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Grade 12th passPhysical Chemistry

For the reaction: 2HI(g) = H_2(g) + I_2(g)
The degree of dissociation(a) of HI_(g) is related to equilibrium constant, K_p by the expression

{ 2 (2Kp)^1÷2 } ÷ {1+ 2 (K_p)^1÷2 }

{1 + 2 (K_p)^1÷2 } ÷ 2

{ (1 + 2K_p) ÷ 2 }^1÷2

{ 2K_p ÷ (1 + 2K_p) }^1÷2
Please please give answer step by step
NOTE : PLEASE give step by step answer...

Yasir Rather

7 Years agoGrade 12th pass

1 Answer

Arun

Approved Tutor Answer7 Years ago

Dear student

2HI → H2 + I2

t=teq. C(1-a) Ca/2 Ca/2

Kp = (Ca/2)^2/(C(1-a))^2 => a^2/4(1-a)^2

a/(1-a) = 2root(Kp)

a = 2root(Kp) – 2root(Kp)*a

a(1 + 2root(Kp)) = 2root(Kp)

a = 2root(Kp)/(1 + 2root(Kp))

Regards

Arun (askIITians forum expert)

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