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Equilibrium in chemical processes - class-XI
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Gaseous $ N _{2}O _{4} $ dissociates into gaseous $ NO _{2} $ according to the reaction $ N _{2}O _{4} (g) \rightleftharpoons 2NO _{2}(g)$ at 300 K and 1 atm pressure, the degree of dissociation of $ N _{2}O _{4} $ is 0.2. If one mole of $ N _{2}O _{4} $ gas is contained in a vessel, then the density of the equilibrium mixture is :
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A
3.11 g/L
💡 Explanation:
$N _2O _4\longrightarrow 2NO _2$
at $t=0$, moles of $N _2O _4=1$, moles of $NO _2=0$
at $t=equilibrium$, mole of $N _2O _4=1-a$, mole of $NO _2=2a$
$a$=degree of dissociation.
Molecular weight of mixture$=\cfrac {(1-a)\times\text{molar mass of }N _2O _4+2a\times \text{molar mass of }NO _2}{(1-a+2a)}$
$=\cfrac {(1-0.2)(28+64)+2\times 0.2\times (14+32)}{1+0.2}$
$M=76.66$
$P=1 atm,T=300K,$
$d=PM/RT$
$=\cfrac {1\times 76.66}{0.082 \times 300}=3.11 gm/lit\ $ .