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     2Y ------(1)
             Z    P  +  Q---(2)
             reactions (1) and (2) having same degree of dissociation  
​             Kp1Kp219
To find: Ratio of total pressures
Solve: Suppose α  is the degrre of dissociation
                                      X     2Y                        Z    P  +  Q
​Initial concentration =    1           0                          1        0        0
Final concentration =  (1-α)        α                        (1-α)    α       α  

Total number of moles at equilibrium: For reaction 1st= 1-α+2α=1+α
                                                             For reaction 2nd =1-α+α+α=1+α
Suppose P1 and P2 is the total pressure for reaction 1 and 2 respectively.
PX= 1-α1+α×P1      PY=2α1+α×P1PZ=1-α1+α×P2      PP=α1+α×P2      PQ=α1+α×P2

Divide (3) by (4), we get


Therefore, the ratio of total pressures, P1:P2 = 1:6



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