Tutorial Week 6 Answers 2018 PDF

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 Tutorial Week 6 Reactions and Synthesis: Thermodynamics of Inorganic Reactions

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1. The calculated enthalpy change associated with the reaction: Ca(s) + F2(g)

Ca2+(g) + 2F-(g)

is +1415 kJmol-1 Given that the reaction to form Ca2+ ions and F- ions is highly endothermic, why does calcium react with fluorine to form the ionic compound CaF2(s)?

The process: Ca2+(g) + 2F-(g)

CaF2(s) is highly exothermic, -D฀LHo = -2643 kJ

2. The value for the Madelung constant depends upon the type of packing in the structure. a) How is the Madelung constant calculated? b) Does a high Madelung constant lead to a greater or lower lattice energy? Why?

a) The Madelung constant is calculated by considering the attractive and repulsive interactions operating in an ionic solid between an ion and the successive shells of ions that surround it. The Madelung constant (A) depends upon the

6 Xz% ions*at*a*distance*d""(1) 12 Mz+ ions*at*a*distance*√2*d (2) 8 Xz% ions*at*a*distance*√3*d"(3) 6 M z+ ions*at*a*distance*2*d …"etc

0"to"1"attractive 0 to 2 repulsive 0"to"3"attractive etc.

Refer to lecture slides for details.

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3 2 3

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b) The calculated lattice energy is proportional to the Madelung constant in both the Born-Landé and and Born-Mayer equations. The higher the value of the Madelung constant the greater the lattice energy.

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3. The Born-Landé and Born-Mayer equations take into account the contribution of Born forces to the lattice enthalpy in addition to the Coulombic interactions. What is the origin of the Born forces?

The fact that ions are not infinitesimally small means that the lattice energy cannot be satisfactorily estimated from a simple Coulombic-type expression. Born forces arise from electron-electron and nucleus-nucleus repulsions between ions that have a finite size (i.e. not point charges).

4. The Kapustinskii equation allows the estimation of the lattice enthalpy but unlike the Born-Landé and Born-Mayer equations it does not include the Madelung constant. a) Explain why the Kapustinskii equation does not include the Madelung constant. b) Would you expect the Born-Mayer equation or the Kapustinskii equation to give a more accurate estimation of the lattice enthalpy?

a) Kapustinskii noted that if the various Madelung constants, obtained for different types of structures, are divided by the number of ions per formula unit then a similar value is obtained i.e. A/Nion = c where c is a constant, therefore: A = Nionc. We can use Nion, multiplied by an appropriate constant (c), in place of the Madelung constant, A.

The Kapustinskii equation was developed by proposing a hypothetical rock-salt (NaCl-type) that is energetically equivalent to the true structure of any ionic solid. The equation is based upon a rock-salt Madelung constant. The k฀ term of the Kapustinskii equation, D฀LHo = Nion(½฀zAzB½฀/d)(1 – d*/d)k฀, includes all the contributing constants.

b) The Born-Mayer equation makes fewer approximations than the Kapustinskii equation but there are significant assumptions and approximations in both. The Born-Mayer should be more accurate but sometimes different factors negate each other resulting in a better agreement between the experimental value and the value estimated using the Kapustinskii equation.

 5. BaCl2 crystallises in two crystalline forms one of which has the fluorite (CaF2) structure.

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a) Using the Born-Haber cycle calculate the lattice enthalpy. b) Using the Born-Landé equation estimate the lattice enthalpy. c) Using the Born-Mayer equation estimate the lattice enthalpy. d) Using the Kapustinskii equation estimate the lattice enthalpy. a) D฀fHo(BaCl2) = D฀atomHo(Ba) + D฀disHo (Cl2) + D฀ion1stHo (Ba) + D฀ion2ndHo (Ba) + 2D฀egHo (Cl)+[-D฀LHo(BaCl2)] Rearranging the equation: D฀LHo(BaCl2) = D฀atomHo(Ba) + D฀disHo (Cl2) + D฀ion1stHo (Ba) + D฀ion2ndHo (Ba) + 2D฀egHo (Cl) - D฀fHo(BaCl2) = 180 + 244 + 509 + 971 -710 - (-859) = 2053 kJ Note that for parts (b-d), d is equal to the sum of the radii of the Ba2+ and Cl- ions. For Born-Lande and Born- Mayer we use 4-coordinate Cl- and 8-coordinate Ba2+. We do this because we know it has the fluorite structure; the value of the radius we use depends upon the coordination number. The calculation of D฀LHo using the Kapustinskii equation does not require any knowledge of the structure, therefore we always use Note that in the Kapustinskii equation the value of the constant k฀ requires units of pm for d.

