Question

In the decomposition reaction $AB_5(g) \rightleftharpoons AB_3(g) + B_2(g)$, at equilibrium in a 10 litres closed vessel at 227°C, 2 moles 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
(a) 0.25
(b) 4.0
(c) 0.04
(d) 2.5

Answer 1

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Answer 2

Given:

Volume of container = 10L

Moles of AB₃ = 2 moles

Moles of B₂ = 5 moles

Moles of AB₅ = 4 moles

To find:

Kc for the formation of AB₅.

Solution:

first, we have to find a concentration of reactants and products.

[AB₃] = $\frac{2}{10}$ = 0.2 M

[B₂] = $\frac{5}{10}$ = 0.5 M

[AB₅] = $\frac{4}{10}$ = 0.4 M

for given reaction,

k = $\frac{[AB_{3} ][B_{2} ]}{[AB_{5} ]}$

k = (0.2 × 0.5)/0.4

k = 0.25

so for formation of AB₅

kc = $\frac{1}{k}$

kc = $\frac{1}{0.25}$

kc = 4.0

Answer:

Option (b)

kc for formation of AB₅ = 4.0