Question

+ B(g) ↔ A2B(g) Kc = ? A2B(g) + B(g) ↔ A2B2(g) Kc = 16.4 2A(g) + 2B(g) ↔ A2B2(g) Kc = 28.2 The equilibrium constant is given for two of the reactions below. Determine the value of the missing equilibrium constant. 2A(g) + B(g) ↔ A2B(g) Kc = ? A2B(g) + B(g) ↔ A2B2(g) Kc = 16.4 2A(g) + 2B(g) ↔ A2B2(g) Kc = 28.2 0.00216 462 1.72 11.8 0.582

Answer 1

Answer:

The value of the missing equilibrium constant ( of the first equation) is 1.72

Explanation:

First equation: 2A + B ↔ A2B   Kc = TO BE DETERMINED

 ⇒ The equilibrium expression for this equation is written as: [A2B]/[A]²[B]

Second equation: A2B + B ↔ A2B2   Kc= 16.4

⇒ The equilibrium expression is written as: [A2B2]/[A2B][B]

Third equation:  2A + 2B ↔ A2B2     Kc = 28.2

⇒ The equilibrium expression is written as: [A2B2]/ [A]²[B]²

If we add the first to the second equation

2A + B + B ↔ A2B2   the equilibrium constant Kc will be X(16.4)

But the sum of these 2 equations, is the same as the third equation ( 2A + 2B ↔ A2B2)   with Kc = 28.2

So this means: 28.2 = X(16.4)

or X = 28.2/16.4

X = 1.72

with X = Kc of the first equation

The value of the missing equilibrium constant ( of the first equation) is 1.72