b) Born-Landé equation: D฀LHo = NAA(½฀zAzB½฀e2/4p฀e฀od)(1 – 1/n) D฀LHo = 6.022 x 1023 x 2.519 x 2 x (1.602 x 10-19)2 /4 x 3.142 x 8.854 x 10-12 x 317 x 10-12)(1 – 1/10) = 2208 x 0.90 = 1987 kJ

c) Born-Mayer equation D฀LHo = NAA(½฀zAzB½฀e2/4p฀e฀od)(1 – d*/d) D฀LHo = 6.022 x 1023 x 2.519 x 2 (1.602 x 10-19)2/(4 x 3.142 x 8.854 x 10-12 x 317 x 10-12) x (1 34.5 x 10-12/317 x 10-12) = 2207 x (1- 0.1088) = 1967 kJ

d) Kapustinskii equation (note the use of 6-coordinate radii here) D฀LHo = Nion(½฀zAzB½฀/d)(1 – d*/d)k฀ = (3 x 2/316)(1 – 34.5/316) x 1.21 x 105 = 2046 kJ

 6. In regards to the Ellingham Diagram below answer the following questions.

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a) What chemical reactions are represented by the lines in the diagram? b) What does the gradient of each line represent? c) Why do some lines show discontinuities? d) Is carbon capable of reducing silica? If so, under what conditions? e) At what temperature will carbon reduce NiO? What will be the product of the oxidation of carbon? f) Is Li capable of reducing Al2O3? If so, under what conditions?

a) The reaction of a metal with one half of a mole of O2 to form the metal oxide. Each line is a plot of the relationship D฀G = D฀H - TD฀S for this reaction. b) –D฀S for the reaction c) Above the boiling point of the metal, the reaction represented by the graph involves two gaseous reactants as opposed to only one gaseous reactant below the boiling point. When two gaseous reactants, as opposed to one, combine to give a solid, the entropy change becomes a more negative value, which leads to a steeper gradient. d) Yes, at a temperature above 1720 K. e) 420 K, CO2 f) Consider the two relevant reactions indicated in the diagram 2Li + 1/2O2

Li2O

D฀G (Li2O)

(1)

Al + 1/2O2

1/3Al2O3

D฀G (Al2O3)

(2)

In order to obtain the reaction we are interested in, we need to reverse the second reaction and add the two reactions 2Li + 1/2O2

Li2O

D฀G (Li2O)

(1)

1/3Al2O3

Al + 1/2O2

-D฀G (Al2O3)

-(2)

Overall 2Li + 1/3Al2O3

Li2O + Al D฀rG = D฀G (Li2O) – D฀G (Al2O3)

The reaction may proceed when D฀rG is negative i.e. when D฀G (Li2O) < D฀rG (Al2O3)) From the graph this is at T < 1770 K. Simple inspection of the graph makes it possible to identify the more stable metal oxide.

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 Data and equations for Tutorial 1

D฀atom Ho(Ba) = +180 kJ mol-1; D฀disHo (Cl2) = +244 kJ mol-1; D฀ion1stHo (Ba) = 509 kJ mol-1; D฀ion2ndHo (Ba) = 971 kJ mol-1; D฀egHo (Cl) = -355 kJ mol-1; D฀fHo(BaCl2) = -859 kJ mol-1. NA = 6.022 x 1023; e = 1.602 x 10-19 C; e฀o = 8.854 x 10-12 J-1C2m-1; d* = 34.5 pm; A (CaF2) = 2.519, Born exponent values, n: Ba2+ 12, Cl- 9; ionic radius of Ba2+: 135 (6-cordinate), 142 (8-coordinate) pm; radius Cl-: 181 (6-coordinate), 175 (4-coordinate) pm; k฀ = 1.21 x 105 kJ pm mol-1. Born-Landé equation: D฀LHo = NAA(½฀zAzB½฀e2/4p฀e฀od)(1 – 1/n) Born-Mayer equation: D฀LHo = NAA(½฀zAzB½฀e2/4p฀e฀o d)(1 – d*/d) Kapustinskii equation: D฀LHo = Nion(½฀zAzB½฀/d)(1 – d*/d)k฀ Radius of Mg2+ 72 pm (6–coordinate), Radius of Ba2+ 135 pm (6–coordinate) Thermochemical radius of CO32- 178 pm. Ellingham Diagram

CourtesyofDrSBest

 Supplementary Questions

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S1. Provide a brief description of the structures of NaCl, zinc blende, wurtzite, CsCl and CaF2 in terms of the close packing of ions and the occupation of interstitial sites. What factors affect the type of packing that is found?

