Question

At temperature T, a compound AB2 (g) diassociates according to the reaction :
2AB2(g) [reverisble sign ] 2AB(g) + B2.
with degree of dissociation, x, which is small compared to unity. Deduce the expression for x in terms of equillibrium constant Kp and the total pressure.

Answer 1

We have,                     2AB2(g)                 ↔       2AB(g)                  +           B2(g)

Mole before dissociation     1                                 0                                       0

Mole after dissociation     1 – x                              x                                       x/2

Total mole at equilibrium (∑n)    =             1 – x + x + x/2

=             1 + x/2

Now,                KP            =             {[nB2×(nAB)2]/[nAB2]2}×[P/∑n] Δn

Or,                  KP            =             {[(x/2)(x)2]/[(1 – x)2]}×[P/(1 + x/2)]

Or,                   KP            =             x2P/2        

[As,        x is small, 1 – x ≈ 1 and 1 + x/2 ≈ 1]

Or,                     x              =  (2KP/P)1/3 (C)      

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Answer 2

Answer:

Please refer to the picture attached!

Explanation:

PS To find out x:

$kp \: = \frac{p {x}^{3}}{2} \\ \\ = > {x}^{3} = \frac{2kp}{p} \\ \\ = > x = \sqrt[3]{ \frac{2kp}{p} }$