Question

Equimolar mixture of two gases $A_2$ and $B_2$ is taken in a rigid vessel at temperature $300K$. The gases react according to the given equations:
$A_2\left(g\right)\rightleftharpoons 2A\left(g\right)K_{P_1}=?$
$B_2\left(g\right)\rightleftharpoons 2B\left(g\right)K_{P_2}=?$
$A_2\left(g\right)+B_2\left(g\right)\rightleftharpoons 2AB\left(g\right)K_{P_3}=2$
If the initial pressure in the container was 2 atm and final pressure developed at equilibrium is 2.75 atm in which equilibrium partial pressure of gas AB was 0.5 atm, calculate the ratio of $\frac{K_{P_2}}{K_{P_1}}$
(Given: degree of dissociation of $B_2$ is greater than $A_2$)

Answer 1

Since initially we have taken equimolar mixture of $A_2$ and $B_2$, and initial pressure is 2 atm, so let initial pressure of each $A_2$ and $B_2$ is 1 atm.
So, $A_2\left(g\right)\rightleftharpoons 2A\left(g\right)$
$t=010$
$att1-x-z2x$
again, $B_2\left(g\right)\rightleftharpoons 2B\left(g\right)$
$t=010$
$att1-y-z2y$
So for $A_2\left(g\right)+B_2\left(g\right)\rightleftharpoons 2AB\left(g\right)$
$Atequl1-x-z1-y-z2z$
Given final pressure developed at equilibrium is 2.75
So, $1-x-z+1-y-z+2x+2y+2z=2.75$
$\Rightarrow x+y=0.75$........(1)
Again partial pressure of gas $AB$ is 0.5 atm
$\therefore 2z=0.5\Rightarrow z=0.25$......(2)
Again according to the equation,
$\frac{P_{AB}^2}{P_{A_2}{P}_{B_2}}=2$
$\Rightarrow \frac{(\frac{1}{2}{)}^2}{(\frac{3}{4}-x)(\frac{3}{4}-y)}=2$
$\Rightarrow (3-4x)(3-4y)=2$
$\Rightarrow (3-4(\frac{3}{4}-y\left)\right)(3-4y)=2sincex+y=0.75$
$\Rightarrow 12y-16y^2=2$
$\Rightarrow y=\frac{1}{2}or\frac{1}{4}$
But degree of dissociation of $B_2$ is greater than $A_2$, so y > x
$\therefore y=\frac{1}{2}\Rightarrow x=\frac{1}{4}$
So, $K_{P_1}=\frac{P_A^2}{P_{A_2}}=\frac{(2x{)}^2}{1-x-z}=\frac{(0.5{)}^2}{0.5}=0.5$
$K_{P_2}=\frac{P_B^2}{P_{B_2}}=\frac{(2y{)}^2}{1-y-z}=\frac{(1{)}^2}{0.25}=4$
$\therefore \frac{K_{P_2}}{K_{P_1}}=\frac{4}{0.5}=8$