Experiment 8Enthalpy of Solution,Lattice Enthalpy andEnthalpy of Hydration

lycelle espanol. alma pabilane.

Introduction

Introduction (So ano ngayon?)

• Ionic substances > Oppositely charged ions [e.g. NaCl!!]> Ionic bond: Very strong

• Solubility in water> Water: dipolarity [she’s very friendly kay

ionic cpds] ion-dipole interaction> Lattice E, Hydration E, Heat of solution

Introduction (Definition of Terms)

• Enthalpy of solution – heat involved in the solution formation of solute and solvent

• Enthalpy of hydration – amount of heat involved when one mole of an ion is dissolved in large amount of water forming infinite dilute solution

• Lattice enthalpy – heat required to separate a mole of solid (large interaction) into a gas (small interaction) of its ions; released when forming crystals from gaseous ions; crystal energy;

Objectives

• Determine enthalpy of solutes using calorimetric method

• Determine differential heats of solution of solute and solvent

Experimental

Materials and Reagents

• Microcalorimeter• Thermometer• Watch glass• Beaker• Wash Bottle• Syringe• KCl (s)

• KBr (s)• KI (s)• NaCl (s)• Anhydrous CaCl2 (s)

Procedure

calorimeter. determine calorimeter constant. tap and ice water (exp 6)

syringe. 20 mL distilled water to calorimeter. 15 mins. measure temp.

weigh 1.0± 0.01 g of salt in a watch glass. add to calorimeter with water. Measure temp every 5 sec. 10 readings.

Procedure

repeat above procedure for 1.50, 2.0, 2.50, and 3.0 g of salt.

repeat. 18, 16, 14, 12 mL water, 1.0 g of salt.

do everything for each salt: CaCl2, NaCl, KCl, KBr, KIi. varying weight of salt ii. varying volume of water

Experimental Results

Results

WKI

(g)VH2O

(mL)nKI nwater

Ti

(oC)Teq

(oC)Qsoln

(J)

ΔHsoln

(kJ/mol)

1.01 20 0.00608 1.1063 30 28.2 150.10 24.67

1.50 20 0.00904 1.1063 29.3 27.2 175.36 19.41

2.00 20 0.01205 1.1063 28 25.5 208.83 17.33

2.50 20 0.01506 1.1063 27.8 24.2 300.75 19.97

3.00 20 0.01807 1.1063 27 23.7 275.75 15.26

Partial Molar Heat of Solution of KI

Results

WKI

(g)VH2O

(mL)nKI nwater

Ti

(oC)Teq

(oC)Qsoln

(J)

ΔHsoln

(kJ/mol)

1.00 20 0.00624 1.1063 30 28.2 150.63 0.13616

1.00 18 0.00624 0.99565 27.1 26 82.863 0.08323

1.00 16 0.00624 0.88502 27 25.9 73.678 0.08325

1.00 14 0.00624 0.77439 26.5 25 87.943 0.11356

1.00 12 0.00624 0.66376 26 24 100.54 0.15653

Partial Molar Heat of Solution of Water

Results

0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.020

5

10

15

20

25

30

f(x) = − 607.721025321941 x + 26.6571155653826R² = 0.671554933613385

Partial Molar Heat of Solution of KI

Series1 Linear (Series1)

moles of KI

Hea

t of S

oluti

on (k

J/m

ol)

Results

0.6 0.7 0.8 0.9 1 1.1 1.20

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

f(x) = − 0.0642329871969782 x + 0.171393735261018R² = 0.120499040475812

Partial Molar Heat of Solution of Water

Heat of SolutionLinear (Heat of Solution)

moles of water

Hea

t of S

oluti

on (k

J/m

ol)

ResultsSalt CaCl2 NaCl KCl KBr KI

W of salt (g) 1.01 1.0 1.0 1.0 1.01

V H2O (mL) 22.15 20 20 20 20

Moles of salt 9.55x10-3 2.566x10-2 1.34x10-2 8.34x10-2 6.08x10-3

Moles H20 1.229 1.105 1.106 1.11 1.11

Ti (oC) 27 29 30.7 32 30

Teq(oC) 28.5 28.5 28.1 30 28.2

Qsoln (J) -341.987 45.145 277.483 1032 150.628

∆Hsolution

(kJ/mol) -35.8 1.75936 20.71 122.43 24.774

∆Hlattice

(Born-Haber) 2197[7] 788.0 kJ[6] 708.8 kJ[6] 679.5 kJ[6] 893.76 kJ[6]

