Question

At a temp, $T$ ,a compound $A{{B}_{4\left( g \right)}}$ dissociates as $2A{{B}_{4}}_{\left( g \right)}$ ⇌ $A{{2}_{\left( g \right)}}\text{ }+4{{B}_{2\left( g \right)}}$ with a degree of dissociation $x$ which is small compared with unity. The expression of $Kp$ in terms of $x$ and total pressure $P$ is:
A. $8{{P}^{3}}{{x}^{5}}$
B. $256{{P}^{3}}{{x}^{5}}$
C. $4Px{}^\text{2}$
D. None of these

Answer 1

We know the temperature as well as pressure of the surroundings provided in the given statement. Then we have to write down the total dissociation reaction along with stoichiometric coefficient. After that we have determined equilibrium constant for dissociation reaction. Degree of dissociation is the fraction of solute which has been dissociated. Thus density of mixture is defined as mass of mixture with per unit volume of mixture.

Complete step-by-step answer: The degree of dissociation can be defined as the amount of solute that is dissociated into ion/radical per mole. Thus now, considering given conditions:
$2A{{B}_{4}}$ ⇌ ${{A}_{2}}~~+~~4{{B}_{2}}$
​ $c$ $0~~~~~~~~~~0$
At equilibrium:
$c-cx~~~~~~~\dfrac{cx}{2}~~~~~2cx$
Thus the total number of moles at equilibrium $=\text{ }c\left( 1\text{ }+\text{ }\dfrac{3x}{2} \right)\text{ }\approx \text{ }c$
Since our $x\text{ }<<\text{ }1$
Hence, now by considering partial pressure of each species we get;
${{p}_{AB4}}~=~\left( \dfrac{P\cdot c\left( 1-x \right)}{c} \right)~=~P\left( 1-x \right)~=~P$
Similarly for ${{p}_{A2}}$ and ${{p}_{B2}}$ using partial pressure of each species we get;
${{p}_{A2}}~=~~\left[ \dfrac{\text{ }P\text{ }\left( \dfrac{cx}{2} \right)}{c}\text{ }=\text{ }\dfrac{{{P}_{x}}}{2} \right]$
and
${{P}_{B2}}~=~\left[ \text{ }\dfrac{P\cdot 2cx}{c}\text{ }=\text{ }2{{P}_{x}}\text{ } \right]$
Using here relation ${{K}_{p}}\text{ }=\text{ }\left[ \text{ }\dfrac{\left( {{A}_{2}} \right)\left( {{B}_{2}} \right)}{\left( A{{B}_{4}} \right){}^\text{2}\text{ }} \right]$ we get;
${{K}_{p}}\text{ }=\text{ }\left[ \text{ }\dfrac{\left( {{p}_{A2}} \right)\text{ }\left( {{p}_{B2}} \right)}{\left( {{p}_{AB4}} \right){}^\text{2}} \right]$
and thus by substituting the values we get;
${{K}_{p}}\text{ }=\text{ }\left[ \text{ }\dfrac{\left( {{p}_{A2}} \right)\text{ }\left( {{p}_{B2}} \right)\text{ }}{\left( {{p}_{AB4}} \right){}^\text{2}} \right]$
${{K}_{p}}\text{ }=\text{ }8{{P}^{3}}{{x}^{5}}$

Therefore, Option A is correct answer that is expression of ${{K}_{p}}$ in term of x as well as total pressure $P$ is $8{{P}^{3}}{{x}^{5}}$.

Note: Note that it is important to know concentration of reactants as well as product which is raised to power of its stoichiometric coefficient before substituting it into equation of dissociation constant. Therefore $R$ is universal gas constant it have various value in a different unit as well as we can use these value according to our convenience but we required here the value of $R$ which is $0.082\text{ }\dfrac{L-atm}{K-mol}$.