Question

2ab2 react to give 2ab +b2 degree of dissociation of ab2 is x what will be the equation for x in term of kp and equilibrium pressure p

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)      

Answer 2

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The final stage of respiration occur through a series of oxidization -reduction electrons transfer reaction that yield the energy to drive oxidative.