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Born-Haber cycle: NaCl
The Born-Haber cycle can by divided into stages like the ones in the table below. Start by filling in the table
using:
+
Na(g)  Na (g) + e
−
−
−
stage
−
+
Cl(g) + e  Cl (g)
Na (g) + Cl (g)  NaCl(s)
standard enthalpy change(s)
∆H
elements to gaseous atoms
θ
at Na
& ∆H
equation(s)
Na(s)  Na(g)
θ
at Cl
½Cl2(g)  Cl(g)
Na(g)  Na (g) + e
1 EA of Cl
st
Cl(g) + e  Cl (g)
gaseous ions to solid lattice
lattice enthalpy (formation)
Na (g) + Cl (g)  NaCl(s)
elements to compound
∆H f NaCl
1 IE of Na
atoms to negative ions
+
−
st
atoms to positive ions
−
−
+
θ
−
Na(s) + ½Cl2(g)  NaCl(s)
Problem Solving Gridlock 1
Each row, column and 2 x 2 box contains information about the first four stages listed. Use your problem
solving skills and the answers in the table above to fill in the blank boxes.
stage
standard enthalpy change(s)
elements to gaseous
atoms
atoms to positive ions
atoms to negative ions
gaseous ions to solid
lattice
st
st
1 IE of Na
lattice enthalpy
(formation)
1 EA of Cl
∆H
θ
at Na
& ∆H
standard enthalpy change(s)
θ
at Cl
lattice enthalpy
(formation)
st
1 EA of Cl
∆H
θ
at Na
& ∆H
θ
at Cl
1st IE of Na
gaseous ions to solid
lattice
elements to gaseous
atoms
atoms to positive ions
atoms to negative ions
stage
Gridlocks – did you unlock the grid?
Gridlock 2
Each row, column and 2 x 2 box contains information about each of the last four stages listed.
stage
atoms to positive ions
atoms to negative ions
+
equations
gaseous ions to solid
lattice
−
−
+
elements to compound Na (g) + Cl (g)  NaCl(s)
−
+
Na(g)  Na (g) + e
Na(s) + ½Cl2(g) 
NaCl(s)
Cl(g) + e  Cl (g)
−
+
Na(g)  Na (g) + e
θ
−
Na (g) + Cl (g)  NaCl(s)
Na(s) + ½Cl2(g) 
NaCl(s)
−
Cl(g) + e  Cl (g)
st
∆H f NaCl
−
1 EA of Cl
lattice enthalpy
(formation)
st
1 IE of Na
equations
−
standard enthalpy change(s)
Gridlock 3
You will need to work out which stages are in this gridlock.
stage
equations
atoms to negative ions
elements to gaseous
atoms
Na(s) + ½Cl2(g) 
NaCl(s)
gaseous ions to solid
lattice
elements to compound
Cl(g) + e  Cl (g)
−
lattice enthalpy
(formation)
Na(s)  Na(g)
−
Cl(g) + e  Cl (g)
½Cl2(g)  Cl(g)
Na(s) + ½Cl2(g) 
NaCl(s)
+
−
Na (g) + Cl (g)  NaCl(s)
equations
−
∆H
θ
at Na
& ∆H
−
θ
at Cl
−
+
Na (g) + Cl (g)  NaCl(s)
Na(s)  Na(g)
½Cl2(g)  Cl(g)
θ
∆H f NaCl
st
1 EA of Cl
standard enthalpy change(s)