(C) Let the initial pressure of $N_2O_5$ be $P_0 = 300 \ mm \ Hg$. Let $\alpha$ be the fraction of $N_2O_5$ decomposed. The reaction is: $N_2O_5 \righ

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(C) Let the initial pressure of $N_2O_5$ be $P_0 = 300 \ mm \ Hg$. Let $\alpha$ be the fraction of $N_2O_5$ decomposed. The reaction is: $N_2O_5 \rightleftharpoons 2NO_2 + \frac{1}{2}O_2$. At $t=0$: $P_0, 0, 0$. At equilibrium: $P_0(1-\alpha), 2P_0\alpha, \frac{1}{2}P_0\alpha$. Total pressure at equilibrium $P_t = P_0(1-\alpha) + 2P_0\alpha + 0.5P_0\alpha = P_0(1 + 1.5\alpha)$. Given $P_t = 480 \ mm \ Hg$ and $P_0 = 300 \ mm \ Hg$. $480 = 300(1 + 1.5\alpha)$. $1.6 = 1 + 1.5\alpha$. $0.6 = 1.5\alpha$. $\alpha = \frac{0.6}{1.5} = 0.4$.

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

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At $T$ $K$,consider the following gaseous reaction,which is in equilibrium: $N_2O_5 \rightleftharpoons 2NO_2 + \frac{1}{2}O_2$. What is the fraction of $N_2O_5$ decomposed at constant volume and temperature,if the initial pressure is $300 \ mm \ Hg$ and pressure at equilibrium is $480 \ mm \ Hg$? (Assume all gases as ideal)

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A
$0.2$
B
$0.6$
C
$0.4$
D
$0.8$

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6-1.Equilibrium (Chemical Equilibrium)

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Degree of dissociation and Vapour density

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At $200 \, ^\circ C$,the observed vapour density $(V.D.)$ of $PCl_5$ is $60$. The degree of dissociation $(\alpha)$ of $PCl_5$ is ..........$\%$ (Given: Molar mass of $PCl_5 = 208.5 \, g/mol$).

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The vapour density of completely dissociated $NH_4Cl$ would be

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The vapour density of a mixture containing $NO_2$ and $N_2O_4$ is $27.6$. The mole fraction of $N_2O_4$ in the mixture 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 $\alpha << 1$):

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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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At $200 \, ^\circ C$,the observed vapour density $(V.D.)$ of $PCl_5$ is $60$. The degree of dissociation $(\alpha)$ of $PCl_5$ is ..........$\%$ (Given: Molar mass of $PCl_5 = 208.5 \, g/mol$).

The vapour density of completely dissociated $NH_4Cl$ would be

The vapour density of a mixture containing $NO_2$ and $N_2O_4$ is $27.6$. The mole fraction of $N_2O_4$ in the mixture is

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 $\alpha << 1$):

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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