Question

In the decomposition reaction AB$_5$(g)$\rightleftharpoons$ AB$_3$(g) + B$_2$(g), at equilibrium in a 10-litre closed vessel at 227 C, 2 moles of AB$_3$, 5 moles of B$_2$ and 4 moles of AB$_5$ are present. The equilibrium constant K$_c$ for the formation of AB$_5$(g) is

• A. 0.25
• B. 4
• C. 0.04
• D. 2.5

Answer 1

Answer: (B)

4

Answer 2

Answer (b) 4.0

AB$_5$(g)$\rightleftharpoons$ AB$_3$(g) + B$_2$(g)

4 mol $\rightleftharpoons$ 2 mol + 5 mol (At equilibrium)

[AB5] = 4/10 = 0.4

[AB3] = 2/10 = 0.2

[B2] = 5/10 = 0.5

Formation of AB5

AB$_3$(g) + B$_2$(g) → AB$_5$

Kc = [AB5]/[[AB3][B2]]

Kc = 4