(A) The initial moles are $2$ moles of $AB_2$. Let the degree of dissociation be $x$.For the reaction: $2AB_{2(g)} \rightleftharpoons 2AB_{(g)} + B_{2

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(A) The initial moles are $2$ moles of $AB_2$. Let the degree of dissociation be $x$.For the reaction: $2AB_{2(g)} \rightleftharpoons 2AB_{(g)} + B_{2(g)}$Initial moles: $2, 0, 0$Moles at equilibrium: $2(1-x), 2x, x$Total moles at equilibrium $= 2-2x+2x+x = 2+x$.Partial pressures at equilibrium:$p_{AB_2} = \frac{2(1-x)}{2+x} P \approx P$ (since $x \ll 1$)$p_{AB} = \frac{2x}{2+x} P \approx xP$$p_{B_2} = \frac{x}{2+x} P \approx \frac{x}{2} P$$K_p = \frac{(p_{AB})^2 (p_{B_2})}{(p_{AB_2})^2} = \frac{(xP)^2 (xP/2)}{P^2} = \frac{Px^3}{2}$

Class 11 Chemistry 1 Equilibrium: Chemical Equilibrium Degree of dissociation and Vapour density

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At temperature $T$,the dissociation equilibrium of $AB_{2(g)}$ is given as: $2AB_{2(g)} \rightleftharpoons 2AB_{(g)} + B_{2(g)}$. If the degree of dissociation $x$ is very small compared to $1$,then the expression for the equilibrium constant $K_p$ in terms of $x$ and total pressure $P$ is:

A
$Px^3 / 2$
B
$Px^2 / 3$
C
$Px^3 / 3$
D
$Px^2 / 2$

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Degree of dissociation and Vapour density

Phosphorus pentachloride dissociates as follows,in a closed reaction vessel:
$PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$
If total pressure at equilibrium of the reaction mixture is $P$ and degree of dissociation of $PCl_5$ is $x,$ the partial pressure of $PCl_3$ will be:

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The vapour density of $N_2O_4$ in the equilibrium $N_2O_4 \rightleftharpoons 2NO_2$ is $40$. The degree of dissociation is:

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$3.2 \text{ moles of hydrogen iodide are heated in a closed vessel at } 444^{\circ}C \text{ until equilibrium is reached. If the degree of dissociation at this temperature is } 22\%, \text{ then the number of moles of hydrogen iodide remaining at equilibrium will be:}$

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At temperature,$T$,a compound $AB_{2(g)}$ dissociates according to the reaction; $2AB_{2(g)} \rightleftharpoons 2AB_{(g)} + B_{2(g)}$ with a degree of dissociation $x$,which is small compared with unity. The expression for $K_p$,in terms of $x$ and the total pressure,$P$ is

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If $D_T$ and $D_0$ are the theoretical and observed vapour densities at a definite temperature and $\alpha$ is the degree of dissociation of a substance,then $\alpha$ in terms of $D_0, D_T$ and $n$ (number of moles of product formed from $1 \, \text{mole}$ of reactant) is calculated by the formula:

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At temperature $T$,the dissociation equilibrium of $AB_{2(g)}$ is given as: $2AB_{2(g)} \rightleftharpoons 2AB_{(g)} + B_{2(g)}$. If the degree of dissociation $x$ is very small compared to $1$,then the expression for the equilibrium constant $K_p$ in terms of $x$ and total pressure $P$ is:

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Phosphorus pentachloride dissociates as follows,in a closed reaction vessel:$PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$If total pressure at equilibrium of the reaction mixture is $P$ and degree of dissociation of $PCl_5$ is $x,$ the partial pressure of $PCl_3$ will be:

The vapour density of $N_2O_4$ in the equilibrium $N_2O_4 \rightleftharpoons 2NO_2$ is $40$. The degree of dissociation is:

$3.2 \text{ moles of hydrogen iodide are heated in a closed vessel at } 444^{\circ}C \text{ until equilibrium is reached. If the degree of dissociation at this temperature is } 22\%, \text{ then the number of moles of hydrogen iodide remaining at equilibrium will be:}$

At temperature,$T$,a compound $AB_{2(g)}$ dissociates according to the reaction; $2AB_{2(g)} \rightleftharpoons 2AB_{(g)} + B_{2(g)}$ with a degree of dissociation $x$,which is small compared with unity. The expression for $K_p$,in terms of $x$ and the total pressure,$P$ is

If $D_T$ and $D_0$ are the theoretical and observed vapour densities at a definite temperature and $\alpha$ is the degree of dissociation of a substance,then $\alpha$ in terms of $D_0, D_T$ and $n$ (number of moles of product formed from $1 \, \text{mole}$ of reactant) is calculated by the formula:

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Phosphorus pentachloride dissociates as follows,in a closed reaction vessel:
$PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$
If total pressure at equilibrium of the reaction mixture is $P$ and degree of dissociation of $PCl_5$ is $x,$ the partial pressure of $PCl_3$ will be: