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Q.

1.0 mole of AB₅ (g) is placed in a closed container under one atmosphere and at 300K. It is heated to 600K, when 20% by mass of it dissociates as
AB₅(e) $\to$ AB(g)+ 2B₂(g).
The resultant pressure is

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a

1.2atm

b

2.8 atm

c

1.4 atm

d

2. 4atrn

answer is D.

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Detailed Solution

$\underset{1 - α}{{AB}_{5} (g}) ⇌ \underset{α}{AB (g}) + \underset{2 α}{2 B_{2} (g)}$
Total moles at equilibrium = I + 2 $α$
$\begin{array}{l} = 1 + \frac{20 \times 2}{100} = 1 .4 \\ \frac{p_{1}}{p_{2}} = \frac{n_{1} T_{1}}{n_{2} T_{2}}; \frac{1}{p_{2}} = \frac{1 \times 300}{1 .4 \times 600} \\ ∴ p_{2} = 2 .8 atm \end{array}$

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