JEE MainChemistry

Reviewed by official_key

+4 marks−1 if incorrectNumericalpyq

Equilibrium: JEE Main Chemistry Question with Solution

Water decomposes at

2300\text{ K}

\mathrm{H}_2\mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}_2(\mathrm{g}) + \frac{1}{2}\mathrm{O}_2(\mathrm{g})

The percent of water decomposing at

2300\text{ K}

and

1\text{ bar}

is

\underline{\quad\quad\quad\quad\quad\quad\quad}

(Nearest integer). Equilibrium constant for the reaction is

2 \times 10^{-3}

at

2300\text{ K}

.

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

Type the value - units or words beside it are fine.

Step-by-step solutionView

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: $\mathrm{H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2}O_2(g)}$ , $T = 2300\text{ K}$ , $P = 1\text{ bar}$ , $K_p = 2 \times 10^{-3}$ .

02find

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

03visualise

Initial moles: $n_{\mathrm{H_2O}} = 1$ , $n_{\mathrm{H_2}} = 0$ , $n_{\mathrm{O_2}} = 0$ . At equilibrium: $n_{\mathrm{H_2O}} = 1 - \alpha$ , $n_{\mathrm{H_2}} = \alpha$ , $n_{\mathrm{O_2}} = \frac{\alpha}{2}$ . Total moles at equilibrium: $n_{\text{total}} = 1 - \alpha + \alpha + \frac{\alpha}{2} = 1 + \frac{\alpha}{2}$ .

04strategise

Since $K_p = 2 \times 10^{-3} \ll 1$ , the degree of dissociation $\alpha \ll 1$ . Therefore, $1 - \alpha \approx 1$ and $n_{\text{total}} \approx 1$ . At total pressure $P = 1\text{ bar}$ , partial pressures are $p_i = \frac{n_i}{n_{\text{total}}} P \approx n_i$ . Thus, $K_p = \frac{p_{\mathrm{H_2}} \cdot (p_{\mathrm{O_2}})^{1/2}}{p_{\mathrm{H_2O}}} \approx \frac{\alpha \cdot (\alpha/2)^{1/2}}{1} = \frac{\alpha^{3/2}}{\sqrt{2}}$ .

05execute

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

✓verify

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

Hints that build this answer step by step

If the initial amount of $\mathrm{H_2O}$ is $1\text{ mol}$ and the degree of dissociation is $\alpha$ , what are the equilibrium mole amounts of $\mathrm{H_2O}$ , $\mathrm{H_2}$ , and $\mathrm{O_2}$ ?
$n(\mathrm{H_2O}) = 1 - \alpha$ , $n(\mathrm{H_2}) = \alpha$ , $n(\mathrm{O_2}) = \frac{\alpha}{2}$
What is the expression for $K_p$ in terms of $\alpha$ and total pressure $P$ assuming $\alpha \ll 1$ ?
$K_p \approx \frac{\alpha^{3/2}}{\sqrt{2}} P^{1/2}$
Given $K_p = 2 \times 10^{-3}$ and $P = 1\text{ bar}$ , what is the value of $\alpha$ ?
$\alpha = 0.02$
What is the percent of water decomposing?
$2\%$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Chemistry
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
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: $\mathrm{H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2}O_2(g)}$ , $T = 2300\text{ K}$ , $P = 1\text{ bar}$ , $K_p = 2 \times 10^{-3}$ .
find: Find the percentage of water decomposed, $\%\alpha = \alpha \times 100$ , to the nearest integer.
visualise: Initial moles: $n_{\mathrm{H_2O}} = 1$ , $n_{\mathrm{H_2}} = 0$ , $n_{\mathrm{O_2}} = 0$ . At equilibrium: $n_{\mathrm{H_2O}} = 1 - \alpha$ , $n_{\mathrm{H_2}} = \alpha$ , $n_{\mathrm{O_2}} = \frac{\alpha}{2}$ . Total moles at equilibrium: $n_{\text{total}} = 1 - \alpha + \alpha + \frac{\alpha}{2} = 1 + \frac{\alpha}{2}$ .
strategise: Since $K_p = 2 \times 10^{-3} \ll 1$ , the degree of dissociation $\alpha \ll 1$ . Therefore, $1 - \alpha \approx 1$ and $n_{\text{total}} \approx 1$ . At total pressure $P = 1\text{ bar}$ , partial pressures are $p_i = \frac{n_i}{n_{\text{total}}} P \approx n_i$ . Thus, $K_p = \frac{p_{\mathrm{H_2}} \cdot (p_{\mathrm{O_2}})^{1/2}}{p_{\mathrm{H_2O}}} \approx \frac{\alpha \cdot (\alpha/2)^{1/2}}{1} = \frac{\alpha^{3/2}}{\sqrt{2}}$ .
execute: Equating to $K_p$ : $\frac{\alpha^{3/2}}{\sqrt{2}} = 2 \times 10^{-3} \implies \alpha^{3/2} = 2\sqrt{2} \times 10^{-3} = 2^{3/2} \times (10^{-2})^{3/2}$ . Taking the $(2/3)$ power on both sides gives $\alpha = 2 \times 10^{-2} = 0.02$ . Percentage decomposition is $0.02 \times 100\% = 2\%$ .
verify: Check approximation: $\alpha = 0.02$ means $1 - \alpha = 0.98$ and $1 + \alpha/2 = 1.01$ , error is about $1\text{--}2\%$ , well within rounding to the nearest integer.

Equilibrium: JEE Main Chemistry Question with Solution

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

We think you should solve this next

For reaction…

At $298\mathrm{~K}$ , the solubility of silver chloride in water is…

Consider the following equation…

Students also ask

Equilibrium: JEE Main Chemistry Question with Solution

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

01given

02find

03visualise

04strategise

05execute

✓verify

For reaction…

At 298 K298\mathrm{~K}298 K, the solubility of silver chloride in water is…

Consider the following equation…

Students also ask

At $298\mathrm{~K}$ , the solubility of silver chloride in water is…