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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

## 1 Answers

10 years ago

Dear student,

I will give u a hint regarding this:

# The degree of dissociation of PCl5 at a certain temperature and under atmospheric pressure is 0.2. Calculate the pressure at which it will be half dissociated at the same temperature.

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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