JEE MainChemistry

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Degree of Dissociation of Water Vapor from Equilibrium Constant

Water decomposes at 2300K H_2O(g) → H_2(g) + 1/2O_2(g) The percent of water decomposing at 2300K and 1bar is (Nearest integer). Equilibrium constant for the reaction is 2 × 10^(-3) at 2300K.

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Correct answer

The percentage of water decomposed at 2300 K and 1 bar is 2%.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

Reaction: $H_{2} O (g) ⇌ H_{2} (g) + \frac{1}{2} O_{2} (g)$ , $T = 2300 K$ , $P = 1 bar$ , $K_{p} = 2 \times 10^{- 3}$ .

02Find

Find the percentage of water decomposed, $% α = α \times 100$ , to the nearest integer.

03Visualise

Initial moles: $n_{H_{2} O} = 1$ , $n_{H_{2}} = 0$ , $n_{O_{2}} = 0$ . At equilibrium: $n_{H_{2} O} = 1 - α$ , $n_{H_{2}} = α$ , $n_{O_{2}} = \frac{α}{2}$ . Total moles at equilibrium: $n_{total} = 1 - α + α + \frac{α}{2} = 1 + \frac{α}{2}$ .

04Strategise

Since $K_{p} = 2 \times 10^{- 3} ≪ 1$ , the degree of dissociation $α ≪ 1$ . Therefore, $1 - α \approx 1$ and $n_{total} \approx 1$ . At total pressure $P = 1 bar$ , partial pressures are $p_{i} = \frac{n_{i}}{n_{total}} P \approx n_{i}$ . Thus, $K_{p} = \frac{p_{H_{2}} \cdot (p_{O_{2}})^{1 / 2}}{p_{H_{2} O}} \approx \frac{α \cdot (α / 2)^{1 / 2}}{1} = \frac{α^{3 / 2}}{\sqrt{2}}$ .

05Execute

Equating to $K_{p}$ : $\frac{α^{3 / 2}}{\sqrt{2}} = 2 \times 10^{- 3} ⟹ α^{3 / 2} = 2 \sqrt{2} \times 10^{- 3} = 2^{3 / 2} \times (10^{- 2})^{3 / 2}$ . Taking the $(2 / 3)$ power on both sides gives $α = 2 \times 10^{- 2} = 0.02$ . Percentage decomposition is $0.02 \times 100 % = 2 %$ .

✓Verify

Check approximation: $α = 0.02$ means $1 - α = 0.98$ and $1 + α / 2 = 1.01$ , error is about $1 – 2 %$ , well within rounding to the nearest integer.

Hints that build this answer step by step

If the initial amount of $H_{2} O$ is $1 mol$ and the degree of dissociation is $α$ , what are the equilibrium mole amounts of $H_{2} O$ , $H_{2}$ , and $O_{2}$ ?
$n (H_{2} O) = 1 - α$ , $n (H_{2}) = α$ , $n (O_{2}) = \frac{α}{2}$
What is the expression for $K_{p}$ in terms of $α$ and total pressure $P$ assuming $α ≪ 1$ ?
$K_{p} \approx \frac{α^{3 / 2}}{\sqrt{2}} P^{1 / 2}$
Given $K_{p} = 2 \times 10^{- 3}$ and $P = 1 bar$ , what is the value of $α$ ?
$α = 0.02$
What is the percent of water decomposing?
$2 %$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Chemistry
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Why can we write partial pressure as moles directly?

Partial pressure is mole fraction times total pressure. Here total pressure is 1 bar, and since alpha is very small, total moles is approximately 1. Hence mole fraction equals moles.

Answer

The percentage of water decomposed at 2300 K and 1 bar is 2%.

Why each option works or fails

Step-by-step solution

Given: Reaction: $H_{2} O (g) ⇌ H_{2} (g) + \frac{1}{2} O_{2} (g)$ , $T = 2300 K$ , $P = 1 bar$ , $K_{p} = 2 \times 10^{- 3}$ .
Find: Find the percentage of water decomposed, $% α = α \times 100$ , to the nearest integer.
Visualise: Initial moles: $n_{H_{2} O} = 1$ , $n_{H_{2}} = 0$ , $n_{O_{2}} = 0$ . At equilibrium: $n_{H_{2} O} = 1 - α$ , $n_{H_{2}} = α$ , $n_{O_{2}} = \frac{α}{2}$ . Total moles at equilibrium: $n_{total} = 1 - α + α + \frac{α}{2} = 1 + \frac{α}{2}$ .
Strategise: Since $K_{p} = 2 \times 10^{- 3} ≪ 1$ , the degree of dissociation $α ≪ 1$ . Therefore, $1 - α \approx 1$ and $n_{total} \approx 1$ . At total pressure $P = 1 bar$ , partial pressures are $p_{i} = \frac{n_{i}}{n_{total}} P \approx n_{i}$ . Thus, $K_{p} = \frac{p_{H_{2}} \cdot (p_{O_{2}})^{1 / 2}}{p_{H_{2} O}} \approx \frac{α \cdot (α / 2)^{1 / 2}}{1} = \frac{α^{3 / 2}}{\sqrt{2}}$ .
Execute: Equating to $K_{p}$ : $\frac{α^{3 / 2}}{\sqrt{2}} = 2 \times 10^{- 3} ⟹ α^{3 / 2} = 2 \sqrt{2} \times 10^{- 3} = 2^{3 / 2} \times (10^{- 2})^{3 / 2}$ . Taking the $(2 / 3)$ power on both sides gives $α = 2 \times 10^{- 2} = 0.02$ . Percentage decomposition is $0.02 \times 100 % = 2 %$ .
Verify: Check approximation: $α = 0.02$ means $1 - α = 0.98$ and $1 + α / 2 = 1.01$ , error is about $1 – 2 %$ , well within rounding to the nearest integer.

Degree of Dissociation of Water Vapor from Equilibrium Constant

What feels right?

The percentage of water decomposed at 2300 K and 1 bar is 2%.

Why each option works or fails

Solution

01Given

02Find

03Visualise

04Strategise

05Execute

✓Verify

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