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Rate law question with KI

  • Thread starter mesa
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  • #1
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Okay, so I did a lab recently where we are to determine the Rate Law for a mixture of KI, starch, Na2S2O3, H202 and enough distilled water to make sure total solution volume is equal for all experiments. So when this thing starts to go we get I2 +2S2O4--->2I + S4O6, so when doing the calculations do we use 1/2M of KI for determining the rate?

Now where does the I2 come from? When KI hits the solution doesn't it dissociate into negatively charged I ions? I wish they gave more detail.
 

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  • #2
646
18
Okay, I think I see where the I2 comes from. So the buffer put's out H+ ions that react with the I- ions that came from the dissociation of KI allowing them to lose an electron and form I2, then the I2 reacts with the thiosulfate to reform I- ions until all thiosulphate is consumed allowing the I2 to form without disruption and combine with the starch changing the color. Is this right?

Now to calculate the rate law that would be rate=k[I-]^x[H2O2]^y?
Then calculate the M of each in solution and set rate = the amount of thiosulphate/time for the reaction to occur?
Or in other words : thiosulphateM/time(s)=k[I-]^x[H2O2]^y
and I- is equal to the molarity of KI in solution, is this right?
 
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  • #3
chemisttree
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Okay, so I did a lab recently where we are to determine the Rate Law for a mixture of KI, starch, Na2S2O3, H202 and enough distilled water to make sure total solution volume is equal for all experiments. So when this thing starts to go we get I2 +2S2O4--->2I + S4O6, so when doing the calculations do we use 1/2M of KI for determining the rate?

Now where does the I2 come from? When KI hits the solution doesn't it dissociate into negatively charged I ions? I wish they gave more detail.
I've heard that H2O2 reacts with iodide under acidic conditions to produce iodine.
 
  • #4
Borek
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I've heard that H2O2 reacts with iodide under acidic conditions to produce iodine.
Just yesterday I have watched this video:
 
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  • #5
646
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Just yesterday I have watched this video:
Hey Borek!
Okay I have a question on her video.

Did her solution turn yellow because she started producing small amounts of I2 early on?
My thinking is the KI dissociates into K+ and I- ions in the water and the NaHSO4 into Na and HSO4. Then the HSO4 decompose into H+ and SO4- so the hydrogen ions would take electrons away from the I- allowing it to form some iodine. Is this right?
 
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  • #6
646
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I've heard that H2O2 reacts with iodide under acidic conditions to produce iodine.
Okay so the buffer I used released H+ ions that took away electrons from the I- ions allowing it to form I2, is that right?
 
  • #7
chemisttree
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Okay so the buffer I used released H+ ions that took away electrons from the I- ions allowing it to form I2, is that right?
When H+ takes away electrons from something, it generates Ho. That's hydrogen gas. Did your solution bubble a lot? No? Hmmmm... Must be something else oxidizing that iodide.

I've heard that H2O2 reacts with iodide under acidic conditions to produce iodine. Modify your thought process to include that observation.
 
  • #8
646
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I've heard that H2O2 reacts with iodide under acidic conditions to produce iodine. Modify your thought process to include that observation.
So the H+ combines with the H2O2 and I- to make H2O and I2?
 
  • #9
AGNuke
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[tex]I_2 + 2S_2O_3^{2-} \xrightarrow{} 2I^- + S_4O_6^{2-}[/tex]
[tex]H_2O_2 + I^-\xrightarrow[]{acidic} H_2O + I_3^- \xrightarrow{} I_2[/tex]

H2O2 is a strong Oxidizing agent. It serves O (Oh Oh! :devil:) to other molecules and oxidize them, and itself gets happily reduced to ye ol' sweet water.

The temporary yellow coloration is due to presence of I3-.
 

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