∆Hlattice

(Kapustinskii) 2567.30 J 849.72 J 753.84 J 719.986 J 671.7229 J

∆Hhydration

(kJ/mol) -33.2 0.85148 0.77455 0.842416 0.696496

Discussion

Discussion

• IONIC COMPOUNDS> have very large attractive forces

between oppositely charged ions in crystal lattice

> high MP and BP> difficult to separate> e.g. ionic salts: NaCl, KCl, KI

Discussion

• SOLUBILITY> a characteristic property of most ionic compounds

• SOLUTION> ions in solution are nearly

independent of one another

Discussion

• Then, how can ionic compounds dissolve at room temperature?

• This is due to interactions between the ions and the molecules of water. (or other polar solvents)

> ion-dipole force!

Discussion

• WATER is dipolar

> Negative (O) ends of water are strongly attracted to (+) ions> Positive (H) ends of water are strongly attracted to (-) ions

• As a result, water molecules cluster around an ion within their vicinity

• HYDRATION

Discussion

From:http://chemed.chem.wisc.edu/chempaths/GenChem-Textbook/Hydration-of-Ions-943.html

Discussion

• Process of Solution (E Solution) (i)> 1st: crystalline salt gaseous ions

(LATTICE E) (ii)> 2nd: water molecules surround the ions

(HYDRATION E) (iii)

> ΔHsoln = ΔHlattice + ΔHhydration

* NOTE : HESS’sss Law

Discussion

ΔG = ΔH - TΔS• SPONTANEOUS

> (-) ΔG (Change in Gibb’s free E)> (-) ΔH (Enthalpy Change) exothermic> (+) ΔS (Entropy Change)> high T (Temperature)

Discussion

• ENTROPY S> S - disorderliness - number of possible arrangements> solid < liquid < gas> usually drives endothermic reaction tobecome spontaneous

Discussion

• Entropy> increases when crystalline salt is

converted to gaseous state ions more disordered

> decreases during hydration water molecules’ arrangement become limited

> to achieve – ΔG , entropy INCREASE in forming gaseous ions MUST outweigh

entropy DECREASE due to hydration

Heat of Solution for Common Salts

cation\anion OH- F- Cl- Br- I- CO32

- NO3- SO4

2-

Li+ -21.2 +4.5 -37.2 -49.1 -63.3 -17.6 -2.7 -30.2

Na+ -42.7 +0.3 +3.9 -0.6 -7.6 -24.6 +20.5 -2.3

K+ -55.2 -17.7 +17.2 +20.0 +20.5 -32.6 +34.9 +23.8

NH4+ - +5.0 +15.2 +16.2 +13.4 - +25.8 +6.2

Mg2+ +2.8 -17.7 -155 -186 -214 -25.3 -85.5 -91.2

Ca2+ -16.2 +13.4 -82.9 -110 -120 -12.3 -18.9 -17.8

Sr2+ -46.0 +10.9 -52.0 -71.6 -90.4 -3.4 +17.7 -8.7

Al3+ - -209 -332 -360 -378 - - -318

Born-Haber Cycle

• The Born-Haber thermochemical cycle is named after the two German physical chemists, Max Born and Fritz Haber, who first used it in 1919.

• formation of an ionic compound from the reaction of an alkali metal (Li, Na, K, Rb, Cs) with a gaseous halogen (F2, Cl2)

• used to calculate Lattice E (in the expt)

Born-Haber Cycle

Born-Haber Cycle

• ΔHsublimation - heat absorbed when 1 mol of atom is vaporized

• ΔHionization (IE) - remove an electron from an isolated atom or molecule

• ΔHdissociation (BDE) - energy involved in bond cleavage

• ΔHelectron affinity (EA) - energy released when an electron is added to a neutral atom or molecule to form a negative ion

• ΔHlattice - energy absorbed when 1 mol crystal is formed

• ΔHformation - energy absorbed when 1 mol of atoms is formed

Kapustinskii equation

• Anatolii Fedorovich Kapustinskii (1906-1960)> Russian. Physical chemist. InorgChem.

• to calculate:> Lattice Energy [difficult to determine experimentally] > ionic radii [given the lattice energy]

• generalized form of the Born-Lande equation [averaged Madelung constant] , does not depend on structureSEE YA REAL SOON!

Calculations

Formulas Used1. moles: n = (mass)(1/MW) = (V)(ρ)(1/MW)

2. Qcal + Qtap + Qcal = 0

3. Qcal = Ccal∆T

4. Qwater = (VH2O)(ρH2O)(spHeat) (4.184 J/goC)(∆T)

5. ΔHsoln = Qsoln + Qwater

6. Qcal + Qsoln + Qwater = 0

7. ΔHsoln = Qsoln /n

8. Kapustinskii (next slide)9. ∆Hsublimation + ∆Hionization + ∆HI-I bond enthalpy +

∆Helectron affinity – ∆Hformation - ∆Hlattice = 0

10. ∆Hhydration = ∆Hsoln - ∆Hlattice

Formulas Used

Kapustinskii Equation• K= • d = 3.45E-11 m• V = number of ions in empirical formula• z = charge of cation/anion• r = ionic radius

Calorimeter Constant

From equations 2 and 3, we get,Ccal = - [(Qcold + Qtap )]/(Ti,tap – Teq)

= {-[(10mL)(0.9997g/mL)(20-10oC)(4.184J/goC) + (10mL)(0.995646g/mL)(20-30oC)(4.184J/goC)]}

(20-30oC)= 0.16961936 J/oC

Sample Calculations

• Partial Molar ΔHsoln of KI (20mL, 1.50 g)

nKI = 1.50 g(1mol KI/166.0028 g) = 0.009036 mol

nH2O= [20mL(0.996175g/mL)] = 1.10686 mol

(18g/mol)Qcal =(0.16961936 J/oC)(27.2–29.3oC) = - 0.3562 J

Qwater=(20mL)(0.99586g/mL)(4.184J/goC)(27.2–29.3oC)

= - 174.99961 JQsoln = -[(-0.3562) + (-174.99961)] = 175.36 J

ΔHsoln = 175.36 J/0.009036 mols = 19.41 kJ/mol

Sample Calculations

• Partial Molar ΔHsoln of water(20mL, 1.0 g)

nKI = 1.0 g(1mol KI/166.0028 g) = 0.006024mol

nH2O= [20mL(0.996175g/mL)] = 1.10686 mol

(18g/mol)Qcal =(0.16961936 J/oC)(28.2 – 30oC) = - 0.3053 J

Qwater=(20mL)(0.995646g/mL)(4.184J/goC)(28.2–30oC)

= - 149.96818 JQsoln = -[(-0.3562) + (-174.99961)] = 150.27 J

ΔHsoln = 150.27 J/1.10686 mols = 0.136 kJ/mol

Sample Calculations

• The Born-Haber cycle for ∆Hlattice

K(s) → K(g) ∆Hsublimation = +89.24 kJ

K(g) → K+ (g)+e- ∆Hionization = +418.8 kJ

0.5 [I2(g) → 2I(g)] ∆HI-I bond enthal = +151(0.5) kJ

0.5[2I(g)+2e- → 2I-(g)] ∆He affinity = -295.16(0.5) kJ

K+(g)+ I-

(g) → KI(s) -∆Hlattice

KI(s) → K(s) + I2(g) ∆Hformation, KI(s) = -457.8 kJ

Sample Calculations

∆Hlattice = ∆Hsublimation + ∆Hionization + ∆HI-I bond enthalpy

+∆Helectron affinity – ∆Hformation, KI(s)

= [89.24 + 418.8 + (151)(0.5) + (-295.16)(0.5)] – (-457.8) kJ= 893.76 kJ

-∆Hlattice = -893.76 kJ

Sample Calculations

• The Kapustinskii Equation for ∆Hlattice

∆Hlattice =

= -671.7224 J

∆Hhydration of KI

∆Hhydration = ∆Hsolution - ∆Hlattice

= 24.774 – (-671.7229) J= 696.496 J

ConclusionRecommend

ations

Every dipole has its moment.

