JEE Main 2023 — Chemistry Physical Chemistry

2023integermedium

Water decomposes at $2300K$

${H}_{2}O(g)\rightarrow {H}_{2}(g)+\frac{1}{2}{O}_{2}(g)$

The percent of water decomposing at $2300K$ and 1 bar is _____ (Nearest integer). Equilibrium constant for the reaction is $2\times {10}^{-3}$ at $2300K$

Official previous-year question

Held on 29 Jan 2023 · Verified 6 Jul 2026.

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Solution

Decomposition of water gives:

${H}_{2}O(g)\rightleftharpoons {H}_{2}(g)+\frac{1}{2}{O}_{2}(g)$

${P}_{0}[1-\alpha ]{P}_{0}\alpha \frac{{P}_{0}\alpha }{2}$ partial pr. at eq.

Thus, ${P}_{0}[1+\frac{\alpha }{2}]=1\ldots (1)$

${K}_{p}=\frac{({P}_{{H}_{2}}){({P}_{{O}_{2}})}^{1/2}}{{P}_{{H}_{2}O}}$

$\frac{({P}_{0}\alpha ){(\frac{{P}_{0}\alpha }{2})}^{1/2}}{{P}_{0}[1-\alpha ]}=2\times {10}^{-3}$

since $\alpha$ is negligible w.r.t 1 thus, $1-\alpha \approx 1$

$\Rightarrow \frac{\alpha \sqrt{\alpha }}{\sqrt{2}}=2\times {10}^{-3}$

$\Rightarrow {\alpha }^{3/2}={2}^{3/2}\times {10}^{-3}$

$\Rightarrow \alpha ={2}^{3/2\times 2/3}\times {10}^{-3\times 2/3}$

$\Rightarrow \alpha =2\times {10}^{-2},%\alpha =2%$

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JEE Main Physical Chemistry in other years

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JEE Main 2023 — Chemistry Physical Chemistry

2023integermedium

Water decomposes at $2300K$

${H}_{2}O(g)\rightarrow {H}_{2}(g)+\frac{1}{2}{O}_{2}(g)$

The percent of water decomposing at $2300K$ and 1 bar is _____ (Nearest integer). Equilibrium constant for the reaction is $2\times {10}^{-3}$ at $2300K$

Official previous-year question

Held on 29 Jan 2023 · Verified 6 Jul 2026.

Did you get this right?

Solution

Decomposition of water gives:

${H}_{2}O(g)\rightleftharpoons {H}_{2}(g)+\frac{1}{2}{O}_{2}(g)$

${P}_{0}[1-\alpha ]{P}_{0}\alpha \frac{{P}_{0}\alpha }{2}$ partial pr. at eq.

Thus, ${P}_{0}[1+\frac{\alpha }{2}]=1\ldots (1)$

${K}_{p}=\frac{({P}_{{H}_{2}}){({P}_{{O}_{2}})}^{1/2}}{{P}_{{H}_{2}O}}$

$\frac{({P}_{0}\alpha ){(\frac{{P}_{0}\alpha }{2})}^{1/2}}{{P}_{0}[1-\alpha ]}=2\times {10}^{-3}$

since $\alpha$ is negligible w.r.t 1 thus, $1-\alpha \approx 1$

$\Rightarrow \frac{\alpha \sqrt{\alpha }}{\sqrt{2}}=2\times {10}^{-3}$

$\Rightarrow {\alpha }^{3/2}={2}^{3/2}\times {10}^{-3}$

$\Rightarrow \alpha ={2}^{3/2\times 2/3}\times {10}^{-3\times 2/3}$

$\Rightarrow \alpha =2\times {10}^{-2},%\alpha =2%$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

Browse all Chemistry topics

JEE Main 2023 — Chemistry Physical Chemistry

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Solution

Your note

JEE Main Physical Chemistry in other years

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Organic Chemistry

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JEE Main 2023 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry