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

Hint:

4 mol → 2 mol + 5 mol (At equilibrium)

Answer 1

The Correct Option is B

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\(_5\)

K_c = [AB₅]/[[AB₃][B₂]]

K_c = 4

Answer 2

\(AB_5(g) \rightleftharpoons AB_3(g) + B_2(g)\)

We need to use the equilibrium concentrations of the substances involved.

Given:
- Volume of the closed vessel = 10 liters
- Temperature = 227°C
- Moles at equilibrium:
 - \(n_{AB_3} = 2\) moles
 - \(n_{B_2} = 5\) moles
 - \(n_{AB_5} = 4\) moles

Now, let's determine the equilibrium concentrations:

\([AB_5]_{eq} = \frac{n_{AB_5}}{V} = \frac{4}{10} = 0.4 \, \text{M}\)

\([AB_3]_{eq} = \frac{n_{AB_3}}{V} = \frac{2}{10} = 0.2 \, \text{M}\)

\([B_2]_{eq} = \frac{n_{B_2}}{V} = \frac{5}{10} = 0.5 \, \text{M}\)

Now, substitute these values into the expression for Kc :

\(K_c = \frac{(0.4)}{(0.2)(0.5)}\)

\(K_c = \frac{0.4}{0.1}\)

\(K_c = 4\)

So, the correct option is (B): 4