Phosphine decomposes at 300 K and 3 atm as | vedclass

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(C) The dissociation reaction is: $4PH_{3(g)} \rightleftharpoons P_{4(g)} + 6H_{2(g)}$ Let the initial moles of $PH_3$ be $1$. At equilibrium,moles ar

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$13.8$

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(C) The dissociation reaction is: $4PH_{3(g)} ightleftharpoons P_{4(g)} + 6H_{2(g)}$ Let the initial moles of $PH_3$ be $1$. At equilibrium,moles are: $PH_3 = (1 - \alpha)$,$P_4 = \alpha/4$,$H_2 = 6\alpha/4 = 1.5\alpha$. Total moles at equilibrium $= (1 - \alpha) + 0.25\alpha + 1.5\alpha = 1 + 0.75\alpha$. Given $\alpha = 0.3$,total moles $= 1 + 0.75(0.3) = 1.225$. The relation between vapour density $(D)$ and observed vapour density $(d)$ is $\frac{D}{d} = \frac{	ext{Total moles at equilibrium}}{	ext{Initial moles}} = 1 + n\alpha$,where $n$ is the change in moles per mole of reactant. Here,$n = \frac{1+6-4}{4} = 0.75$. $D = \frac{M_{PH_3}}{2} = \frac{34}{2} = 17$. $d = \frac{D}{1 + 0.75 	imes 0.3} = \frac{17}{1.225} \approx 13.87$.

Phosphine decomposes at $300 \ K$ and $3 \ atm$ as
$4PH_{3(g)} \rightleftharpoons P_{4(g)} + 6H_{2(g)}$
Calculate the vapour density of phosphine if it dissociates to the extent of $30 \%$.

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Phosphine decomposes at $300 \ K$ and $3 \ atm$ as $4PH_{3(g)} \rightleftharpoons P_{4(g)} + 6H_{2(g)}$ Calculate the vapour density of phosphine if it dissociates to the extent of $30 \%$.