Question

The degree of dissociation is 0.4 at 400K and 1atm for gaseous reaction. \[\text{PC}{{\text{l}}_{5}}\text{ }\leftrightarrows\text{ PC}{{\text{l}}_{3}}\text{ + C}{{\text{l}}_{2}}\]. Assuming the ideal behaviour of all gases. Calculate the density of the equilibrium mixture at 400K and 1atm pressure.
(A) \[4.53\text{ g/L}\]
(B) \[9.26\text{ g/L}\]
(C) \[1.25\text{ g/L}\]
(D) \[7.28\text{ g/L}\]

Answer 1

Hint: Degree of dissociation is defined as fraction of the mole of reactant that got dissociated. It is represented by \[\alpha \].
\[\alpha \text{ = }\dfrac{\text{amount of reactant dissociated}}{\text{amount of reactant present initially}}\]

Complete step by step answer: In order to find density, we will first find the number of moles in the reaction. We know that \[\alpha = 0.4\].
We will calculate both, the number of moles present initially and the number of moles present at equilibrium.
\[\begin{matrix}
{} \\
\text{Number of moles initially -} \\
\text{Number of moles at equilirbium - } \\
\end{matrix}\begin{matrix}
PC{{l}_{5}} \\
1 \\
(1-0.4) \\
\end{matrix}\begin{matrix}\leftrightarrows
PC{{l}_{3}} \\
1 \\
0.4 \\
\end{matrix}+\begin{matrix}
C{{l}_{2}} \\
1 \\
0.4 \\
\end{matrix}\]
\[\begin{align}
& \text{Total = 0}\text{.6 + 0}\text{.4 + 0}\text{.4} \\
& \text{ = 1}\text{.4} \\
& \text{ n = 1}\text{.4} \\
\end{align}\]
We know that the ideal gas equation is \[\text{PV = nRT}\]
But we have to find density, for which we need volume. Therefore, \[\text{V = }\dfrac{\text{nRT}}{\text{P}}\]
We have been given that \[\text{P = 1 atm}\] and
So, we take \[\text{R = 0}\text{.0821 L atm }{{\text{K}}^{-1}}\text{ mo}{{\text{l}}^{-1}}\].
Substituting these values in the given equation we get,
\[\begin{align}
& \text{V = }\dfrac{\text{nRT}}{\text{P}} \\
&\implies \text{ }\dfrac{1.4\text{ }\times 0.0821 \times 400\text{ }}{1\text{ atm}} \\
& \text{ V = 45}\text{.976 L} \\
\end{align}\]
Now, we will calculate the mass of \[\text{PC}{{\text{l}}_{5}}\]. Mass always remains constant.
We have, \[\text{P = 31}\text{.0 Cl = 35}\text{.5}\]
\[\begin{align}
& \text{total = 31}\text{.0 + 5}\times \text{35}\text{.5} \\
&\implies \text{ 31}\text{.0 + 177}\text{.5} \\
&\implies \text{ 208}\text{.5 g} \\
\end{align}\]
Now, we know the formula for density.
\[\begin{align}
& \text{Density = }\dfrac{\text{mass}}{\text{volume}} \\
&\implies \text{ }\dfrac{208.9}{45.976} \\
&\implies \text{ 4}\text{.54 g }{{\text{L}}^{-1}}\text{ }\approx \text{ 4}\text{.53 g }{{\text{L}}^{-1}} \\
\end{align}\]

Hence, option A is correct.

Additional information: Reactants and products coexist in equilibrium so that reactant conversion to product is always less than 100%. Equilibrium reactions may involve the decomposition of a covalent reactant or ionization of ionic compounds into their ions in polar solvents.

Note: Since, we were given that \[\alpha \text{ = 0}\text{.4}\], to find the total number of moles, we subtracted it from 1. Also, remember that mass remains constant.