NaCl: FCC (CCP) arrangement of Cl- ions with Na+ in all octahedral sites. Zinc blende (ZnS): FCC (CCP) arrangement of S 2- ions with Zn2+ in half the tetrahedral sites. Wurtzite (ZnS): HCP arrangement of S2- ions with Zn2+ in half the tetrahedral sites. CsCl: Simple cubic arrangement of Cl- ions with Cs+ in all cubic sites. CaF2: Simple cubic arrangement of F - ions with Ca2+ in half the cubic sites OR FCC arrangement of Ca2+ with F- in all the tetrahedral sites.

Important factors affecting the type of structure adopted are i) the cation to anion ratio ii) the radius ratio (cation:anion) iii) geometric preferences of the cations and anions

S2. Crystalline ionic solids represent examples of highly ordered materials. i) What is the driving force for the formation of an ionic solid? ii) What factors impact upon the stability of an ionic solid?

i) Whilst the formation of ions from elements normally has a net energy cost associated with it, an ionic solid in which cations are surrounded by anions and vice versa is a very stable arrangement. The process in which ions are combined to form a solid is highly exothermic. ii) Factors impacting upon the stability of an ionic compound include: • The arrangement of the ions • The charge on the ions • The size of the ions

7  S3. The lattice enthalpies of MgCO3 and BaCO 3 may be estimated using the Kapustinskii equation. The thermochemical radius of the carbonate anion is estimated as 178 pm. The 6-coordinate radii of Mg2+ and Ba2+ are 72 and 135 pm respectively.

a) What is a thermochemical radius? b) Using the Kapustinskii equation estimate the lattice enthalpy of MgCO3 and BaCO 3. c) Would you expect MgCO3 or BaCO3 to be more stable with respect to decomposition? Why?

a) A thermochemical radius is a radius assigned to a non-spherical molecular ion such as the nitrate ion. It is possible to calculate a radius known as the thermochemical radius for the non-spherical ion, which may then be used in calculations involving other systems that include the same ion. b) D฀LHo = Nion(½฀zAzB½฀/d)(1 – d*/d)k฀ For MgCO3 = (2 x 4/250) (1 – 34.5/250) x 1.21 x 105 = 3338 kJ For BaCO3 = (2 x 4/313) (1 – 34.5/313) x 1.21 x 105 = 2752 kJ c) The decomposition reaction is MCO3(s)

MO(s) + CO2 (g)

The difference in lattice enthalpy between MgO and MgCO3 is 457 kJ The difference in lattice enthalpy between BaO and BaCO3 is 277 kJ The high stability of MgO relative to MgCO3 is a driving force for the decomposition of MgCO3 at a relatively low temperature. There is a relatively small difference in energy between the lattice energies for BaCO3 and BaO which means an enhanced stability for BaCO3 with respect to decomposition. BaCO3 would be more stable to decomposition than MgCO3.

8  S4. For the following compounds indicate whether they would be considered as a basic oxide, an acidic

oxide or an amphoteric oxide. SO2, BaO, Li2O, Ga2O3 For the acidic oxides and basic oxides give equations that illustrate their acidic or basic behaviour. For the amphoteric oxides give equations that illustrate both acidic and basic behaviour.

SO2 – acidic oxide SO2(g) + H2O(l)

H2SO3(aq) Gives an acidic solution when mixed with water.

SO2(g) + HO-

HSO3-(aq) Reacts with a base.

BaO – basic oxide BaO(s) + H2O(l)

Ba2+ + 2HO-(aq) Gives a basic solution when mixed with water.

BaO(s) + 2H3O+(aq)

Ba2+ + 3H2O(l) Reacts with an acid

Li2O – basic oxide Li2O(s) + H2O(l)

2Li+(aq) + 2HO-(aq) Gives a basic solution when mixed with water.

Li2O(s) + 2H3O+(aq)

2Li+(aq) + 3H2O(l) Reacts with an acid

Ga2O3 – amphoteric oxide Ga2O3(s) + 6H3O+(aq) + 3H2O(l)

2[Ga(H2O)6]3+ (aq) Reacts with an acid

Ga2O3(s) + 2OH-(aq) + 3H2O(l)

2[Ga(OH)4 ]-(aq) Reacts with a base...