- #1
jbowers9
- 89
- 1
I'm baaaaack...
Is this the correct way to procede
1. The problem statement, all variables and given/known d
The Born Haber cycle for the process was given as follows:
........U
......NH4Cl(s) → NH4+ + Cl-
......↑......↓.....P(NH3)
......↑......NH3 + H+ + Cl-
......↑......↓.....-IH + ECl
1/2N2(g) + 2H2(g) + 1/2Cl2(g) → NH3 + H + Cl
where;
lattice energy/mole NH4CL U = 640kj/mol, ionization energy/mole hydrogen atoms IH = 1305kj/mol, electron affinity/mole ECL = 387kj/mol, heat of formation for NH3 = -45.6kj/mol, heat of formation for NH4Cl = -314.2
-U......NH4+ + Cl- → NH4Cl(s)
-∆ƒHºmNH4Cl...NH4Cl(s) → 1/2N2(g) + 2H2(g) + 1/2Cl2(g)
∆ƒHºmNH3...2N2(g) + 3/2H2(g) → NH3(g)
∆ƒHºmH......1/2H2(g) → H(g)
∆ƒHºmCl......1/2Cl2(g) → Cl(g)
IH - ECl...NH3(g) + H(g) + Cl(g) → NH3(g) + H+ + Cl-
- P(NH3)...NH3(g) + H+ + Cl- → NH4+ + Cl-
All added entities should sum to zero if the cyclical process is reversible, no?
You're ending where you began.
U - ∆ƒHºmNH4Cl + ∆ƒHºmNH3 + ∆ƒHºmH + ∆ƒHºmCl + IH - ECl =
P(NH3) = 2761 kJ/mol
Is this the correct way to procede
1. The problem statement, all variables and given/known d
The Born Haber cycle for the process was given as follows:
........U
......NH4Cl(s) → NH4+ + Cl-
......↑......↓.....P(NH3)
......↑......NH3 + H+ + Cl-
......↑......↓.....-IH + ECl
1/2N2(g) + 2H2(g) + 1/2Cl2(g) → NH3 + H + Cl
where;
lattice energy/mole NH4CL U = 640kj/mol, ionization energy/mole hydrogen atoms IH = 1305kj/mol, electron affinity/mole ECL = 387kj/mol, heat of formation for NH3 = -45.6kj/mol, heat of formation for NH4Cl = -314.2
Homework Equations
-U......NH4+ + Cl- → NH4Cl(s)
-∆ƒHºmNH4Cl...NH4Cl(s) → 1/2N2(g) + 2H2(g) + 1/2Cl2(g)
∆ƒHºmNH3...2N2(g) + 3/2H2(g) → NH3(g)
∆ƒHºmH......1/2H2(g) → H(g)
∆ƒHºmCl......1/2Cl2(g) → Cl(g)
IH - ECl...NH3(g) + H(g) + Cl(g) → NH3(g) + H+ + Cl-
- P(NH3)...NH3(g) + H+ + Cl- → NH4+ + Cl-
The Attempt at a Solution
All added entities should sum to zero if the cyclical process is reversible, no?
You're ending where you began.
U - ∆ƒHºmNH4Cl + ∆ƒHºmNH3 + ∆ƒHºmH + ∆ƒHºmCl + IH - ECl =
P(NH3) = 2761 kJ/mol
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