(A) The reaction is $PCl_5 \rightleftharpoons PCl_3 + Cl_2$. Initial moles: $a, 0, 0$. At equilibrium: $a(1-\alpha), a\alpha, a\alpha$. Total moles at

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(A) The reaction is $PCl_5 \rightleftharpoons PCl_3 + Cl_2$. Initial moles: $a, 0, 0$. At equilibrium: $a(1-\alpha), a\alpha, a\alpha$. Total moles at equilibrium = $a(1-\alpha) + a\alpha + a\alpha = a(1+\alpha)$. Given $\alpha = 0.25$. Total moles = $a(1+0.25) = 1.25a$. Mole fraction of $Cl_2$ $(X_{Cl_2})$ = $\frac{a\alpha}{a(1+\alpha)} = \frac{0.25}{1.25} = \frac{1}{5} = 0.2$. Partial pressure of $Cl_2$ = $X_{Cl_2} \times P_{total} = 0.2 \times 2.0 \ atm = 0.4 \ atm$.

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

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For the thermal dissociation of $PCl_5$ as $PCl_5 \rightleftharpoons PCl_3 + Cl_2$,if '$a$' moles of $PCl_5$ are taken and at equilibrium,the degree of dissociation of $PCl_5$ is $0.25$ and the total pressure is $2.0 \ atm$,then the partial pressure of $Cl_2$ at equilibrium will be: (in $atm$)

A
$0.4$
B
$1.0$
C
$0.5$
D
$0.25$

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1 Equilibrium: Chemical Equilibrium

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Degree of dissociation and Vapour density

At $STP$,the density of air is $0.001293 \ g \ mL^{-1}$. Its vapour density is $---$.

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For the reaction $A \rightleftharpoons \frac{1}{2} B + C$,the degree of dissociation $\alpha$ in terms of vapour density $D_t$ (theoretical) and $D_o$ (observed) is given by:

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$4 \ mol$ of $PCl_5$ is heated in a closed vessel at a constant temperature. If the degree of dissociation for $PCl_5$ is $0.5$,what will be the total number of moles at equilibrium?

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The vapour density of a gas is $11.2$. The volume occupied by $11.2 \ g$ of the gas at $STP$ will be.....$L$

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For the equations $(i) \ HCN + CCl_4 \rightarrow \alpha_1$ and $(ii) \ HCN + C_6H_6 \rightarrow \alpha_2$,which degree of dissociation is greater,$\alpha_1$ or $\alpha_2$?

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For the thermal dissociation of $PCl_5$ as $PCl_5 \rightleftharpoons PCl_3 + Cl_2$,if '$a$' moles of $PCl_5$ are taken and at equilibrium,the degree of dissociation of $PCl_5$ is $0.25$ and the total pressure is $2.0 \ atm$,then the partial pressure of $Cl_2$ at equilibrium will be: (in $atm$)

Similar Questions

At $STP$,the density of air is $0.001293 \ g \ mL^{-1}$. Its vapour density is $---$.

For the reaction $A \rightleftharpoons \frac{1}{2} B + C$,the degree of dissociation $\alpha$ in terms of vapour density $D_t$ (theoretical) and $D_o$ (observed) is given by:

$4 \ mol$ of $PCl_5$ is heated in a closed vessel at a constant temperature. If the degree of dissociation for $PCl_5$ is $0.5$,what will be the total number of moles at equilibrium?

The vapour density of a gas is $11.2$. The volume occupied by $11.2 \ g$ of the gas at $STP$ will be.....$L$

For the equations $(i) \ HCN + CCl_4 \rightarrow \alpha_1$ and $(ii) \ HCN + C_6H_6 \rightarrow \alpha_2$,which degree of dissociation is greater,$\alpha_1$ or $\alpha_2$?

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