Q6 / 87 · NTA Abhyas (JEE Main) 2020 · Chemistry · Chemical Equilibrium

At a definite temperature, the equilibrium constant for a reaction,   A + B ⇌ 2 C  , was found to be 81. Starting with 1 mole A and 1 mole B, the mole fraction of C at equilibrium is :

  • Correct answer
    A
    9 1 1
  • B
    1 1 1
  • C
    2 1 1
  • D
    7 1 1

Solution

A + B ⇌ 2 C 1 1 0 1 - x 1 - x 2 x

∴      8 1 = 4 x 2 1 - x 2

or      2 x 1 - x = 9

∴      x = 9 1 1

Thus,    A + B ⇌ 2 C 2 1 1 2 1 1 18 1 1

( Total mole = 2 1 1 + 2 1 1 + 18 1 1 = 22 1 1 )

∴   Mole fraction of C= 18 1 1 2 2 1 1 = 9 1 1

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

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