Draw the Born-Haber cycle and use the data below for the formation of calcium chloride, to (ii) calculate the electron affinity of chlorine: RHat = +190 kJ/mol Ca(s) Ca(g) DHJ = +1730 kJ/mol 2e- Ca2* (9) + Ca(g) DHat = +121 kJ/mol Cl2(g) 2C(g) | Ca2* (g) CaCl2(s) DHLE = -2184 kJ/mol 2c(g) + DH = -795 kJ/mol Ca(s) + Cl2(g) CaCl2(s)

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Please could someone help me with this question? I need it to be set out like the second image, but for CaCl2 and not MgCl2. Many thanks

Draw the Born-Haber cycle and use the data below for the formation of calcium chloride, to
(ii)
calculate the electron affinity of chlorine:
DHạt
= +190 kJ/mol
Ca(g)
Ca(s)
2e-
DHJE = +1730 kJ/mol
Ca2* (g)
+
Ca(g)
DHạt = +121 kJ/mol
Cl2(g)
2Cl(g)
DHLE
= -2184 kJ/mol
Ca2* (g)
2C1 (9)
CaCl2(s)
+
CaCl2(s)
DHF = -795 kJ/mol
Ca(s)
Cl2(g)
+
Transcribed Image Text:Draw the Born-Haber cycle and use the data below for the formation of calcium chloride, to (ii) calculate the electron affinity of chlorine: DHạt = +190 kJ/mol Ca(g) Ca(s) 2e- DHJE = +1730 kJ/mol Ca2* (g) + Ca(g) DHạt = +121 kJ/mol Cl2(g) 2Cl(g) DHLE = -2184 kJ/mol Ca2* (g) 2C1 (9) CaCl2(s) + CaCl2(s) DHF = -795 kJ/mol Ca(s) Cl2(g) +
Mg (g) + 2e + 2CI(g)
2 x first electron
affinity(CI)
= 2x-349 = -698
2x AH (CI) = 2 x 122 = +244
%3D
%3D
2+
Mg (g) + 2e + Cl2 (g)
since two chlorines
are involved, all the
enthalpies related to
chlorine are doubled
24
Mg (g) + 2C1 (g)
second IE(Mg) = +1451
Mg (g) + e+ Cl(g)
as one magnesium
forms Mg2+ you must
use the first and second
IEs of magnesium
(not 2 x first IE)
first IE(Mg) = +738
ALH (MgCl2) = -2524
%3D
%3D
Mg(g) + Cl,(g)
the lattice formation
AH (Mg) = +148
Mg (s) + Clalg)
enthalpy of MgCl2 is
|-2524 kJ mol1
A;H (MgCl2)
= -641
MgCl2 (s)
Figure 2 The Born-Haber cycle for magnesium chloride, MgCl, All enthalpies
are in kJ mol
Transcribed Image Text:Mg (g) + 2e + 2CI(g) 2 x first electron affinity(CI) = 2x-349 = -698 2x AH (CI) = 2 x 122 = +244 %3D %3D 2+ Mg (g) + 2e + Cl2 (g) since two chlorines are involved, all the enthalpies related to chlorine are doubled 24 Mg (g) + 2C1 (g) second IE(Mg) = +1451 Mg (g) + e+ Cl(g) as one magnesium forms Mg2+ you must use the first and second IEs of magnesium (not 2 x first IE) first IE(Mg) = +738 ALH (MgCl2) = -2524 %3D %3D Mg(g) + Cl,(g) the lattice formation AH (Mg) = +148 Mg (s) + Clalg) enthalpy of MgCl2 is |-2524 kJ mol1 A;H (MgCl2) = -641 MgCl2 (s) Figure 2 The Born-Haber cycle for magnesium chloride, MgCl, All enthalpies are in kJ mol
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