In the decomposition reaction
${AB}_{5} (g) ⇌ {AB}_{3} (g) + B_{2} (g)$ , at equilibrium in a 10 litre closed vessel at 227°C, 2moles of AB₃, 5 moles of B₂ and 4 moles of AB₅, are present. The equilibrium constant K_c for the formation of AB₅(g) is

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0.25

0.04

2.5

4.0

answer is A.

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Detailed Solution

$\begin{array}{l} {AB}_{5} ⇌ {AB}_{3} + B_{2} \\ K_{c} = \frac{[{AB}_{3}] [B_{2}]}{[{AB}_{5}]} = \frac{\frac{2}{10} \times \frac{5}{10}}{\frac{4}{10}} = \frac{10}{100} \times \frac{10}{4} = 0 .25 \end{array}$

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