# c20 - Bond Energy calcuation Born Haber cycle and Lattice...

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Unformatted text preview: Bond Energy calcuation; Born Haber cycle and Lattice Energy and Polarity The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond energy. H2 (g) Cl2 (g) HCl (g) O2 ( g ) N2 (g) H (g) + H (g) Bond Energy ∆ H0 = 436.4 kJ Cl (g) + Cl (g) ∆ H0 = 242.7 kJ H (g) + Cl (g) ∆ H0 = 431.9 kJ O (g) + O (g) ∆ H0 = 498.7 kJ N (g) + N (g) ∆ H0 = 941.4 kJ Bond Energies Single bond < Double bond < Triple bond 9.10 O N O N Average bond energy in polyatomic molecules H2O (g) OH (g) H (g) + OH (g) ∆ H0 = 502 kJ H (g) + O (g) ∆ H0 = 427 kJ 502 + 427 = 464 kJ Average OH bond energy = 2 9.10 Bond Energies (BE) and Enthalpy changes in reactions Imagine reaction proceeding by breaking all bonds in the reactants and then using the gaseous atoms to form all the bonds in the products. ∆ H0 = total energy input – total energy released = Σ BE(reactants) – Σ BE(products) 9.10 H2 (g) + Cl2 (g) 2HCl (g) 2H2 (g) + O2 (g) 2H2O (g) 9.10 Use bond energies to calculate the enthalpy change for: H2 (g) + F2 (g) 2HF (g) ∆ H0 = Σ BE(reactants) – Σ BE(products) Type of bonds broken Number of bonds broken Bond energy (kJ/mol) Energy change (kJ) H F H F 1 1 Number of bonds formed 436.4 156.9 Bond energy (kJ/mol) 436.4 156.9 Energy change (kJ) Type of bonds formed H F 2 568.2 1136.4 ∆ H0 = 436.4 + 156.9 – 2 x 568.2 = -543.1 kJ 9.10 Bond Energy • ΔH rxn = Σ bonds broke – Σ bonds made Reactants Products CH4 + 2O2 CO2 + 2H2O Bonds broken: 4mol C-H get this correct! 4 x 413 kJ/mol 2mol O=O 2 x 498 kJ/mol Total = 2648 kJ/mol Bond Energy • ΔH rxn = Σ bonds broke – Σ bonds made CH4 + 2O2 CO2 + 2H2O Bonds made: 2mol C=O get this correct! 2 x 745 kJ/mol 4mol O-H 4 x 463 kJ/mol Total = 3342 kJ/mol Bond Energy • ΔH rxn = Σ bonds broke – Σ bonds made CH4 + 2O2 CO2 + 2H2O ΔH rxn = 2648 kJ – 3342 kJ = - 694 kJ Bond Energy • ΔH rxn = Σ bonds broke – Σ bonds made CH4 + 2O2 CO2 + 2H2O ΔH rxn = 2648 kJ – 3342 kJ = - 694 kJ Two ways to the these types of problems: 1. Break all bonds and then make all bonds 2. Break only those that change and make only those that change Bond Energy Two ways to the these types of problems: 1. Break all bonds and then make all bonds. That is what we just did. 2. Break only those that change and make only those that change. I do not teach the second way because it is far to easy to make a mistake using this method! Electrostatic (Lattice) Energy Lattice energy (E) is the energy required to completely separate one mole of a solid ionic compound into gaseous ions. Q+ is the charge on the cation Q- is the charge on the anion r is the distance between the ions cmpd MgF2 MgO LiF LiCl lattice energy 2957 Q= +2,-1 3938 Q= +2,-2 1036 853 - E = k Q+ Qr Lattice energy (E) increases as Q increases and/or as r decreases. 