Chemistry Q130 – AP-EAMCET 2026

The equilibrium constants

K_{p_{1}}

and

K_{p_{2}}

for the gaseous reactions,

X ⇌ 2 Y

and

Z ⇌ P + Q

respectively are in the ratio of

1 : 9

. If the degree of dissociation of

X

and

Z

are equal, then the ratio of their total pressures at these equilibria is

Clue 1 of 3: Equilibrium Constant ( $K_{p}$ ) Definition

The equilibrium constant $K_{p}$ for a gaseous reaction is expressed in terms of the partial pressures of reactants and products at equilibrium.

🧠 Full Solution Path

Step 1: Express

K_{p_{1}}

for the first reaction (

X ⇌ 2 Y

)✦ Active

For the first reaction, $X ⇌ 2 Y$ , let's assume we start with 1 mole of $X$ . If the degree of dissociation is $α$ , then at equilibrium, the moles of $X$ remaining will be $(1 - α)$ and moles of $Y$ formed will be $2 α$ . The total moles at equilibrium ( $n_{1}$ ) will be the sum of these.

\begin{array}{lccc} Reaction: & X & ⇌ & 2 Y \\ Initial moles: & 1 & 0 \\ Eq. moles: & 1 - α & 2 α \\ Total moles (n_{1}) = (1 - α) + 2 α = 1 + α \end{array}

The partial pressures are $P_{X} = \frac{1 - α}{1 + α} P_{1}$ and $P_{Y} = \frac{2 α}{1 + α} P_{1}$ . The equilibrium constant $K_{p_{1}}$ is given by:

K_{p_{1}} = \frac{(P_{Y})^{2}}{P_{X}} = \frac{{(\frac{2 α}{1 + α} P_{1})}^{2}}{\frac{1 - α}{1 + α} P_{1}} = \frac{4 α^{2} P_{1}^{2}}{(1 + α)^{2}} \times \frac{1 + α}{(1 - α) P_{1}} = \frac{4 α^{2} P_{1}}{(1 - α) (1 + α)} = \frac{4 α^{2} P_{1}}{1 - α^{2}}

💡 Teacher's Secret Hint

Remember to square the partial pressure of Y in the numerator for the $K_{p}$ expression due to the stoichiometric coefficient. Also, correctly calculate the total number of moles at equilibrium.

Step 2: Express

K_{p_{2}}

for the second reaction (

Z ⇌ P + Q

)○ Expand

Similarly, for the second reaction, $Z ⇌ P + Q$ , let's assume we start with 1 mole of $Z$ . If the degree of dissociation is $α$ , then at equilibrium, moles of $Z$ remaining will be $(1 - α)$ and moles of $P$ and $Q$ formed will each be $α$ . The total moles at equilibrium ( $n_{2}$ ) will be the sum of these.

\begin{array}{lcccc} Reaction: & Z & ⇌ & P & + & Q \\ Initial moles: & 1 & 0 & 0 \\ Eq. moles: & 1 - α & α & α \\ Total moles (n_{2}) = (1 - α) + α + α = 1 + α \end{array}

The partial pressures are $P_{Z} = \frac{1 - α}{1 + α} P_{2}$ , $P_{P} = \frac{α}{1 + α} P_{2}$ , and $P_{Q} = \frac{α}{1 + α} P_{2}$ . The equilibrium constant $K_{p_{2}}$ is given by:

K_{p_{2}} = \frac{P_{P} P_{Q}}{P_{Z}} = \frac{(\frac{α}{1 + α} P_{2}) (\frac{α}{1 + α} P_{2})}{\frac{1 - α}{1 + α} P_{2}} = \frac{α^{2} P_{2}^{2}}{(1 + α)^{2}} \times \frac{1 + α}{(1 - α) P_{2}} = \frac{α^{2} P_{2}}{(1 - α) (1 + α)} = \frac{α^{2} P_{2}}{1 - α^{2}}

💡 Teacher's Secret Hint

Notice that the total moles at equilibrium for both reactions are the same ( $1 + α$ ). This will simplify the ratio calculation significantly.

Step 3: Calculate the ratio of total pressures (

P_{1} : P_{2}

)○ Expand

We are given that the ratio of equilibrium constants $K_{p_{1}} : K_{p_{2}}$ is $1 : 9$ , which means $\frac{K_{p_{1}}}{K_{p_{2}}} = \frac{1}{9}$ . Now, we substitute the expressions for $K_{p_{1}}$ and $K_{p_{2}}$ derived in the previous steps into this ratio.

\frac{\frac{4 α^{2} P_{1}}{1 - α^{2}}}{\frac{α^{2} P_{2}}{1 - α^{2}}} = \frac{1}{9}

We can cancel the common terms $α^{2}$ and $(1 - α^{2})$ from the numerator and denominator.

\frac{4 P_{1}}{P_{2}} = \frac{1}{9}

Now, solve for the ratio $\frac{P_{1}}{P_{2}}$ :

\frac{P_{1}}{P_{2}} = \frac{1}{9 \times 4} = \frac{1}{36}

Thus, the ratio of their total pressures is $1 : 36$ .

💡 Teacher's Secret Hint

Pay close attention to cancelling common terms like $α^{2}$ and $(1 - α^{2})$ from the numerator and denominator of the ratio. This avoids needing to know the actual value of $α$ .

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Chemistry Question 130 – AP-EAMCET 2026

Clue 1 of 3: Equilibrium Constant (Kp) Definition

🧠 Full Solution Path

Clue 1 of 3: Equilibrium Constant ( $K_{p}$ ) Definition