Question

The equilibrium constant is given for two of the reactions below. Determine the value of the missing equilibrium constant. A(g) + 2B(g) ↔ AB2(g) Kc = 59 AB2(g) + B(g) ↔ AB3(g) Kc = ? A(g) + 3B(g) ↔ AB3(g) Kc = 478

Answer 1

Answer:

The correct answer is 8.10

Explanation:

Given:

A(g) + 2B(g) ↔ AB₂(g)   Kc = 59 ---- Eq. 1

A(g) + 3B(g) ↔ AB₃(g)   Kc = 478 ----- Eq. 2

We have to rearrange the chemical equations in order to obtain:

AB₂(g) + B(g) ↔ AB₃(g) Kc = ?

AB₂(g) is a reactant, so we have to use the reverse reaction of Eq. 1, in this case Kc= 1/59. Since AB₃(g) is a product, we use the forward reaction of Eq.2, and the constant is the same: Kc= 478.  The following is the sum of rearranged chemical equations, and the compounds in bold and italic are canceled:

 AB₂(g)       ↔   A(g) + 2B(g)          Kc₁= 1/59

A(g) + 3B(g) ↔   AB₃(g)                  Kc₂= 478

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AB₂(g) + B(g) ↔ AB₃(g)

If we add reactions at equilibrium, the equilibrium constants Kc are mutiplied as follows:

Kc = Kc₁ x Kc₂ = 1/59 x 478 = 478/59 = 8.10

The value of the missing equilibrium constant is 8.10.