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Thermochemistry Assignment Solutions for Chem 612 - Fall, 2008, Study notes of Chemistry

The solutions to Assignment 1 for Chem 612, a university course offered in the Fall of 2008. The assignment involves calculating thermochemical properties of MgCl2, including enthalpies of formation, sublimation, ionization, dissociation, and electron affinity, as well as the lattice energy. The document also discusses the differences between ionic and polar covalent bonds, and provides Lewis structures for various anions and molecules.

What you will learn

  • What are the Lewis structures for the cyanate and fulminate ions, and why is the fulminate ion less stable?
  • Why is the difference in electronegativity an unreliable criterion for determining the nature of a bond?
  • What are the thermochemical properties of MgCl2?

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Chem 612 โ€“ Fall, 2008
Assignment 1 - Solutions
This assignment is due at the beginning of class on Wednesday, September 10, 2008.
1. The standard enthalpy of formation, โˆ†Hof, is โ€“642 kJ/mol for MgCl2. The enthalpy of
sublimation of Mg(s), โˆ†Hosub, is 146 kJ/mol, and the first and second ionization energies, I1
and I2, of Mg(g) are 737.7 kJ/mol and 1451 kJ/mol, respectively. The dissociation energy of
Cl2, D, is 243.4 kJ/mol, and the electron affinity of Cl(g) is โ€“349 kJ/mol. Given these data and
additional data below, give answers to the following.
a. Write thermochemical equations, including state designations, corresponding to all of the
data given above and also the lattice energy, U.
Mg(s) + Cl2(g) 6MgCl2(s)โˆ†Hof = โ€“642 kJ/mol
Mg(s) 6 Mg(g)โˆ†Hosub = 146 kJ/mol
Mg(g) 6 Mg+(g) + eโ€“I1 = 737.7 kJ/mol
Mg+(g) 6Mg2+(g) + eโ€“I2 = 1451 kJ/mol
Cl2(g) 6 2 Cl(g)D = 243.4 kJ/mol
Cl(g) + eโ€“ 6 Clโ€“(g)A = โ€“349 kJ/mol
Mg2+(g) + 2 Clโ€“(g) 6 MgCl2(s)U = 2522 kJ/mol
b. Sketch the Born-Haber cycle for MgCl2(s), similar to the cycle shown in class for NaCl(s).
Mg2+(g) + 2 Cl-(g)
MgCl2(s)Mg(s) + Cl2(g)
โˆ†Hosub D
Mg(g) + 2 Cl(g)
I1 + I22A
โˆ†Hof
-U
c. Calculate the lattice energy, U, for MgCl2(s) to verify the value of 2522 kJ/mol given in
the table shown in class. Show your work.
โˆ†Hof = โˆ†Hosub + I1 + I2 + D + 2A โ€“U
U = โˆ†Hosub + I1 + I2 + D + 2A โ€“ โˆ†Hosub โ€“ โˆ†Hof
= [146 + 737.7 +1451 + 243.4 + (2)(โ€“349) + 642]kJ/mol = 2522 kJ/mol
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Download Thermochemistry Assignment Solutions for Chem 612 - Fall, 2008 and more Study notes Chemistry in PDF only on Docsity!

Chem 612 โ€“ Fall, 2008 Assignment 1 - Solutions

This assignment is due at the beginning of class on Wednesday, September 10, 2008.

  1. The standard enthalpy of formation, โˆ† H of, is โ€“642 kJ/mol for MgCl 2. The enthalpy of sublimation of Mg( s ), โˆ† H osub, is 146 kJ/mol, and the first and second ionization energies, I 1 and I 2 , of Mg( g ) are 737.7 kJ/mol and 1451 kJ/mol, respectively. The dissociation energy of Cl 2 , D , is 243.4 kJ/mol, and the electron affinity of Cl( g ) is โ€“349 kJ/mol. Given these data and additional data below, give answers to the following.

a. Write thermochemical equations, including state designations, corresponding to all of the data given above and also the lattice energy, U.

Mg( s ) + Cl 2 ( g ) 6 MgCl 2 ( s ) โˆ† H of = โ€“642 kJ/mol Mg( s ) 6 Mg( g ) โˆ† H osub = 146 kJ/mol Mg( g ) 6 Mg+( g ) + e โ€“^ I 1 = 737.7 kJ/mol Mg+( g ) 6 Mg2+( g ) + e โ€“^ I 2 = 1451 kJ/mol Cl 2 ( g ) 6 2 Cl( g ) D = 243.4 kJ/mol Cl( g ) + e โ€“^6 Clโ€“( g ) A = โ€“349 kJ/mol Mg2+( g ) + 2 Clโ€“( g ) 6 MgCl 2 ( s ) U = 2522 kJ/mol

b. Sketch the Born-Haber cycle for MgCl 2 ( s ), similar to the cycle shown in class for NaCl( s ).

Mg2+( g ) + 2 Cl-( g )

Mg( s ) + Cl2( g ) MgCl2( s )

โˆ† H osub D

Mg( g ) + 2 Cl( g )

I 1 + I 2 2 A

โˆ† H of

  • U

c. Calculate the lattice energy, U , for MgCl 2 ( s ) to verify the value of 2522 kJ/mol given in the table shown in class. Show your work.

