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Chemical Equilibrium

Section: Chemistry

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Question : 8 of 17

Marks: +1, -0

The following equilibrium is established at STP.

B_2(g) \rightleftharpoons 2 B(g)

Atoms of B occupy 20 % of total volume at STP. The total pressure of the system is 1 bar .

K_{p}

=22.7\,\text{L}

[AP EAPCET 21 May 2025 S1]

$0.05$
$0.1$
$0.5$
$0.025$

Solution:

Volume occupied by

B

atom

=20\%

of total volume

=20\% \;\text{ of }\; 22.7\,\text{L}\;=4.54\,\text{L}

\;\text{ Remaining volume }\;\;=(22.7-4.54)\,\text{L}

\;=18.16\,\text{L}

Mole fraction of

B=\chi_B=\frac{4.54}{22.7}=0.2

Mole fraction of

B_2=\chi_{B_2}=\frac{18.16}{22.7}=0.8

Partial pressure of

B=p_B=0.2 \times 1=0.2

bar

Partial pressure of

\;B_2=p_{B_2}=0.8 \times 1=0.8 \;\text{ bar }\;

\;\;\; K_p=\frac{p_B^2}{p_{B_2}}=\frac{0.2^2}{0.8}=\frac{0.04}{0.8}=0.05

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