(B) For the equilibrium $PCl_5 \rightleftharpoons PCl_3 + Cl_2$,the equilibrium constant $K_p$ is given by:$K_p = \frac{\alpha^2}{1 - \alpha^2} P$Give

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$\alpha \propto \frac{1}{\sqrt{P}}$

(B) For the equilibrium $PCl_5 \rightleftharpoons PCl_3 + Cl_2$,the equilibrium constant $K_p$ is given by:$K_p = \frac{\alpha^2}{1 - \alpha^2} P$Given $\alpha Thus,$K_p \approx \alpha^2 P$.Rearranging for $\alpha$,we get $\alpha^2 \approx \frac{K_p}{P}$,which implies $\alpha \approx \sqrt{\frac{K_p}{P}}$.Therefore,$\alpha \propto \frac{1}{\sqrt{P}}$.

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Degree of dissociation and Vapour density

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The degree of dissociation of $PCl_5$ $(\alpha)$ obeying the equilibrium,$PCl_5 \rightleftharpoons PCl_3 + Cl_2$,is approximately related to the pressure at equilibrium by (given $\alpha << 1$):

A
$\alpha \propto P$
B
$\alpha \propto \frac{1}{\sqrt{P}}$
C
$\alpha \propto \frac{1}{P^2}$
D
$\alpha \propto \frac{1}{P^4}$

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6-1.Equilibrium (Chemical Equilibrium)

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Degree of dissociation and Vapour density

For a $10^{-2} \ M \ HCN$ solution with $[H^+] = 10^{-3} \ M$,find the value of the degree of dissociation in percentage. (in $\%$)

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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:

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The vapour density of undecomposed $N_2O_4$ is $46$. When heated,the vapour density decreases to $24.5$ due to its dissociation into $NO_{2(g)}$. The percentage dissociation of $N_2O_4$ is:

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The degree of dissociation of $PCl_5$ $(\alpha)$ obeying the equilibrium $PCl_5 \rightleftharpoons PCl_3 + Cl_2$ is related to the pressure at equilibrium $(P)$ by:

Medium

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At temperature $T$,compound $AB_{2(g)}$ dissociates as $AB_{2(g)} \rightleftharpoons AB_{(g)} + \frac{1}{2} B_{2(g)}$ having degree of dissociation $x$ (small compared to unity). The correct expression for $x$ in terms of $K_p$ and $p$ is

MediumJEE MAIN 2025

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The degree of dissociation of $PCl_5$ $(\alpha)$ obeying the equilibrium,$PCl_5 \rightleftharpoons PCl_3 + Cl_2$,is approximately related to the pressure at equilibrium by (given $\alpha << 1$):

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For a $10^{-2} \ M \ HCN$ solution with $[H^+] = 10^{-3} \ M$,find the value of the degree of dissociation in percentage. (in $\%$)

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 undecomposed $N_2O_4$ is $46$. When heated,the vapour density decreases to $24.5$ due to its dissociation into $NO_{2(g)}$. The percentage dissociation of $N_2O_4$ is:

The degree of dissociation of $PCl_5$ $(\alpha)$ obeying the equilibrium $PCl_5 \rightleftharpoons PCl_3 + Cl_2$ is related to the pressure at equilibrium $(P)$ by:

At temperature $T$,compound $AB_{2(g)}$ dissociates as $AB_{2(g)} \rightleftharpoons AB_{(g)} + \frac{1}{2} B_{2(g)}$ having degree of dissociation $x$ (small compared to unity). The correct expression for $x$ in terms of $K_p$ and $p$ 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: