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Chemical Equilibrium

Section: Chemistry

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Question : 1 of 17

Marks: +1, -0

At

T(K)

, consider the following gaseous reaction, which is in equilibrium.

\mathrm{N}_2\mathrm{O}_5 \rightleftharpoons 2\mathrm{NO}_2 + \frac{1}{2}\mathrm{O}_2

What is the fraction of

\mathrm{N}_2\mathrm{O}_5

decomposed at constant volume and temperature, if the initial pressure is 300 mm Hg and pressure at equilibrium is 480 mm Hg ? (Assume all gases as ideal)

[AP EAPCET 24 May 2025 S1]

$0.2$
$0.6$
$0.4$
$0.8$

Solution:

The given reaction is,

\mathrm{N}_2\mathrm{O}_5 \rightarrow 2\mathrm{NO}_2 + \frac{1}{2}\mathrm{O}_2

Let

p_0=

initial pressure of

\mathrm{N}_2\mathrm{O}_5 x =

pressure of

\mathrm{N}_2\mathrm{O}_5

decomposed. Initial state

= \mathrm{N}_2\mathrm{O}_5

has pressure

p_0

Equilibrium state

= \mathrm{N}_2\mathrm{O}_5

has pressure

=p_0 - x

\mathrm{NO}_2 = 2x

pressure,

\mathrm{O}_2 = \frac{1}{2} x

pressure

p_{\text{equilibrium}} = (p_0 - x) + 2x + \frac{1}{2}x

480 = p_0 + \frac{3}{2} x

\Rightarrow p_0 = 300\text{ mmHg}

180 = \frac{3}{2} x, x = 120\text{ mmHg}

Fraction decomposed

= \frac{\text{ pressure of }\mathrm{N}_2\mathrm{O}_5\text{ decomposed }}{\text{ initial pressure of }\mathrm{N}_2\mathrm{O}_5}

= \frac{x}{p_0} = \frac{120}{300} = \frac{2}{5} = 0.4

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