Q24 / 90 · JEE Main 2016 (10 Apr Online) · Chemistry

If $100 mole$ of $H_{2} O_{2}$ decompose at $1 bar$ and $300 K$ , the work is done ( $kJ$ ) by one mole of $O_{2} (g)$ as it expands against $1 bar$ pressure is:

$2 H_{2} O_{2} (l) ⇌ 2 H_{2} O (l) + O_{2} (g)$

$(R = 8 .3 J K^{- 1} {mol}^{- 1})$

Correct answer
A
$124.50$
B
$249.00$
C
$498.00$
D
$62.25$

Solution

Irreversible process

W = P ∆ V = nRT

\underset{}{\underset{100 mol}{2 H_{2} O_{2}}} (l) ⇌ \underset{}{2 \underset{}{H_{2} O} (l)} + \underset{\begin{array}{l} 50 mol \end{array}}{O_{2}} (g)

∴

100 mol H_{2} O_{2}

on decomposition will give

50 mol O_{2}

= 50 \times 8.3 \times 300

= 124500 J

=124.50 kJ

Hence, the correct answer is option: 'A'.

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