(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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The vapor density of a gas is $11.2$. At $N.T.P.$,$11.2 \ g$ of this gas occupies a volume of ............. $litres$.

For the reaction $A_{3(g)} \rightleftharpoons 3A_{(g)}$,the initial moles of $A_3$ is $a$. If $\alpha$ is the degree of dissociation of $A_3$,then the total number of moles at equilibrium will be:

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