(C) Consider the dissociation reaction: $A \rightleftharpoons nB$ Initial moles: $1 \quad 0$ Moles at equilibrium: $1 - \alpha \quad n\alpha$ Total

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$\alpha = \frac{D_T - D_0}{(n - 1)D_0}$

(C) Consider the dissociation reaction: $A \rightleftharpoons nB$ Initial moles: $1 \quad 0$ Moles at equilibrium: $1 - \alpha \quad n\alpha$ Total moles at equilibrium: $1 - \alpha + n\alpha = 1 + \alpha(n - 1)$ Since vapour density is inversely proportional to the number of moles,we have: $\frac{D_T}{D_0} = \frac{\text{Total moles at equilibrium}}{\text{Initial moles}} = \frac{1 + \alpha(n - 1)}{1}$ Rearranging for $\alpha$: $\frac{D_T}{D_0} = 1 + \alpha(n - 1)$ $\frac{D_T}{D_0} - 1 = \alpha(n - 1)$ $\frac{D_T - D_0}{D_0} = \alpha(n - 1)$ $\alpha = \frac{D_T - D_0}{(n - 1)D_0}$

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

Medium

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:

A
$\alpha = \frac{D_0 - D_T}{(1 - n)D_T}$
B
$\alpha = \frac{D_T - D_0}{(n - 1)D_T}$
C
$\alpha = \frac{D_T - D_0}{(n - 1)D_0}$
D
$\alpha = \frac{D_0 - D_T}{(n - 1)D_T}$

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6-1.Equilibrium (Chemical Equilibrium)

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Degree of dissociation and Vapour density

Ammonia is heated in a closed vessel in the presence of a catalyst at $15 \ atm$ pressure and a temperature range of $27 \ ^\circ C$ to $347 \ ^\circ C$. Under these conditions,ammonia undergoes partial dissociation according to the reaction $2NH_3 \rightleftharpoons N_2 + 3H_2$. When the pressure is increased to $50 \ atm$ while keeping the volume of the vessel constant,calculate the percentage of $NH_3$ dissociated.

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Ammonium carbonate,when heated to $200\, ^oC$,dissociates into a mixture of $NH_3$ and $CO_2$ vapors with a vapor density of $13.0$. What is the degree of dissociation of ammonium carbonate?

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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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For the reaction,$N_2 + O_2 \rightleftharpoons 2NO$,the equilibrium constant is $K_c = 2$. If both reactants have the same initial moles,what is the degree of dissociation of $N_2$ and $O_2$?

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For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,the correct relation between degree of dissociation $(\alpha)$ of $N_2O_{4(g)}$ and equilibrium constant,$K_p$ is $(P=$ total pressure of mixture $)$

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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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Ammonium carbonate,when heated to $200\, ^oC$,dissociates into a mixture of $NH_3$ and $CO_2$ vapors with a vapor density of $13.0$. What is the degree of dissociation of ammonium carbonate?

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.......$\%$

For the reaction,$N_2 + O_2 \rightleftharpoons 2NO$,the equilibrium constant is $K_c = 2$. If both reactants have the same initial moles,what is the degree of dissociation of $N_2$ and $O_2$?

For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,the correct relation between degree of dissociation $(\alpha)$ of $N_2O_{4(g)}$ and equilibrium constant,$K_p$ is $(P=$ total pressure of mixture $)$

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