Consider the following chemical equilibrium of the gas phase reaction at a constant temperature : A(g) B(g)+C(g) If p being the total pressure, Kp is the pressure equilibrium constant and α is the degree of dissociation, then which of the following is true at equilibrium ?

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q55 – JEE-MAIN 2025

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Home Chemistry EquilibriumQuestion #439

Consider the following chemical equilibrium of the gas phase reaction at a constant temperature :

A (g) ⇌ B (g) + C (g)

p

being the total pressure,

K_{p}

is the pressure equilibrium constant and

α

is the degree of dissociation, then which of the following is true at equilibrium ?

Clue 1 of 3: Le Chatelier's Principle

Consider how changes in pressure affect the equilibrium position for reactions involving a change in the number of moles of gas.

🥷

Ninja Strategy•Le Chatelier's Principle and Limiting Cases

First, use Le Chatelier's principle to quickly assess the general effect of pressure on $α$ . Then, use limiting cases ( $p ≫ K_{p}$ or $K_{p} ≫ p$ ) to check the extreme behaviors of $α$ without full derivation.

🧠 Full Solution Path

Step 1: Set up the equilibrium expression✦ Active

For the reaction $A (g) ⇌ B (g) + C (g)$ , let the initial moles of $A$ be $n_{0}$ and the degree of dissociation be $α$ . At equilibrium, the moles of $A$ , $B$ , and $C$ will be $n_{0} (1 - α)$ , $n_{0} α$ , and $n_{0} α$ respectively. The total moles at equilibrium will be $n_{T} = n_{0} (1 + α)$ .

K_{p} = \frac{P_{B} \cdot P_{C}}{P_{A}} = \frac{(\frac{n_{0} α}{n_{0} (1 + α)} p) \cdot (\frac{n_{0} α}{n_{0} (1 + α)} p)}{(\frac{n_{0} (1 - α)}{n_{0} (1 + α)} p)}

K_{p} = \frac{α^{2} p}{1 - α^{2}}

💡 Teacher's Secret Hint

Remember that partial pressure is mole fraction times total pressure.

Step 2: Analyze the relationship between

α

p

, and

K_{p}

○ Expand

Rearranging the $K_{p}$ expression to solve for $α$ :

K_{p} (1 - α^{2}) = α^{2} p

K_{p} - K_{p} α^{2} = α^{2} p

K_{p} = α^{2} (p + K_{p})

α^{2} = \frac{K_{p}}{p + K_{p}}

α = \sqrt{\frac{K_{p}}{p + K_{p}}}

💡 Teacher's Secret Hint

This derived expression is crucial for evaluating the options.

Step 3: Evaluate the given options○ Expand

1. When $p$ increases, the denominator $(p + K_{p})$ increases, so $\frac{K_{p}}{p + K_{p}}$ decreases, and thus $α$ decreases. This is consistent with Le Chatelier's principle as the reaction proceeds with an increase in the number of moles of gas. So, option 1 is true.

2. When $p$ increases, $α$ decreases, so option 2 is false.

3. If $p ≫ K_{p}$ , then $α \approx \sqrt{\frac{K_{p}}{p}}$ . Since $p$ is very high, $α$ will be very small (close to 0), not 1. So, option 3 is false.

4. If $K_{p} ≫ p$ , then $α \approx \sqrt{\frac{K_{p}}{K_{p}}} = 1$ . So, $α$ becomes close to unity, not much less than unity. So, option 4 is false.

💡 Teacher's Secret Hint

Always check limiting cases to confirm the behavior of $α$ .

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Chemistry Question 55 – JEE-MAIN 2025

Clue 1 of 3: Le Chatelier's Principle

🧠 Full Solution Path