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JIPMER 2014
Exam
JIPMER
Year
2014
Subject
Chemistry
Chapter
Chemical Equilibrium

JIPMER 2014 · Chemistry · Chemical Equilibrium

The dissociation equilibrium of a gas A B_2 can be represented as 2...

The dissociation equilibrium of a gas AB2A B_2 can be represented as 2AB2(g)⇌2AB(g)+B2(g)2 A B_2(g) \rightleftharpoons 2 A B(g)+B_2(g) The degree of dissociation is ' xx ' and is small compared to 1 . The expression relating the degree of dissociation ( xx ) with equilibrium constant KpK_p and total pressure pp is

Options

  1. A(2Kp/p)\left(2 K_p / p\right)
  2. B(2Kp/p)1/3\left(2 K_p / p\right)^{1 / 3}
  3. C(2Kp/p)1/2\left(2 K_p / p\right)^{1 / 2}
  4. D(Kp/p)\left(K_p / p\right)

Correct answer

B. (2Kp/p)1/3\left(2 K_p / p\right)^{1 / 3}

Step-by-step solution

(2AB2(g)⇌2AB(g)+B2(g)Initial moles 100At equilibrium 2(1−x)2xx\begin{array}{cccc} & 2 A B_2(g) & \rightleftharpoons & 2 A B(g) & + & B_2(g) \\ \text{Initial moles } & 1 & & 0 & & 0 \\ \text{At equilibrium } & 2(1-x) & & 2 x & & x \end{array} where, x=\quad x= degree of dissociation  Total moles at equilibrium =2−2x+2x+x&=(2+x)\begin{aligned}\text{ Total moles at equilibrium } & =2-2 x+2 x+x \& =(2+x)\end{aligned} So, ρAB2=2(1−x)p(2+x)ρAB=2xp(2+x)\begin{aligned} & \text{ So, } \quad \rho_{A B_2}= \frac{2(1-x) p}{(2+x)} \\ & \rho_{A B}=\frac{2 x p}{(2+x)}\end{aligned} pB2=xp(2+x)Kp=(pAB)2(pB2)(pAB2)2\begin{aligned} p_{B_2} & =\frac{x p}{(2+x)} \\ K_p & =\frac{\left(p_{A B}\right)^2\left(p_{B_2}\right)}{\left(p_{A B_2}\right)^2}\end{aligned} =\frac{\left(\frac{2 x p}{2+x}\right)^2 \left(\frac{x}{2+x} p\right)}{\left(\frac{2(1-x)}{(2+x)} p\right)^2}=x3p(2+x)(1−x)2\frac{x^3 p}{(2+x)(1-x)^2} [∵x≪1 and 2, so  (1−x)=1,(2+x)=2]=x3p2\begin{aligned} {[\because x \ll 1 \text{ and } 2 \text{,\text{ so } }} & (1-x)=1,(2+x)=2] \\ & =\frac{x^3 p}{2}\end{aligned} x=\left(\frac{2 K_p}{p}$\right)^{1 / 3}
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