What Is the Equilibrium Constant for the Reaction 2HI <==> H2 + I2?

[chiakimaron]

Please help me the following questions
In an experiment to determine Kc , the equilibrium constant for the reversible reaction
2HI<==> H2 + I2
0.21 g of hydrogen iodide was heated at 530℃ in a bulb of volume 100 cm^3 until equilibrium was reached . This bulb was broken under potassium iodide solution , and the iodine present was found to be sufficient to react with 4.0 cm^3 of 0.1 M sodium thiosulphate solution .
(a) Why was it adequate to absorb the iodine at room temperature , although in this case its amount at 530℃ was actually required?
(b) From the information given calculate Kc , the equilibrium constant , at 530℃ .

 

[lightarrow]

Please help me the following questions
In an experiment to determine Kc , the equilibrium constant for the reversible reaction
2HI<==> H2 + I2
0.21 g of hydrogen iodide was heated at 530? in a bulb of volume 100 cm^3 until equilibrium was reached . This bulb was broken under potassium iodide solution , and the iodine present was found to be sufficient to react with 4.0 cm^3 of 0.1 M sodium thiosulphate solution .
(a) Why was it adequate to absorb the iodine at room temperature , although in this case its amount at 530? was actually required?
(b) From the information given calculate Kc , the equilibrium constant , at 530? .

(a) H2 is not water soluble while HI and I2 are, so when you break the bulb inside the solution, H2 immediately escapes or forms a separate, gaseous phase, while I2 dissolves in water and reacts with I- forming I3-, so H2 and I2 will not be present anylonger in the same phase and so they don't have time to react as faster as before, after the temperature has lowered.
(b) There is something strange with the data you have written: at 530°C, using thermodinamical data tabulated, I compute an equilibrium constant of 203.4 for the reaction 2HI <--> H2 + I2 (it's 14.3 for the reaction HI <--> 0.5H2 + 0.5I2), from which I compute 1.59*10^(-3) mol of I2 at equilibrium at that temperature inside the bulb, which corresponds to 3.18*10^(-3) mol of thiosulphate (I assumed the reaction: 2S2O3-- + I2 --> 2I- + S2O4--) that, at a concentration of 0.1M, corresponds to 31.8 cm^3 of thiosulphate solution (and not to 4 cm^3 as your data).