(iii) Draw the Born-Haber cycle and use the data below for the formation of calcium chloride, to calculate the electron affinity of chlorine: Ca() Ca@) AHat = +190 kJ/mol Ca) Ca (o 2e- AHIE = +1730 kJ/mol Claa) Cla) AHat = +121 kJ/mol Ca"() + CaClz) AHLE = -2184 kJ/mol Ca) + Cl2@) CaClae) AH: = -795 kJ/mol
(iii) Draw the Born-Haber cycle and use the data below for the formation of calcium chloride, to calculate the electron affinity of chlorine: Ca() Ca@) AHat = +190 kJ/mol Ca) Ca (o 2e- AHIE = +1730 kJ/mol Claa) Cla) AHat = +121 kJ/mol Ca"() + CaClz) AHLE = -2184 kJ/mol Ca) + Cl2@) CaClae) AH: = -795 kJ/mol
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Transcribed Image Text:(iii)
Draw the Born-Haber cycle and use the data below for the formation of calcium chloride, to
calculate the electron affinity of chlorine:
Cate)
Ca(g)
AHat = +190 kJ/mol
2+
Ca ()
AHIE = +1730 kJ/mol
Ca@)
2e-
AHat = +121 kJ/mol
Ca (o)
CaCl26)
AHLE = -2184 kJ/mol
2+
%3!
Ca(s) +
Cla)
CaClae)
AHF = -795 kJ/mol
Transcribed Image Text:(i) Match the following enthalpy changes with the reactions below: enthalpy change of
combustion, enthalpy change of formation, enthalpy change of solution, enthalpy change of
hydration, enthalpy change of atomisation, enthalpy change for ionisation, enthalpy change for
electron affinity:
1
3H2 (9) +
O2 (o)
CH:CH2OH () AH =
20 (s) +
2NHa (a)
So? (a)
AH =
+
(NH.)2SO4 (9)
AH =
Na'(g)
Na (ag)
O2 (a)
CO2 (9)
AH =
C (e)
AH =
Na (g)
Na (s)
ΔΗ-
Na'(g)
e
Na (g)
cI (g)
Nacl (s)AH =
Na' (g)
(ii) Identify the different enthalpy changes and construct the Born-Haber cycle for the formation of
sodium choride (NaCI) using the information below. Calculate the missing enthalpy change value
Nacl (s)
AH° = ?
Na'@)
AH° = +121 kJmol"
Cl2 (4)
Cl (g)
AH° = -364 kJmol"
(g)
AH° = +108 kJmol"
Na (s)
Na (a)
AH° = +500 kJmol"
Na (a)
Na' o +
e
Cl2 (g)
NaCl (9)
AH° = -411 kJmol"
Na (e)
+
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