(D) For the reaction: $N_2O_4 \rightleftharpoons 2NO_2$Initial moles: $1$ mole of $N_2O_4$ and $0$ moles of $NO_2$.Moles at equilibrium: If $\alpha$ i

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(D) For the reaction: $N_2O_4 \rightleftharpoons 2NO_2$Initial moles: $1$ mole of $N_2O_4$ and $0$ moles of $NO_2$.Moles at equilibrium: If $\alpha$ is the degree of dissociation,then $(1 - \alpha)$ moles of $N_2O_4$ remain and $2\alpha$ moles of $NO_2$ are formed.Total moles at equilibrium $= (1 - \alpha) + 2\alpha = 1 + \alpha$.

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

Medium

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:

A
$3$
B
$1$
C
$(1 - \alpha)^2$
D
$(1 + \alpha)$

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1 Equilibrium: Chemical Equilibrium

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Degree of dissociation and Vapour density

At $T \, ^\circ C$,the vapour density of $PCl_5$ is $104.25$. The degree of dissociation of $PCl_5$ is ....$\%$

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Calculate the degree of dissociation $\alpha$ for $0.1 \, N \, CH_3COOH$ (dissociation constant $K_a = 1 \times 10^{-5}$).

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If $PCl_5$ undergoes $80\%$ dissociation at $250 \, ^\circ C$,what is its vapour density at this temperature?

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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:

Similar Questions

At $T \, ^\circ C$,the vapour density of $PCl_5$ is $104.25$. The degree of dissociation of $PCl_5$ is ....$\%$

For the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$; if $K_C = 2$,then the degree of dissociation of $O_2$ is:

Given is a concentrated solution of a weak electrolyte $A_{x}B_{y}$ of concentration '$c$' and dissociation constant '$K$'. The degree of dissociation is given by:

Calculate the degree of dissociation $\alpha$ for $0.1 \, N \, CH_3COOH$ (dissociation constant $K_a = 1 \times 10^{-5}$).

If $PCl_5$ undergoes $80\%$ dissociation at $250 \, ^\circ C$,what is its vapour density at this temperature?

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