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0.96g of Hydrogen iodide was heated to 450 degree centigrade till the equilibrium was reached.It was then quickly cooled and the amount of iodine formed required 15.7ml of N/10 Na2S2O3. Calculate the degree of dissociation of HI at 450 degree centigrade
Dear student,
I will give u a hint regarding this:
PCl5 dissociates as: PCl5 ? PCl3 Cl2 If α is the degree of dissociation at certain temperature under atmospheric pressure, then Initial concentration: PCl5 = 1 PCl3 = 0 Cl2 = 0 At Equilibrium: PCl5 = 1 – α PCl3 = α Cl2 = α Total number of moles at equilibrium = 1 – α + α + α = 1 + α Partial pressures of PCl5, PCl3 and Cl2 will be: p (PCl3) = α p / 1 + α p (Cl2) = α p / 1 + α p (PCl5) = (1 – α) p / 1 + α Kp = p (PCl3) X p (Cl2) / p (PCl5) Kp = [(α p / 1 + α) X (α p / 1 + α)]/[ (1 – α) p / (1 + α)] = α2 p / (1 – α) 2 Substituting p = 1 atm and α = 0.2 Kp = (0.2) 2 X 1 / (1 – (0.2)) 2 = 0.041 When α = 1/2 = 0.5, let pressure is p’ Kp = α2 p’ / (1 – α) 2 0.041 = (0.5) sup>2 p’ / (1 – (0.5)) 2 p’ = (0.041) [1 – (0.5) 2] / (0.5) 2 p’ = 0.125 atm
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