9r F < r C.3 l Born-Haber Cycle for Determining Lattice Energy o ∆ Ho = ∆ H1 + ∆ Ho + ∆ Ho + ∆ Ho + ∆ Ho overall 2 3 4 5 9.3 9.3 Born Haber Cycle • Li(s) Li(g) Δ Hsub =155.2 kJ/mol • 1/2F2(g) F(g) Δ HBE = 75.3 kJ/mol • Li(g) Li+1(g) + e- Δ H IE = 520 kJ/mol • F(g) + e- F-1(g) Δ H EA = -328 kJ/mol • Li+1(g) + F-1(g) LiF(g) = Δ H –LE ? • Li(s) + 1/2F(g) LiF(s) Δ Hoverall = -594.1 kJ/mol Born Haber Cycle • Li(s) Li(g) Δ Hsub =155.2 kJ/mol • 1/2F2(g) F(g) Δ HBE = 75.3 kJ/mol • Li(g) Li+1(g) + e- Δ H IE = 520 kJ/mol • F(g) + e- F-1(g) Δ H EA = -328 kJ/mol • Li+1(g) + F-1(g) LiF(g) = Δ H –LE ? • Li(s) + 1/2F(g) LiF(s) Δ Hoverall = -594.1 kJ/mol • ? = -1017 kJ/mol Lattice energy for LiF Born Haber Cycle • Li(s) Li(g) Δ Hsub =155.2 kJ/mol • 1/2F2(g) F(g) Δ HBE = 75.3 kJ/mol • Li(g) Li+1(g) + e- Δ H IE = 520 kJ/mol • F(g) + e- F-1(g) Δ H EA = -328 kJ/mol • Li+1(g) + F-1(g) LiF(g) = Δ H –LE ? • Li(s) + 1/2F(g) LiF(s) Δ Hoverall = -594.1 kJ/mol • ? = -1017 kJ/mol Lattice energy for LiF overall or f • Molecular Polarity Molecular Water Boiling point Boiling = 100 ˚C 100 Methane Boiling point Boiling = -161 ˚C -161 Why do water and Why methane differ so much in their boiling points? points? Why do ionic compounds dissolve in water? Bond Polarity Bond Polarity +δ -δ HCl is POLAR because HCl POLAR it has a positive end and a negative end. and Cl has a greater share Cl in bonding electrons than does H. than H Cl• • •• •• Cl has slight negative charge (-δ ) and H Cl (has slight positive charge (+ δ ) Bond Polarity Bond • Three molecules with Three polar, covalent bonds. polar, • Each bond has one atom Each with a slight negative charge (-δ ) and and (another with a slight positive charge (+ δ ) positive Bond Polarity Bond Polarity This model, calc’d using CAChe software for molecular calculations, shows that H is + (red) and Cl is - (yellow). Calc’d charge is + or - 0.20. +δ -δ H Cl •• •• • • Bond Polarity Due to the bond polarity, the H Due —Cl bond energy is —Cl GREATER than expected for a “pure” covalent bond. “pure” BOND ENERGY “pure” bond 339 kJ/mol calc’d real bond 432 kJ/mol 432 Difference = 92 kJ. This difference is Difference measured mproportional to the difference in easured ELECTRONEGATIVITY, χ . ELECTRONEGATIVITY proportional to the difference in Electronegativity, χ χ is a measure of the ability of an atom in a molecule to attract electrons to itself. electrons Concept proposed by Linus Pauling 1901-1994 Linus Pauling, 1901Linus 1994 The only person to receive two unshared The Nobel prizes (for Peace and Chemistry). Chemistry areas: bonding, electronegativity, protein structure protein Electronegativity Electronegativity Figure 9.14 4 3.5 3 2.5 2 1.5 1 0.5 0 O N C H F Cl S Si P Electronegativity, χ See Figure 9.14 • F has maximum χ. has • Atom with lowest χ is the center Atom atom in most molecules. atom • Relative values of χ determine Relative BOND POLARITY (and point of BOND attack on a molecule). attack 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Bond Polarity Bond Polarity Which bond is more polar (or Which DIPOLAR)? DIPOLAR)? O—H O—F O—H χ 3.5 - 2.1 3.5 - 4.0 ∆ 1.4 0.5 OH is more polar than OF OH +δ H +δ O -δ F -δ O and polarity is “reversed.” Molecules—such as HI and Molecules—such H2O— can be POLAR (or O— POLAR dipolar). Molecular Polarity Molecular Polarity They have a DIPOLE MOMENT. The polar HCl The DIPOLE molecule will turn to align with an electric field. molecule Molecular Polarity Molecular The magnitude of The the dipole is given in Debye units. Named for Peter Debye (1884 1966). Rec’d 1936 Nobel prize for work on x-ray Dipole Moments Dipole Moments Why are some molecules polar Why but others are not? but Molecular Polarity Molecular Molecules will be polar if a) bonds are polar AND AND b) the molecule is NOT “symmetric” All above are NOT polar All NOT Polar or Nonpolar? PCO and H O. Which one is polar? olar or Nonpolar? Compare 2 2 • • O •• C Carbon Dioxide Carbon Dioxide O •• • • • CO2 is NOT polar even though the CO bonds are polar. bonds • CO2 is symmetrical. Positive C Positive atom is reason CO2 and H2O CO react to give H2CO3 -0.75 +1.5 -0.75 Consequences of H2O Consequences of H Polarity Polar or Nonpolar? Polar or Nonpolar? • Consider AB3 molecules: BF3, Cl2CO, and NH3. Molecular Polarity, BF3 Molecular F B F F B atom is atom positive and F atoms are negative. negative. B—F bonds in BF3 are polar. But molecule is symmetrical and NOT polar NOT Molecular Polarity, HBF2 Molecular H B F F B atom is atom positive but H & F atoms are negative. are B—F and B—H bonds in HBF2 are polar. But molecule is NOT symmetrical and is polar. symmetrical Is Methane, CH4, Is Polar? Polar? Methane is symmetrical and is NOT Methane polar. polar. Is CH3F Polar? Is CH C—F bond is very polar. C—F Molecule is not symmetrical and so is polar. and CH4 … CCl4 CH Polar or Not? • Only CH4 and CCl4 are NOT polar. These are the only two molecules that are “symmetrical.” Substituted Ethylene Substituted • C—F bonds are MUCH more polar than C—F C—H bonds. • Because both C—F bonds are on same Because side of molecule, molecule is POLAR. POLAR Substituted Ethylene Substituted • C—F bonds are MUCH more polar than C C—F —H bonds. —H • Because both C—F bonds are on opposing Because ends of molecule, molecule is NOT POLAR. POLAR CHEMICAL BONDING BONDING Cocaine • Use the Born-Haber cycle to calculate the lattice energy of LiCl(s) given the following data: ∆ H(sublimation) Li = 155.2 kJ/mol I1 (Li) = 520 kJ/mol Bond energy (Cl-Cl) = 242.8 kJ/mol EA (Cl) = - 348 kJ/mol ∆ Hf (LiCl(s)) = - 408.8 kJ/mol • • • • • A. B. C. D. E. -40 kJ/mol 40 kJ/mol 736 kJ/mol 857 kJ/mol 1,553 kJ/mol • Use the Born-Haber cycle to calculate the lattice energy of LiCl(s) given the following data: ∆ H(sublimation) Li = 155.2 kJ/mol I1 (Li) = 520 kJ/mol Bond energy (Cl-Cl) = 242.8 kJ/mol EA (Cl) = - 348 kJ/mol ∆ Hf (LiCl(s)) = - 408.8 kJ/mol • • • • • A. B. C. D. E. -40 kJ/mol 40 kJ/mol 736 kJ/mol 857 kJ/mol 1,553 kJ/mol ...
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## This note was uploaded on 02/10/2011 for the course CHM 2046 taught by Professor Veige/martin during the Spring '07 term at University of Florida.

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