Equilibrium NEET 2008: Pressure Ratio of 36:1 Explained

What is the pressure ratio in the equilibrium NEET 2008 question?

The correct answer to the equilibrium NEET 2008 question is C) 36 : 1. The factor of four comes from the second reaction’s product partial-pressure expression, despite equal degrees of dissociation.

This Chemistry, Equilibrium, 2008 item is rated hard by the question bank, not by an official exam difficulty scale. It compares two separate gaseous equilibria:

X (g) ⇌ Y (g) + Z (g)

$A (g) ⇌ 2 B (g)$ Their equilibrium constants satisfy:

K_{p 1} : K_{p 2} = 9 : 1

Equal fractions of X and A dissociate. Find the ratio of their total pressures at equilibrium, defined respectively as:

P_{1} = total equilibrium pressure for X ⇌ Y + Z

P_{2} = total equilibrium pressure for A ⇌ 2B

$P_{1} : P_{2}$ The supplied choices are:

How do you derive the first equilibrium constant from mole fractions?

Start with one mole of X and no products, following the official solution. Each mole of X that dissociates forms one mole each of Y and Z. Calculate their partial pressures separately, multiply them, then divide by the partial pressure of X.

Define the degree of dissociation as the fraction of the initial reactant that dissociates:

α = \frac{moles of initial reactant dissociated}{initial moles of reactant}

These entries show initial moles, change above the arrow, and equilibrium moles:

X : & 1 \overset{- α}{\to} 1 - α Y : & 0 \overset{+ α}{\to} α Z : & 0 \overset{+ α}{\to} α

Add all equilibrium amounts, including the undissociated reactant:

n_{total, 1} = (1 - α) + α + α = 1 + α

Apply the mole-fraction rule to each gas:

partial pressure = mole fraction \times total pressure

p_{X} = \frac{1 - α}{1 + α} P_{1}, p_{Y} = p_{Z} = \frac{α}{1 + α} P_{1}

Substitute these pressures into the first equilibrium constant:

K_{p 1} & = \frac{p_{Y} p_{Z}}{p_{X}} & = \frac{(\frac{α}{1 + α} P_{1}) (\frac{α}{1 + α} P_{1})}{(\frac{1 - α}{1 + α} P_{1})} & = \frac{α^{2} P_{1}}{(1 + α) (1 - α)} & = \frac{α^{2} P_{1}}{1 - α^{2}}

Where does the factor of four enter the second equilibrium constant?

The factor of four comes from squaring the partial pressure of B. Its equilibrium amount is twice that of either product in the first reaction. The coefficient two controls both the amount formed and the exponent in the equilibrium expression.

Take one mole of A initially, with no B. Use the same degree of dissociation because the question specifies equal fractions dissociated:

A : & 1 \overset{- α}{\to} 1 - α B : & 0 \overset{+ 2 α}{\to} 2 α

The total equilibrium amount is:

n_{total, 2} = (1 - α) + 2 α = 1 + α

Convert both amounts to partial pressures using the second total pressure:

p_{A} = \frac{1 - α}{1 + α} P_{2}, p_{B} = \frac{2 α}{1 + α} P_{2}

The full substitution gives:

K_{p 2} & = \frac{p_{B}^{2}}{p_{A}} & = \frac{{(\frac{2 α}{1 + α} P_{2})}^{2}}{(\frac{1 - α}{1 + α} P_{2})} & = \frac{4 α^{2} P_{2}^{2}}{(1 + α)^{2}} \cdot \frac{1 + α}{(1 - α) P_{2}} & = \frac{4 α^{2} P_{2}}{(1 + α) (1 - α)} & = \frac{4 α^{2} P_{2}}{1 - α^{2}}

Both reactions have identical total mole counts, but their product numerators are not interchangeable. The first multiplies two separate product pressures; the second squares the pressure of a product formed in twice the amount:

α \cdot α = α^{2}, (2 α)^{2} = 4 α^{2}

How does dividing the constants give 36 : 1?

Dividing cancels the common dissociation factor, leaving the first pressure divided by four times the second pressure. The given constant ratio then fixes the pressure ratio without requiring a numerical value for the degree of dissociation.

