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JEE Main 201912 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
12 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Chemical Equilibrium

JEE Main 2019 · 12 January · Morning Shift · Chemistry · Chemical Equilibrium

In a chemical reaction, A + 2B K ⇌ 2C + D , the initial...

In a chemical reaction, A+2B⇌K2C+DA + 2B\overset{K}{⇌}2C + D , the initial concentration of BB was 1.51.5 times of the concentration of AA , but the equilibrium concentrations of AandBA and B were found to be equal. The equilibrium constant (K)\left(K\right) for the chemical reaction is:

Options

  1. A11
  2. B44
  3. C14\dfrac{1}{4}
  4. D1616

Correct answer

B. 44

Step-by-step solution

A+2B⇌2C+DA + 2B\mathop{⇌}_{}2\text{C} + \text{D}
Initial conc. of [A] = a , [B] = 1.5a , [C] = 0 and [D] = 0\text{[A] = a , [B] = 1.5a , [C] = 0 and [D] = 0}

At equilibrium [A]=(a−x),[B]=1.5(a−2x),[C]=2xand[D]=x[A] = (a - x) , [B] = 1.5(a - 2x) , [C] = 2x and [D] = x

At equilibrium a−x=1.5a−2xa - x = 1.5a - 2x

0.5a=x0.5a = x

a=2xa = 2x

kC=(2x)2(x)(a−x)(1.5a−2x)2=4x2.x(x)(x)2=4{k}_{C} = \dfrac{{\left(2x\right)}^{2}\left(x\right)}{\left(a - x\right){\left(1.5a - 2x\right)}^{2}} = \dfrac{4x^{2} . x}{\left(x\right){\left(x\right)}^{2}} = 4
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