemistry-Answers-Book2-FINAL.pdf
It is from edexcel's chemistry A2 book so the answer is on its website.
I don't know how to remove gas samples but the question is after a lesson with the title of "Methods of measuring the rate of reaction" which is the first lesson in the Further Kinetics section.
(edited 5 years ago)
Original post by CuriosityYay
The question is:
state a suitable technique to obtain rate data for each of the following reactions:
(a)Mg(s) + 2H+(aq) -> Mg2+(aq) + H2(g)
(b)CH3COOCH2CH3(l) + OH-(aq) -> CH3COO-(aq) + CH3CH2OH(aq)
(c)H2 (g) + I2 (g) -> 2HI (g)
The answer is on page 38 of https://www.pearsonschoolsandfecolleges.co.uk/AssetsLibrary/SECTORS/Secondary/SUBJECT/Science/Misc/Edexcel-AS-and-A-level-answers/Edexcel-Chemistry-Answers-Book2-FINAL.pdf
state a suitable technique to obtain rate data for each of the following reactions:
(a)Mg(s) + 2H+(aq) -> Mg2+(aq) + H2(g)
(b)CH3COOCH2CH3(l) + OH-(aq) -> CH3COO-(aq) + CH3CH2OH(aq)
(c)H2 (g) + I2 (g) -> 2HI (g)
The answer is on page 38 of https://www.pearsonschoolsandfecolleges.co.uk/AssetsLibrary/SECTORS/Secondary/SUBJECT/Science/Misc/Edexcel-AS-and-A-level-answers/Edexcel-Chemistry-Answers-Book2-FINAL.pdf
NaHCO3 is a base and will react with HI. Excess KI is added to dissolve the iodine. (Oxidising agents will qualitatively liberate iodine from KI.) It is titrated with Na2S2O3 of known concentration to determine [I2]
Iodate (V) is an oxidising agent, qualitatively liberating I2 from I-:
IO3- + 5I- + 6H+ -> 3I2 + 3H2O
The I2 is then titrated with standard Na2S2O3,
Original post by BobbJo
NaHCO3 is a base and will react with HI. Excess KI is added to dissolve the iodine. (Oxidising agents will qualitatively liberate iodine from KI.) It is titrated with Na2S2O3 of known concentration to determine [I2]
Iodate (V) is an oxidising agent, qualitatively liberating I2 from I-:
IO3- + 5I- + 6H+ -> 3I2 + 3H2O
The I2 is then titrated with standard Na2S2O3,
Iodate (V) is an oxidising agent, qualitatively liberating I2 from I-:
IO3- + 5I- + 6H+ -> 3I2 + 3H2O
The I2 is then titrated with standard Na2S2O3,
Thanks.
But how will NaHCO3 stop the reaction if it just reacts with the product?
And just to confirm: KI will only dissolve the reactant iodine while iodate(V) will liberate iodide ions into iodine from for example a compound like HI
Original post by CuriosityYay
Thanks.
But how will NaHCO3 stop the reaction if it just reacts with the product?
And just to confirm: KI will only dissolve the reactant iodine while iodate(V) will liberate iodide ions into iodine from for example a compound like HI
But how will NaHCO3 stop the reaction if it just reacts with the product?
And just to confirm: KI will only dissolve the reactant iodine while iodate(V) will liberate iodide ions into iodine from for example a compound like HI
It is also unclear to me how NaHCO3 will stop the reaction. I don't think NaHCO3 will stop the reaction. For this reason, their answer is not very convincing to me.
KI will dissolve the iodine. Any oxidising agent such as iodate(V) will liberate iodine from I- ions.
http://users.metu.edu.tr/chem223/IodometrySB.pdf
Here is a procedure on the titration part
Original post by BobbJo
It is also unclear to me how NaHCO3 will stop the reaction. I don't think NaHCO3 will stop the reaction. For this reason, their answer is not very convincing to me.
KI will dissolve the iodine. Any oxidising agent such as iodate(V) will liberate iodine from I- ions.
http://users.metu.edu.tr/chem223/IodometrySB.pdf
Here is a procedure on the titration part
KI will dissolve the iodine. Any oxidising agent such as iodate(V) will liberate iodine from I- ions.
http://users.metu.edu.tr/chem223/IodometrySB.pdf
Here is a procedure on the titration part
Thank you.
I really didn't get how NaHCO3 would stop the reaction. So yeah, it is probably an error.
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