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Question: The equilibrium constants \({K_{P1}}\) and \({K_{P2}}\) for the reaction \(X \rightleftharpoons 2Y\)...

The equilibrium constants KP1{K_{P1}} and KP2{K_{P2}} for the reaction X⇌2YX \rightleftharpoons 2Y and Z⇌P+QZ \rightleftharpoons P + Q, respectively are in the ratio of 1:91:9. The degree of dissociation of XX and ZZ be equal, then the ratio of total pressure at these equilibria is:
A.1:361:36
B.1:11:1
C.1:31:3
D.1:91:9

Explanation

Solution

To answer this question, you must recall the formula of equilibrium constant of a reaction in terms of the pressure of the reactants and compounds, that is KP{K_P}. It is given by the partial pressure at equilibrium of products divided by the partial pressure at equilibrium of reactants raised to the power of their respective stoichiometric coefficients.

Formula used:
KP1=PY2PX{K_{P_1}} = \dfrac{{{P_Y}^2}}{{{P_X}}} and KP2=PPPQPZ{K_{P_2}} = \dfrac{{{P_P}{P_Q}}}{{{P_Z}}}

Complete step by step solution:
For the reaction, X⇌2YX \rightleftharpoons 2Y, let at equilibrium, xxmoles of compound X are dissociated.
We can write the partial pressures of X and Y as
PX=(1−x1+x)P1{P_X} = \left( {\dfrac{{1 - x}}{{1 + x}}} \right){P_1}
PY=(2x1+x)P1{P_Y} = \left( {\dfrac{{2x}}{{1 + x}}} \right){P_1}
We know that, KP1=PY2PX{K_{P1}} = \dfrac{{{P_Y}^2}}{{{P_X}}}.
Substituting the values, we get,
KP1=4x2P1(1−x2){K_{P1}} = \dfrac{{4{x^2}{P_1}}}{{\left( {1 - {x^2}} \right)}}
Similarly, for the reaction, Z⇌P+QZ \rightleftharpoons P + Q,
The partial pressures are:
PZ=(1−x1+x)P2;PP=(x1+x)P2;PQ=(x1+x)P2{P_Z} = \left( {\dfrac{{1 - x}}{{1 + x}}} \right){P_2};{P_P} = \left( {\dfrac{x}{{1 + x}}} \right){P_2};{P_Q} = \left( {\dfrac{x}{{1 + x}}} \right){P_2}
We know that, KP2=PPPQPZ{K_{P2}} = \dfrac{{{P_P}{P_Q}}}{{{P_Z}}}.
Substituting the values, we get,
KP2=x2(1−x2)P2{K_{P2}} = \dfrac{{{x^2}}}{{\left( {1 - {x^2}} \right)}}{P_2}
So, KP1KP2=4x2P1(1−x2)×(1−x2)x2P2=4P1P2=19\dfrac{{{K_{P1}}}}{{{K_{P2}}}} = \dfrac{{4{x^2}{P_1}}}{{\left( {1 - {x^2}} \right)}} \times \dfrac{{\left( {1 - {x^2}} \right)}}{{{x^2}{P_2}}} = \dfrac{{4{P_1}}}{{{P_2}}} = \dfrac{1}{9} (Given)
∴P1P2=136\therefore \dfrac{{{P_1}}}{{{P_2}}} = \dfrac{1}{{36}}

Thus, the correct option is A.

Note:
The numerical value of an equilibrium constant is obtained by allowing a reaction to proceed to equilibrium and then measuring the concentrations of each substance involved in that reaction. Since the concentrations are measured at the equilibrium point, the equilibrium constant stays the same for a given reaction independent of the initial concentrations of the reactants and products. This information enables us to derive a standard expression that can serve as a model for any given reaction. This basic model form of equilibrium constant is given as
Keq=(aC)c(aA)a(aB)b{K_{eq}} = \dfrac{{{{\left( {{a_C}} \right)}^c}}}{{{{\left( {{a_A}} \right)}^a}{{\left( {{a_B}} \right)}^b}}} for a reaction: aA+bB⇌cCaA + bB \rightleftharpoons cC
If K>1K > 1 then equilibrium favors the formation of products, reaction proceeds in the forward direction.
If K<1K < 1 then equilibrium favors the formation of reactants, reaction proceeds in the backward direction.