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Author Topic: Sodium thiosulfate/Iodine equilibrium  (Read 10418 times)

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dolphinsiu

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Sodium thiosulfate/Iodine equilibrium
« on: April 01, 2005, 09:37:54 AM »

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, 12:38:11 PM by hmx9123 »
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Donaldson Tan

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Re:AL Chemistry(Only Part B)I have tried but why is wrong?
« Reply #1 on: April 01, 2005, 12:06:16 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
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dolphinsiu

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Re:Sodium thiosulfate/Iodine equilibrium
« Reply #2 on: April 01, 2005, 06:26:15 PM »

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?
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Borek

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Re:Sodium thiosulfate/Iodine equilibrium
« Reply #3 on: April 01, 2005, 10:25:05 PM »

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.

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