JEE Main 2022 — Chemistry Physical Chemistry

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$C(s)+{O}_{2}(g)\rightarrow {\mathrm{CO}}_{2}(g)+400\mathrm{kJ}$

$C(s)+\frac{1}{2}{O}_{2}(g)\rightarrow \mathrm{CO}(g)+100\mathrm{kJ}$

When coal of purity $60%$ is allowed to burn in presence of insufficient oxygen, $60%$ of carbon is converted into ' $\mathrm{CO}$ ' and the remaining is converted into ' ${\mathrm{CO}}_{2}$ '.

The heat generated when $0.6\mathrm{kg}$ of coal is burnt is

Official previous-year question

Held on 29 Jul 2022 · Verified 6 Jul 2026.

Options

A
$1600\mathrm{kJ}$
B
$3200\mathrm{kJ}$
C
$4400\mathrm{kJ}$
D
$6600\mathrm{kJ}$

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Solution

$C(S)+{O}_{2}(g)\rightarrow {\mathrm{CO}}_{2}(g)+400\mathrm{kJ}$

$1$ mole

$C(s)+\frac{1}{2}{O}_{2}(g)\rightarrow \mathrm{CO}(g)+100\mathrm{kJ}...(2)$

$0.6\mathrm{Kg}\times 1000$

$=600\mathrm{gm}$

Mass of $60%$ pure coal

= $600\times \frac{60}{100}$ (Pure Carbon)

$=360\mathrm{gm} \mathrm{Number}\mathrm{of}\mathrm{moles}=\frac{360}{12}=30\mathrm{moles}\mathrm{of}\mathrm{pure}\mathrm{carbon}$

Moles of 60% pure Carbon converted into ${\mathrm{CO}}_{2}=(30-30\times \frac{60}{100})$

$=12$ mole and

Moles of 60% pure carbon converted in to $\mathrm{CO}=30\times \frac{60}{100}=18$ mole

Energy generated during $2$ equation $=18\times 100$

$=1800\mathrm{kJ}$

Energy generated during ${1}^{\mathrm{st}}$ reaction.

$=12\times 400$

$=4800$

Total $=1800+4800=6600\mathrm{kJ}$

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Solution

$C(S)+{O}_{2}(g)\rightarrow {\mathrm{CO}}_{2}(g)+400\mathrm{kJ}$

$1$ mole

$C(s)+\frac{1}{2}{O}_{2}(g)\rightarrow \mathrm{CO}(g)+100\mathrm{kJ}...(2)$

$0.6\mathrm{Kg}\times 1000$

$=600\mathrm{gm}$

Mass of $60%$ pure coal

= $600\times \frac{60}{100}$ (Pure Carbon)

$=360\mathrm{gm} \mathrm{Number}\mathrm{of}\mathrm{moles}=\frac{360}{12}=30\mathrm{moles}\mathrm{of}\mathrm{pure}\mathrm{carbon}$

Moles of 60% pure Carbon converted into ${\mathrm{CO}}_{2}=(30-30\times \frac{60}{100})$

$=12$ mole and

Moles of 60% pure carbon converted in to $\mathrm{CO}=30\times \frac{60}{100}=18$ mole

Energy generated during $2$ equation $=18\times 100$

$=1800\mathrm{kJ}$

Energy generated during ${1}^{\mathrm{st}}$ reaction.

$=12\times 400$

$=4800$

Total $=1800+4800=6600\mathrm{kJ}$

JEE Main 2022 — Chemistry Physical Chemistry

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JEE Main 2022 — Chemistry Physical Chemistry

Options

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Solution

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JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

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