# The equilibrium constants KP1 and KP2 for the reactions X↔2Y and Z↔P +Q, respectively are in the ratio of 1 : 9. If the degree of dissociation of X and Z be equal then the ratio of total pressure at these equilibria is

Given: X   $\to$  2Y ------(1)
Z  $\to$  P  +  Q---(2)
reactions (1) and (2) having same degree of dissociation
​             $\frac{{K}_{p1}}{{K}_{p2}}$$\frac{1}{9}$
To find: Ratio of total pressures
Solve: Suppose $\alpha$  is the degrre of dissociation
X   $\to$  2Y                        Z  $\to$  P  +  Q
​Initial concentration =    1           0                          1        0        0
Final concentration =  (1-$\alpha$)        $\alpha$                        (1-$\alpha$)    $\alpha$       $\alpha$

Total number of moles at equilibrium: For reaction 1st= $1-\alpha +2\alpha =1+\alpha$
For reaction 2nd =$1-\alpha +\alpha +\alpha =1+\alpha$
Suppose P1 and P2 is the total pressure for reaction 1 and 2 respectively.

${K}_{P1}=\frac{\left({P}_{Y}{\right)}^{2}}{{P}_{X}}=\frac{\left(\frac{2\alpha }{1+\alpha }{\right)}^{2}×{P}_{1}^{2}}{\frac{\alpha }{1+\alpha }}-----\left(3\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{K}_{P2}=\frac{\left({P}_{P}×{P}_{Q}\right)}{{P}_{Z}}=\frac{\frac{\alpha }{1+\alpha }{P}_{2}×\frac{\alpha }{1+\alpha }{P}_{2}}{\frac{1-\alpha }{1+\alpha }×{P}_{2}}----\left(4\right)$
Divide (3) by (4), we get

$\frac{{K}_{p1}}{{K}_{p2}}$=$\frac{4{P}_{1}^{2}}{{P}_{2}^{2}}$=$\frac{1}{9}$(given)

$\frac{{P}_{1}^{2}}{{P}_{2}^{2}}=\frac{1}{36}\phantom{\rule{0ex}{0ex}}\frac{{P}_{1}}{{P}_{2}}=\frac{1}{6}$
Therefore, the ratio of total pressures, P1:P2 = 1:6

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