\frac{K_{p 1}}{K_{p 2}} & = \frac{α^{2} P_{1} / (1 - α^{2})}{4 α^{2} P_{2} / (1 - α^{2})} & = \frac{α^{2}}{1 - α^{2}} \cdot \frac{1 - α^{2}}{α^{2}} \cdot \frac{P_{1}}{4 P_{2}} & = \frac{P_{1}}{4 P_{2}}

Substitute the stated ratio and multiply by four:

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

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

Final answer: C) 36 : 1. Substituting back confirms the given constant ratio: $\frac{36}{4} = 9$

Each constant contains the first power of total pressure because two pressure factors in the numerator are divided by one in the denominator. Neither expression retains pressure squared.

Why does equal dissociation not justify D) 1 : 1?

Equal dissociation fixes the mole counts on this one-mole starting basis, not the total pressures. The incorrect inference is: equal fractions dissociate, so total moles are equal, so pressures must be equal, giving D) 1 : 1.

Equal mole counts alone do not establish equal pressure. The question does not supply equal vessel volumes and temperatures as a basis for that conclusion. Degree of dissociation measures a fraction reacted, not pressure, and the different equilibrium constants must still be satisfied.

Equal pressures would directly contradict the stated constant ratio:

P_{1} = P_{2} ⟹ \frac{K_{p 1}}{K_{p 2}} = \frac{P_{2}}{4 P_{2}} = \frac{1}{4} \neq 9

Use this order: build equilibrium mole counts, convert them to partial pressures, then compare the equilibrium-constant expressions. Do not infer pressure equality from matching totals.

Which two practice questions check whether you understand the method?

Use equal pressures in Question 1 and equal constants in Question 2 to test the distinction. These are original practice variants, not additional verified PYQs. Both retain the same reactions and the official solution’s pure-reactant basis: initially one mole of X or A, with no products.

What is the constant ratio if dissociation and total pressures are equal?

Answer: 1 : 4. Equal pressures do not remove the factor caused by product stoichiometry.

Question 1: Keep equal degrees of dissociation, but make the two total pressures equal. Determine the ratio of equilibrium constants.

Unlike the original question, the constant ratio is not given as nine to one. Using the common total pressure in the derived expressions: $P_{1} = P_{2} = P$

\frac{K_{p 1}}{K_{p 2}} = \frac{P}{4 P} = \frac{1}{4}

What is the pressure ratio if dissociation and equilibrium constants are equal?

Answer: 4 : 1. The first pressure must be four times the second to make the constants equal.

Question 2: Keep equal degrees of dissociation, but replace the original constant ratio with equal equilibrium constants. Determine the total-pressure ratio. $K_{p 1} = K_{p 2}$

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

For further practice, use Equilibrium Practice Questions NEET: 6 MCQs Explained.

Equal dissociation alone does not fix pressure; equal dissociation plus the constant ratio fixes it for these reactions. Before moving on, reproduce both partial-pressure substitutions without looking at the solution.

Frequently asked questions

What is the answer to the Equilibrium NEET 2008 pressure-ratio question?

The correct answer is C) 36:1. For the two reactions with equal degrees of dissociation, Kp1/Kp2 = P1/(4P2). Substituting the given constant ratio of 9:1 gives P1/P2 = 36.

Why is there a factor of four in Kp for A ⇌ 2B?

Starting with one mole of pure A, dissociation of a fraction α produces 2α moles of B, so its partial pressure is 2αP2/(1+α). The equilibrium expression is Kp2 = pB²/pA. Squaring the factor 2 in B's partial pressure produces the factor of four.

Does equal degree of dissociation mean equal total pressure?

No. Equal degree of dissociation specifies equal fractions of reactant dissociated, not equal total pressures. For these reactions, equal pressures would give Kp1/Kp2 = 1/4, contradicting the stated ratio of 9.

What is the pressure ratio if both equilibrium constants are equal?

For X ⇌ Y + Z and A ⇌ 2B, starting from pure reactants with equal degrees of dissociation, the pressure ratio would be 4:1. Setting Kp1/Kp2 = 1 in P1/(4P2) gives P1 = 4P2. This is a practice variant, not the original question's condition.