(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 1 Equilibrium: Chemical Equilibrium Degree of dissociation and Vapour density

Medium

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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1 Equilibrium: Chemical Equilibrium

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Degree of dissociation and Vapour density

At $NTP$,$5.6 \ L$ of a gas weighs $8 \ g$. The vapour density of the gas is:

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If in the reaction $N_2O_4 \rightleftharpoons 2NO_2$,$\alpha$ is the fraction of $N_2O_4$ that dissociates,then the total number of moles at equilibrium is:

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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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The vapor density of a gas is $16$. If the ratio of specific heat at constant pressure to specific heat at constant volume is $1.4$,find its atomic weight.

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For the following equilibrium $N_2O_4 \rightleftharpoons 2NO_2$ in gaseous phase,$NO_2$ is $50\%$ of the total volume when equilibrium is set up. Hence,the percentage of 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 approximately related to the pressure at equilibrium by (given $\alpha << 1$):

Similar Questions

At $NTP$,$5.6 \ L$ of a gas weighs $8 \ g$. The vapour density of the gas is:

If in the reaction $N_2O_4 \rightleftharpoons 2NO_2$,$\alpha$ is the fraction of $N_2O_4$ that dissociates,then the total number of moles at equilibrium is:

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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