NEETChemistryChemical EquilibriumMCQ+4 / −1
The dissociation equilibrium of a gass AB2 can be represented as :
2AB2(g) 2AB(g) + B2(g)
The degree of dissociation is x and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant Kp and total pressure P is
2AB2(g) 2AB(g) + B2(g)
The degree of dissociation is x and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant Kp and total pressure P is
- A(2Kp/P)1/2
- B(Kp/P)
- C(2Kp/P)
- D(2Kp/P)1/3
View written solutionFree
Correct answer: D
Amount of moles at equilibrium = 2(1 – x) + 2x + x = 2 + x
$${K_p} = {{{{\left[ {{p_{AB}}} \right]}^2}\left[ {{p_{{B_2}}}} \right]} \over {{{\left[ {{p_{A{B_2}}}} \right]}^2}}}$$
= $${{{{\left[ {{{2x} \over {2 + x}} \times P} \right]}^2}\left[ {{x \over {2 + x}} \times P} \right]} \over {{{\left[ {{{2\left( {1 - x} \right)} \over {2 + x}} \times P} \right]}^2}}}$$
= $${{\left[ {{{4{x^3}} \over {2 + x}} \times P} \right]} \over {4{{\left( {1 - x} \right)}^2}}}$$
$ \Rightarrow $ Kp = $${{{4{x^3} \times P} \over 2} \times {1 \over 4}}$$
($ \because $ 1 – x ≈ 1 and 2 + x ≈ 2)
$ \Rightarrow $ x = $${\left( {{{8{K_p}} \over {4P}}} \right)^{{1 \over 3}}}$$
$ \Rightarrow $ x = $${\left( {{{2{K_p}} \over P}} \right)^{{1 \over 3}}}$$
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