The answer to your question is liquid hydrogen
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
<h3>Further explanation</h3>
In the periodic system
The group number in the periodic system is determined by the same valence electrons (main group), while the period is determined by the number of shells
valence electrons :
- Transition metals: ns, (n-1) d, np
- Lanthanides and actinides: ns, (n-2) f, (n-1) d, np
Valence electrons for the main group: Group 1,2,13-18
Lanthanides and actinides: Group 3–16
Transition metals: Group 3–12
1. Valence electron : 4s²
Group 2
Period 4
2. Valence electron : 2s²2p²
Group 14
Period 2
3. Valence electron : 3s²3p⁴
Group 16
Period 3
4. Valence electron : 4s²3d⁸
Group 10
Period 4
5. Valence electron : 5s²5p⁶
Group 18
Period 5
6. Valence electron : 7s²
Group Lanthanides and actinides (6)
Period 7
