Question

The degree of dissociation is 0.4 at 400K and 1atm for gaseous reaction. $$\text{PC}{\text{l}}_5\leftrightarrows \text{ PC}{\text{l}}_3\text{ + C}{\text{l}}_2$$. 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 $$\alpha$$.
$$\alpha \text{ = }{\displaystyle \frac{\text{amount of reactant dissociated}}{\text{amount of reactant present initially}}}$$

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.
$$\begin{array}{c}\\ \text{Number of moles initially -}\\ \text{Number of moles at equilirbium - }\end{array}\begin{array}{c}PC{l}_5\\ 1\\ (1-0.4)\end{array}\begin{array}{c}\leftrightarrows PC{l}_3\\ 1\\ 0.4\end{array}+\begin{array}{c}C{l}_2\\ 1\\ 0.4\end{array}$$
$$\begin{array}{rl}& \text{Total = 0}\text{.6 + 0}\text{.4 + 0}\text{.4}\\ & \text{ = 1}\text{.4}\\ & \text{ n = 1}\text{.4}\end{array}$$
We know that the ideal gas equation is $$\text{PV = nRT}$$
But we have to find density, for which we need volume. Therefore, $$\text{V = }{\displaystyle \frac{\text{nRT}}{\text{P}}}$$
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,
$$\begin{array}{rl}& \text{V = }{\displaystyle \frac{\text{nRT}}{\text{P}}}\\ & ⟹{\displaystyle \frac{1.4\times 0.0821\times 400}{1\text{ atm}}}\\ & \text{ V = 45}\text{.976 L}\end{array}$$
Now, we will calculate the mass of $$\text{PC}{\text{l}}_5$$. Mass always remains constant.
We have, $$\text{P = 31}\text{.0 Cl = 35}\text{.5}$$
$$\begin{array}{rl}& \text{total = 31}\text{.0 + 5}\times \text{35}\text{.5}\\ & ⟹\text{ 31}\text{.0 + 177}\text{.5}\\ & ⟹\text{ 208}\text{.5 g}\end{array}$$
Now, we know the formula for density.
$$\begin{array}{rl}& \text{Density = }{\displaystyle \frac{\text{mass}}{\text{volume}}}\\ & ⟹{\displaystyle \frac{208.9}{45.976}}\\ & ⟹\text{ 4}\text{.54 g }{\text{L}}^{-1}\approx \text{ 4}\text{.53 g }{\text{L}}^{-1}\end{array}$$

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.