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NEET 2008
Exam
NEET
Year
2008
Subject
Chemistry
Chapter
Chemical Equilibrium

NEET 2008 · Chemistry · Chemical Equilibrium

The dissociation equilibrium of a gas AB_2 can be represented as, 2...

The dissociation equilibrium of a gas AB2\mathrm{AB}_2 can be represented as, 2AB2( g)⇌2AB(g)+B2( g)2 \mathrm{AB}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{g})+\mathrm{B}_2(\mathrm{~g}). The degree of dissociation is ' x\mathrm{x} ' and is small compared to 1. The expression relating the degree of dissociation (x)(\mathrm{x}) with equilibrium constant KPK_P and total pressure P\mathrm{P} is

Options

  1. A(KpP)\left(\frac{K_p}{P}\right)
  2. B(2KPP)\left(\frac{2 K_P}{P}\right)
  3. C(2KPP)1/3\left(\frac{2 K_P}{P}\right)^{1 / 3}
  4. D(2KPP)1/2\left(\frac{2 K_P}{P}\right)^{1 / 2}

Correct answer

C. (2KPP)1/3\left(\frac{2 K_P}{P}\right)^{1 / 3}

Step-by-step solution

2AB2⇌2AB+B21001−xxx/2\begin{aligned}& 2 \mathrm{AB}_{2} \rightleftharpoons2 \mathrm{AB}+\mathrm{B}_{2} \\ & 1 \\ & 00 \\ & 1-x \\ & \mathrm{x} \quad \mathrm{x}/ 2 \\ & \end{aligned}Total mole at equi. =1+x2=1+\frac{x}{2} \therefore K_P=(x1+x/2×P)2(x/21+x/2×P)(1−x1+x/2×P)2(\frac{\left(\frac{x}{1+x / 2} \times P\right)^2\left(\frac{x / 2}{1+x / 2} \times P\right)}{\left(\frac{1-x}{1+x / 2} \times P\right)^2} (Here xx is degree of dissociation) or KP=x3P2orx3=2KPPorx=(2KPP)1/3K_P=\frac{x^3 P}{2} or x^3=\frac{2 K_P}{P} or x=\left(\frac{2 K_P}{P}\right)^{1 / 3}
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