Question

60 ml of h2 and 42ml of i2 are heated in a closed vessel . at equilibrium the vessel contains 28 ml hi. calculate degree of dissociation of hi.

a. 0.739

b. 0.719

c. 0.729

d. 0.749

Answer 1

B)0.719, may be the ans

Answer 2

Hii mate,

● Answer -
0.76

● Explanation -
- When H2 is heated with I2, hydroiodic acid is formed.
Reaction : H2 + I2 → 2HI
Initial : 60 42 0
At eqlbm : 60-x 42-x 2x

Given is 2x = 28,
[HI] = x = 28/2 = 14
[H2] = 60 - x = 60 - 14 = 46
[I2] = 42 - x = 42 - 14 = 28

Equilibrium constant is given by -
Kc = 28^2 / (46×28)
Kc = 0.61

- In aqueous form, HI dissociates to H2 & I2,
Reaction : 2HI → H2 + I2
Initial : 1 0 0
At eqlbm : 1-α α/2 α/2

Equilibrium constant is given by -
Kc' = (α/2)(α/2) / (1-α)^2
1/Kc = α^2 / 4(1-α)^2
2(1-α)/α = Kc
2(1-α)/α = 0.61
α = 0.76

Therefore, degree of dissociation is 0.76.


Hope this helps you ...