Conclusion and Recommendations

• Calorimetry as a method of determining the enthalpy of solution– Relatively effective– Error. Error. ERROR!

• ∆Hsolution = nsolute∆Hintegral,solute + nsolvent∆Hintegral,solvent

• ∆Hhydration + ∆Hlattice = ∆Hsolution

Conclusion and Recommendations

• Kapustinskii equation > Relative precision > assumption: does not consider structure (for C157 and InorgChem)*

• Born-Haber cycle> prone to error: use of standard values

Useful Tips for the Exam

• charge of ion (z), lattice energy• atomic radius of ion (r), lattice energy

* Remember the Kapustinskii equation, z is in numerator and r is in denominator.

> Know your periodic table trends: ionic radii, ionization energy, electron affinity etc.> Careful on signs of delta H’s etc.

Some Useful Periodic Table Trends

• Atomic Radius: LR decreasing &Metallic prop TB increasing• Ionization E: LR increasing(remove an electron) TB decreasing• Electron Affinity: Group 2: low(accepting electron) Group 7: high

Noble Gases: 0• E negativity: LR increasing(attraction to TB decreasingform a bond) *direction of arrow : INCREASING

Q and A Portion

Q: How many physical chemists does it take to

change a light bulb?

A:Only one, but he'll change it three times,

plot a straight line through the data, and

then extrapolate to zero concentration.

Questions and Answers

1. How is the heat of solution of KI affected by (a) number of moles of KI (b) number of moles H2O? Explain the significance of each observation. > moles of KI, heat of solution

> more energy is required to separate ions on the lattice.

> heat of solution until a certain # of moles of water then .

> moles water, solvation of KI until all the solute are dissolved.

> moles water, heat of solution

Questions and Answers

2. Explain the significance of the partial molar heats of solution of solute and solvent.

> partial molar heat of solution of the solute/solvent is the energy required to break solute-solute /solvent-solvent interactions.

> PMHS of solute, solute-solute interactions, solubility

> PMHS of solvent, solvent-solvent interactions, the more difficult it is for a solvent to solvate a given solute.

Questions and Answers

3. Plot cation radius against lattice enthalpy for CaCl2 NaCl and KCl. Explain the observed trend.

500 1000 1500 2000 2500 30000

20

40

60

80

100

120

140

160

f(x) = − 0.0121317857351884 x + 144.199993283823R² = 0.335119889696518

Lattice Enthalpy

Catio

n ra

dius

Questions and Answers

• atomic radius, lattice enthalpy . • longer radius

> forms a longer, weaker bond>thus the energy required to break the bond would be less compared to the

energy required to break a shorter bond (stronger bond).

Questions and Answers

4. Plot anion radius against lattice enthalpy for KCl, KBr and KI. Explain the observed trend.

165 170 175 180 185 190 195 200 205 2100

100

200

300

400

500

600

700

800

900

f(x) = − 3.73737984496124 x + 1445.90260465116R² = 0.546130007000808

Lattice Enthalpy

Anio

n ra

dius

Questions and Answers

• As anion radius increases,> Lattice E decreases> Larger anion is more polarisable and

forms weaker bonds than small anions and would require less energy to break, hence the decrease in lattice enthalpy.

Thank you for listening!Good luck! <3 <3 <3

References

1. http://www.rod.beavon.clara.net/solubility.htm2. http://

chemistry.tutorvista.com/inorganic-chemistry/reaction-pathways-born-haber-cycle.html

3. http://www.slideshare.net/janetra/6-the-born-haber-cycle4. http://

chemed.chem.wisc.edu/chempaths/GenChem-Textbook/Hydration-of-Ions-943.html

5. http://www.docbrown.info/page07/delta2H.htm6. http://www.science.uwaterloo.ca/~

cchieh/cact/applychem/lattice.html7. http://chemistry.bd.psu.edu/jircitano/BH.html8. http://chemistry.about.com/b/2007/03/24/periodic-table-t

rends.htm