JEE MainChemistryToughPhysical Chemistry

A2(g) and B2(g) having partial pressures 60 mm of Hg & 42 mm of
Hg respectively, are present in a closed vessel. At equilibrium, partial
pressure of AB(g) is 28 mm of Hg. If all measurements are made under
similar condition, then calculate percentage of dissociation of AB (g).
(Round of answers to nearest integer).

Show answer and solution

A_{2}(g) + B_{2}(g) \rightleftharpoons 2AB(g)

at t=0 60 42 0
at eq. (60-x) (42-x) (2x)
(Partial pressures)

\Rightarrow 2x = 28 \Rightarrow x = 14

K_{p} = \frac{\left( P_{AB} \right)^{2}}{\left( P_{A_{2}} \right)\left( P_{B_{2}} \right)} = \frac{(28)^{2}}{(46)(28)} = \frac{14}{23}

For

\Delta n_{g} = 0;K_{p} = K_{C_{1}} = \frac{14}{23}

Let degree of dissociation for AB is, then

2AB(g) \rightleftharpoons A_{2}(g) + B_{2}(g)\left( K_{C_{2}} = \frac{1}{K_{C_{1}}} = \frac{23}{14} \right)

1 0 0
at t = 0 (1 - x) x x

\frac{\left( \frac{x}{2} \right)\left( \frac{x}{2} \right)}{(1 - x)^{2}} = \frac{23}{14} \rightarrow \alpha = 0.719

Hence percentage of dissociatin

= 0.719 \times 100 = 72\%

Source: Grass-G / CH-23 Chemistry Paper 31 Dec. FST.docx

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