JEE Main 2023 — Chemistry Physical Chemistry

2023integermedium

$A(g)\rightarrow 2B(g)+C(g)$ is a first order reaction. The initial pressure of the system was found to be $800\mathrm{mm}\mathrm{Hg}$ which increased to $1600\mathrm{mm}\mathrm{Hg}$ after $10\mathrm{min}$ . The total pressure of the system after $30$ min will be $\mathrm{mm}\mathrm{Hg}$ . (Nearest integer)

Official previous-year question

Held on 13 Apr 2023 · Verified 6 Jul 2026.

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Solution

The final pressure of each component can be calculated as follows,

$A(g)\rightarrow 2B(g)+C(g) \mathrm{intial}80000 \mathrm{At}\mathrm{time}=t800-p2pp$

$800+2p=1600$

$2p=800$

$p=400$

The first order reaction rate constant can be calculated as follows,

$K=\frac{2.303}{10}\mathrm{log}\frac{800}{400}=\frac{2.303\times \mathrm{log}2}{10}$

For $30$ min,

$K=\frac{2.303}{30}\mathrm{log}\frac{(800)}{(800-y)}$

$\frac{2.303\times \mathrm{log}2}{10}=\frac{2.303}{30}\mathrm{log}(\frac{800}{800-y})$

$\Rightarrow (\frac{800}{800-y})=8$

$100=800–y$

$y=700$

Total pressure after $30$ min

$(800–y)+(2y)+(y)$

$=800+2y$

$=800+1400$

$=2200\mathrm{mm}\mathrm{Hg}$

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Solution

The final pressure of each component can be calculated as follows,

$A(g)\rightarrow 2B(g)+C(g) \mathrm{intial}80000 \mathrm{At}\mathrm{time}=t800-p2pp$

$800+2p=1600$

$2p=800$

$p=400$

The first order reaction rate constant can be calculated as follows,

$K=\frac{2.303}{10}\mathrm{log}\frac{800}{400}=\frac{2.303\times \mathrm{log}2}{10}$

For $30$ min,

$K=\frac{2.303}{30}\mathrm{log}\frac{(800)}{(800-y)}$

$\frac{2.303\times \mathrm{log}2}{10}=\frac{2.303}{30}\mathrm{log}(\frac{800}{800-y})$

$\Rightarrow (\frac{800}{800-y})=8$

$100=800–y$

$y=700$

Total pressure after $30$ min

$(800–y)+(2y)+(y)$

$=800+2y$

$=800+1400$

$=2200\mathrm{mm}\mathrm{Hg}$

JEE Main 2023 — Chemistry Physical Chemistry

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Solution

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JEE Main 2023 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

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