Q39 / 87 · NTA Abhyas (JEE Main) 2020 · Chemistry · Chemical Equilibrium

The dissociation equilibrium of a gas AB 2 can be represented as,
2  AB 2 (g)⇌2 AB(g)+ B 2 (g)
The degree of dissociation is x and is small as compared to 1 . The expression relating the degree of dissociation (x) with equilibrium constant K P and total pressure P is:

  • A
    (2K p /P) 1 2
  • B
    K p /P​
  • C
    2K p /P​
  • Correct answer
    D
    (2K p /P) 1 3

Solution

                                                          2 AB⁡ 2 g ⇌ 2AB⁡ g + B⁡ 2 g⁡
At equilibrium   2(1−x)           2x           x
Total moles = (2 + x)
Partial pressures
P AB 2 = 2(1−x) 2 + x p,  P AB  =  2x 2 + x p,  P B 2 = x 2+x p
K p = p AB 2 pB 2 ( pAB 2 ) 2 = [ ( 2 x 2 + x ) p ] 2 [ x p 2 + x ] [ 2 ( 1 - x ) 2 + x p ] 2
K p = 4 x 3 P 4 2+x (1−x) 2
2+x →2 ( x is small)
1−x →1
K P = 4 x 3 p 8 = x 3 p 2
x= 2kp p 3

Hence, the correct answer is option: 'D'.

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