Atomistry » Vanadium » Energy
Atomistry »
  Vanadium »
    Energy »

Vanadium Ionization Energy

The Vanadium Ionization Energy is the energy required to remove from atom one mole of electrons with subsequent production of positively charged ion of Vanadium.
V -> V+ + e-
This process can be repeated many times, but the energy cost is increased dramatically. The general equation for the Vanadium is:
VN+ -> V(N+1)+ + e-

Ionization Energy

V -> V1+ = 650 kJ/mol = 155.50 Kcal/mol
V1+ -> V2+ = 1414 kJ/mol = 338.28 Kcal/mol
V2+ -> V3+ = 2828 kJ/mol = 676.56 Kcal/mol
V3+ -> V4+ = 4507 kJ/mol = 1078.23 Kcal/mol
V4+ -> V5+ = 6294 kJ/mol = 1505.74 Kcal/mol
V5+ -> V6+ = 12362 kJ/mol = 2957.42 Kcal/mol
V6+ -> V7+ = 14489 kJ/mol = 3466.27 Kcal/mol
V7+ -> V8+ = 16760 kJ/mol = 4009.57 Kcal/mol
V8+ -> V9+ = 19860 kJ/mol = 4751.20 Kcal/mol
V9+ -> V10+ = 22240 kJ/mol = 5320.57 Kcal/mol

Last articles

Xe in 6AYK
Xe in 6QII
Xe in 6ASM
Xe in 5NSW
Xe in 6FY9
Xe in 5O1K
Xe in 5O27
Xe in 5M69
Xe in 5KPU
Xe in 5I63
© Copyright 2008-2020 by atomistry.com
Home   |    Site Map   |    Copyright   |    Contact us   |    Privacy