Equilibrium Question 716
Question: Gaseous $ N_2O_4 $ dissociates into gaseous $ NO_2 $ according to the reaction $ [N_2O_4(g)\rightarrow 2NO_2(g)] $ At 300 K and 1 atm pressure, the degree of dissociation of $ N_2O_4 $ is 0.2. If one mole of $ N_2O_4 $ gas is contained in a vessel, then the density of the equilibrium mixture is:
Options:
A) 1.56 g/L
B) 6.22 g/L
C) 3.11g/L
D) 4.56 g/L
Show Answer
Answer:
Correct Answer: C
Solution:
$\begin{aligned} & \begin{matrix} {} & {} & N_2O_4(g)\rightarrow 2NO_2(g) \\ \end{matrix} \\ & \begin{matrix} t=0 & {} & 1 \\ \end{matrix}\begin{matrix} {} & {} & {} & 0 \\ \end{matrix} \\ & \begin{matrix} t=eq\text{.} & 1-\alpha & {} \\ \end{matrix}\begin{matrix} {} & 2\alpha \\ \end{matrix} \\ \end{aligned} $ Where $ \alpha $ = Degree of dissociation. Mol. wt. of mixture $ =\frac{(1-\alpha )\times {M_{N_2O_4}}+2\alpha \times {M_{NO_2}}}{(1+\alpha )} $
$ =\frac{(1-0.2)92+2\times 0.2\times 46}{(1+0.2)}=76.66 $ Now, As per ideal gas equation $ PV=nRT $
$ PM_{mixture}=dRT $
$ \therefore d=\frac{PM_{mix}}{RT}=\frac{1\times 76.66}{0.0821\times 300}=3.11g/L $
BETA
