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

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.

{} \\\ \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{aligned} & \text{Total = 0}\text{.6 + 0}\text{.4 + 0}\text{.4} \\\ & \text{ = 1}\text{.4} \\\ & \text{ n = 1}\text{.4} \\\ \end{aligned}$$ We know that the ideal gas equation is $$\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{aligned} & \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{aligned}$$ Now, we will calculate the mass of $$\text{PC}{{\text{l}}_{5}}$$. Mass always remains constant. We have, $$\text{P = 31}\text{.0 Cl = 35}\text{.5}$$ $$\begin{aligned} & \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{aligned}$$ Now, we know the formula for density. $$\begin{aligned} & \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{aligned}$$ **Hence, option A is correct.** **Additional information:** Reactants and products coexist in equilibrium so that reactant conversion to product is always less than 100%. Equilibrium reactions may involve the decomposition of a covalent reactant or ionization of ionic compounds into their ions in polar solvents. **Note:** Since, we were given that $$\alpha \text{ = 0}\text{.4}$$, to find the total number of moles, we subtracted it from 1. Also, remember that mass remains constant.