March 28, 2024, 07:53:33 PM
Forum Rules: Read This Before Posting


Topic: Sodium thiosulfate/Iodine equilibrium  (Read 18793 times)

0 Members and 1 Guest are viewing this topic.

dolphinsiu

  • Guest
Sodium thiosulfate/Iodine equilibrium
« on: April 01, 2005, 03:37:54 PM »
An equilibrium is established by putting 2 g of HI(g) in a 500 cm3 glass bulb maintained at a constant temperature.the reaction is represented by follow:
2HI(g)<=>H2(g)+I2(g)
The amount of iodine in equilibriun mixture is determined by titration.23 cm3 of 0.02M Na2S2O3 is needed to react with the iodine.

(a)Write the equation for a reaction of iodine and sodium thiosulphate
My Ans. I2(aq)+2S2O32-(aq)->2I-(aq)+S4O62-(aq)

(b)Determine the value of Kc of the reaction
My Ans. mol.of Na2S2O3 /2 = eqm. mol. of I2 =2.3x10-3mol

eqm.mol. of HI=2/128=0.015625

Kc=(2.3x10-3)^2/0.015625^2=0.0217

But the real ans is 0.0437 .Why?
« Last Edit: April 01, 2005, 06:38:11 PM by hmx9123 »

Offline Donaldson Tan

  • Editor, New Asia Republic
  • Retired Staff
  • Sr. Member
  • *
  • Posts: 3177
  • Mole Snacks: +261/-13
  • Gender: Male
    • New Asia Republic
Re:AL Chemistry(Only Part B)I have tried but why is wrong?
« Reply #1 on: April 01, 2005, 06:06:16 PM »
You made a few mistakes:

1 mole of iodine reacts with 2 mole of thiosulphate ions.
amount of thiosulphate reacted = 23/1000 x 0.02 = 4.6 x 10-4 moles
amount of iodine present at eqbm = 2.3 x 10-4 moles

amount of HI present at eqbm is not 2g. the iodine and hydrogen present at equilibrium comes from the dissociation of HI molecules.

rule of the thumb: always work with SI units, convert 500cm3 to 0.5dm3
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

dolphinsiu

  • Guest
Re:Sodium thiosulfate/Iodine equilibrium
« Reply #2 on: April 02, 2005, 12:26:15 AM »
I am so sorry.The question should be:
23 cm3 of 0.2M Na2S2O3 is needed to react with the iodine.

But my ans is still wrong and it is 0.0435
But real ans. is 0.0437
Please check where my mistake is made:

                    2HI    <=>         H2            +            I2
At start:   0.015625 mol         0 mol                    0 mol
At eqm:   0.011025 mol      2.3x10-4mol       2.3x10-4 mol

Kc=(2.3x10-4)^2/(0.011025)^2=0.0435

But should I care about hte data given as 500cm3 in this question?

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27633
  • Mole Snacks: +1799/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re:Sodium thiosulfate/Iodine equilibrium
« Reply #3 on: April 02, 2005, 04:25:05 AM »
But my ans is still wrong and it is 0.0435
But real ans. is 0.0437

It is 0.5% error, I would suspect some rodunding came into play and your answer is generally OK.

ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Sponsored Links