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NEET 2008
Exam
NEET
Year
2008
Subject
Chemistry
Chapter
Biomolecules

NEET 2008 · Chemistry · Biomolecules

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 (x)(x) 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)200 (initially) 2(1−x)2xx (at equilibrium) \begin{array}{cccr} 2 A B_{2(g)} & \rightleftharpoons & 2 A B_{(g)} & + & B_{2(g)} & \\ 2 & & 0 & & 0 & \text{ (initially) } \\ 2(1-x) & & 2 x & & x & \text{ (at equilibrium) } \end{array}Amount of  moles at equilibrium =2(1−x)+2x+x=2+x\begin{aligned} \text{Amount\text{ of } moles at equilibrium } & =2(1-x)+2 x+x \\ & =2+x\end{aligned})Kp=[pAB]2[pB2][pAB2]2Kp=(2x2+x×P)2×(x2+x×P)(2(1−x)2+x×P)2=4x32+x×P4(1−x)2Kp=4x3×P2×14(∵1−x≈12+x≈2)x=(8Kp4P)1/3⇒x=(2KpP)1/3\begin{aligned} & K_p= \frac{\left[p_{A B}\right]^2\left[p_{B_2}\right]}{\left[p_{A B_2}\right]^2} \\ & K_p= \frac{\left(\frac{2 x}{2+x} \times P\right)^2 \times\left(\frac{x}{2+x} \times P\right)}{\left(\frac{2(1-x)}{2+x} \times P\right)^2}=\frac{\frac{4 x^3}{2+x} \times P}{4(1-x)^2} \\ & K_p= \frac{4 x^3 \times P}{2} \times \frac{1}{4} \quad(\because 1-x \approx 1 & 2+x \approx 2) \\ & x=\left(\frac{8 K_p}{4 P}\right)^{1 / 3} \quad \Rightarrow x=\left(\frac{2 K_p}{P}\right)^{1 / 3}\end{aligned})
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