Equilibrium constant queations

[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).

 

[Blueshift5]

I can help you answer the following questions about the equilibrium constant for the reversible reaction 2HI<==> H2 + I2. To start, let's define what the equilibrium constant (Kc) represents. It is a measure of the ratio of products to reactants at equilibrium for a given chemical reaction. In other words, it tells us how far the reaction has proceeded towards the products or reactants at a specific temperature.

(a) The reason it was adequate to absorb the iodine at room temperature is because the reaction was already at equilibrium when the bulb was broken. This means that the amount of iodine present at 530℃ was enough to react with the sodium thiosulphate solution at room temperature. The equilibrium constant is a constant value for a specific temperature, so the amount of iodine present at a different temperature would not affect the value of Kc.

(b) To calculate Kc at 530℃, we can use the information given in the question. First, we need to determine the moles of iodine present in 4.0 cm^3 of 0.1 M sodium thiosulphate solution. This can be calculated using the formula n = c x V, where n is the number of moles, c is the concentration, and V is the volume. Plugging in the values, we get n = 0.1 x 0.004 = 0.0004 moles of iodine.

Next, we need to determine the initial moles of hydrogen iodide (HI) present in the bulb before it was broken. This can be calculated using the mass of HI (0.21 g) and its molar mass (127.9 g/mol). We get n = 0.21/127.9 = 0.00164 moles of HI.

Now, we can use the stoichiometric ratio between HI and I2 to determine the moles of I2 present at equilibrium. Since the reaction is 2:1, we know that the moles of I2 will be half of the moles of HI. Therefore, at equilibrium, there will be 0.00082 moles of I2.

Finally, we can plug in all these values into the equation for Kc, which is Kc =

[I2]/[HI]^2. Substituting the values, we get Kc = (