โˆ† H of = โˆ† H osub + I 1 + I 2 + D + 2 A โ€“ U

U = โˆ† H osub + I 1 + I 2 + D + 2 A โ€“ โˆ† H osub โ€“ โˆ† H of = [146 + 737.7 +1451 + 243.4 + (2)(โ€“349) + 642]kJ/mol = 2522 kJ/mol

d. Explain why the lattice energy of MgCl 2 is so much greater than that of NaCl ( kJ/mol).

Mg2+^ is twice as highly charged and much smaller. Both factors increase the force of attraction between oppositely charged ions, resulting in a much greater potential energy

e. The values of the lattice energy for MgCl 2 , CaCl 2 , and SrCl 2 are 2522 kJ/mol, 2253 kJ/mol, and 2127 kJ/mol, respectively. Why is this the expected trend?

The cations are progressively larger through the series, increasing the separation between oppositely charged ions. Potential energy decreases with separation of the charges.

f. The difference in lattice energies between MgCl 2 and CaCl 2 is 269 kJ, but the difference between CaCl 2 and SrCl 2 is only 126 kJ/mol. Why?

The ionic radii (Shannon CN6 values) are 72 pm, 100 pm, and 110 pm, respectively for Mg2+, Ca2+, and Sr2+. The lattice energy trend reflects the cation size trend. Both Mg2+ and Ca2+^ have noble-gas configurations of the type ns^2 np^6 ( n = 2, 3), whereas Sr2+^ has a configuration 3 d^104 s^24 p^6. The size difference between Ca2+and Sr2+^ is so much less than between Mg2+^ and Ca2+, because Ca and Sr are separated by the intervening first transition series, through which sizes decline more gradually owing to poor shielding by 3 d electrons. As a result the cation-anion separation in SrCl 2 is not that much greater than in CaCl 2 , and the two salts have very similar lattice energies.

  1. Difference in electronegativity has sometimes been used to decide if a bond between a pair of atoms is better described as polar covalent or ionic. One often cited criterion is that an electronegativity difference, โˆ†ฯ‡, greater than about 1.6โ€“1.7 marks the beginning of ionic bonds. On this basis, the bond in HF (โˆ†ฯ‡ = 1.9) should be more ionic than that in NaCl (โˆ†ฯ‡ = 1.8). While most chemists would call NaCl ionic, few would do the same for HF. The electron density map of HF indicates a polar covalent bond. But aside from electron density mapping, what properties of HF suggest that it is polar covalent and not ionic? Why is difference in electronegativity an unreliable criterion?

All simple ionic compounds, particularly those composed of monatomic ions, are crystalline solids at room temperature. HF, by contrast, is a gas at room temperature. Moreover, it has a dipole moment (ฮผ = 5.93 x 10โ€“30^ C@m), characteristic of a polar covalent molecule. Ionic compounds contain no molecules of the compound, and therefore have no measurable dipole moments. Clearly difference in electronegativity is not a reliable guide to deciding the nature of the bonding between two atoms. The fault lies in assuming that electronegativities are constant from case to case. This is particularly liable to break down in the case of hydrogen. Indeed, as Gillespie and Popelier assert (p. 15), โ€œalmost all bonds are polar bonds, whether they are approximately described as covalent or ionicโ€. โ€œCovalentโ€ and โ€œionicโ€ cannot be precisely and quantitatively defined, except to note that there are purely covalent bonds (e.g., homonuclear diatomic molecules) but no purely ionic bonds.

  1. Draw Lewis structures, including valence electrons on pendant atoms, for the following molecules and ions. Where appropriate, assign formal charges, but do not draw double bonds or invoke hypervalence solely for the purpose of minimizing formal charges. ClF 2 +, I 3 โ€“ , ClF 3 , IF 5 , XeO 3 , Cl 2 O 7 , SCl 4 , B(OH) 3 , B 3 N 3 H 6 (B and N form a ring), Se 2 O 5 2โ€“.

ClF 2 +^ electron count = Cl (7) + 2F (14) + cation charge (-1) = 20

F Cl F

I 3 โ€“^ electron count = 3Cl (21) + anion charge (1) = 22

I I I Central I is hypervalent.

_

ClF 3 electron count = Cl (7) + 3F (21) = 28

Cl is hypervalent.

F Cl F

F

IF 5 electron count = I (7) + 5F (35) = 42

I is hypervalent.

I

F

F F

F F

XeO 3 electron count = Xe (8) + 3O (18) = 26

O Xe O

O

Cl 2 O 7 electron count = 2Cl (14) + 7O (42) = 56

3+

O Cl

O

O

O Cl

O

O

O 3+

SCl 4 electron count = S (6) + 4Cl (28) = 34

S is hypervalent.

Cl

Cl S Cl

Cl

B(OH) 3 electron count = B (3) + 3O (18) + 3H (3) = 24

B B is electron deficient.

O

O O

H

H

H

B 3 N 3 H 6 electron count = 3B (9) + 3N (15) + 6H (6) = 30

B N N

B B N

H

H

H

H

H

H B N N

B B N

H

H

H

H

H

H