Calorimetry - Standard Heats of Formation

Calorimetry is an experimental method in which the heat of a reaction (the system) is determined by measuring temperature changes in the surroundings.  In the lab, heat changes in physical and chemical processes are measured with a calorimeter (a closed container designed specifically for these measurements).  The basics of calorimetry lie with specific heat and heat capacity, so let's examine them first.

Specific Heat and Heat Capacity
The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of that substance by one degree Celsius.  It has the units J/g  C

The heat capacity (C) of a substance is the amount of heat required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. It has the units J/C

Specific heat is an intensive property, where as heat capacity is an extensive property.  The relationship between them is C = ms (s = specific heat of water [4.184 J/gC]; m = mass in grams; C = heat capacity).  Knowing the change in the sample's temperature will tell us the amount of heat (q) that has been gained or lost in the process.  The equations we use to calculate heat change are:

1. q = (m)(s)(Δt)
2. q = (C)(Δt)

Let's look at an example problem:

Constant-Volume Calorimetry

Heat of combustion is usually measured in a steel container called a constant-volume bomb calorimeter.  Because of the way it's built, we can assume that no heat (or mass) is lost to the surroundings during the time it takes to make the measurements.  Therefore we can call the bomb and the water surrounding it an isolated system.  To calculate qcal we need to know the heat capacity of the calorimeter (Ccal) and the temperature rise, that is,  qcal = Ccal Δt

Constant-Pressure Calorimetry

A simpler device is the constant-pressure calorimeter, which is used to determine the heat-changes for noncombustion reactions.  It can be made from something as crude as two styrofoam cups.  This device measures the heat effects of a variety of reactions, such as acid-base neutralization.  Because the pressure is constant, the heat change for the process (qrxn) is equal to the enthalpy change (ΔH) so we treat this calorimeter as an isolated system.

Let's do another example.  This one separates the B's from the C's.

Types of Problems Likely to Be On the Exam! Chapter 6

6.1 The Nature of Energy and Types of Energy
6.2 Energy Changes in Chemical Reactions


Cold packs, whose temperatures are lowered when ammonium nitrate dissolves in water, are carried by athletic trainers when transporting ice is not possible. Which of the following is true of this reaction?
(1) ΔH > 0, process is exothermic
(2) ΔH = 0, since cold packs are sealed
(3) ΔH < 0, process is exothermic
(4) ΔH < 0, process is endothermic
(5) ΔH > 0, process is endothermic



6.3 Introduction to Thermodynamics



Calculate q when 28.6 g of water is heated from 22.0°C to 78.3°C.
(1) 9.37 kJ
(2) 1.61 × 103kJ
(3) 0.385 kJ
(4) 1.61 kJ
(5) 6.74 kJ




6.4 Enthalpy of Chemical Reactions


Galena is the ore from which elemental lead is extracted. In the first step of the extraction process, galena is heated in air to form lead(II) oxide.

2PbS(s) + 3O2(g) → 2PbO(s) + 2SO2(g) ΔH = -827.4 kJ

What mass of galena is converted to lead oxide if 975 kJ of heat are liberated?
(1) 282 g
(2) 406 g
(3) 564 g
(4) 478 g
(5) 203 g


6.1:
6.2: 5
6.3: 5
6.4: 3

Enthalpy, Thermochemical Equations, and Combustion

Enthalpy and the First Law of Thermodynamics

• ΔU = q + w
• At constant pressure
• q = ΔH and W = -PΔV
• or ΔU = ΔH -PΔV (enthalpy change for a process at constant pressure)
• Enthalpy (H) is used to quantify heat flow into or out of a system in a process that occurs at constant pressure
• ΔH = H (products) - H (reactants)
• ΔH = heat given off or absorbed by the system at constant pressure

Thermochemical Equations

• The stoichiometric coefficients always refer to the number of moles of a substance

• If you reverse a reaction, the sign of ΔH changes

• If you mulitply both sides of an equation by some constant factor (m), then ΔH must be multiplied by that same factor (m)

• The physical states of all reactants and products must be specified in thermochemical equations

Energy, Open-Closed Systems, Endo-Exothermic, State Functions, & Work

Energy is the capacity to do work

• Radiant Energy comes from the sun and is the main source of energy on earth
• Thermal Energy is the energy associated with the random motions of atoms and molecules
• Chemical Energy is the energy that is stored in the chemical bonds of substances
• Nuclear Energy is the energy stored within the neutrons and protons in the nucleus of the atom
• Potential Energy is the energy available by virtue of an object's position 

Energy Changes in Chemical Reactions

• Heat is the transfer of thermal energy between  two bodies that are at different temperatures
• Temperature is a measure of the thermal energy
• Temperature DOES NOT equal thermal energy.

Open-Closed Systems 

• Thermochemistry is the study of heat change in chemical reactions
• The system is the specific part of the universe that is of interest in a particular study (the universe is "everything else".)
• Open system = allows the exchange of both energy and mass with surroundings
• Closed system = allows the exchange of but not mass
• Isolated system = allows neither energy nor mass to be exchanged (such as putting something in a vacuum jacket)

Endo-Exothermic Systems

• Exothermic process is any process that gives off heat (transfers thermal energy) from the system to the surroundings

2H2 (g) + O2 (g) => 2H2O (l) + ENERGY

H2O (s) => H2O (l) + ENERGY

• Endothermic process is any process in which heat has to be supplied to the system from the surroundings

ENERGY + 2HgO (s) => 2Hg (l) + O2 (g)

ENERGY + H2O (l) => H2O (g)

State Functions

• Thermodynamics is the scientific study of the interconversion of heat and other kinds of energy
• State functions are thermodynamic properties that are determined by the state of the system (at any time) regardless of how that condition was achieved 
• Energy, pressure, volume, temperature
• First law of thermodynamics - energy can be converted from one form to another, but cannot be created of destroyed

• ΔU = q + w
• ΔU is the change of the internal energy of the system
• q is the heat exchange between the system and the surroundings
• w is the work done on, or by, the system
• W = -PΔV is when a gas expands against a constant external pressure

Work

• Work done by the system on the surroundings
• w = F x d
• W = -PΔV
• PV = F/d*2 x d*3 = F x d = w
• Work is NOT a state function!!
• ΔV = V (final) - V (initial)

Types of Problems Likely to be on the Exam! - Chapter 10

10.1 Molecular Geometry

Give the number of lone pairs around the central atom and the molecular geometry of NH4+.
1) 0 lone pairs, tetrahedral
2) 1 lone pair, trigonal bipyramidal
3) 1 lone pair, seesaw
4) 2 lone pairs, T-shaped
5) 3 lone pairs, trigonal bipyramidal


Give the number of lone pairs around the central atom and the molecular geometry of the tri-iodide ion, I3-.
1) 0 lone pairs, trigonal planar
2) 1 lone pair, tetrahedral
3) 1 lone pair, trigonal pyramidal
4) 2 lone pairs, seesaw
5) 3 lone pairs, linear


According to VSEPR theory, which of the following species should have a tetrahedral molecular geometry?
1) H2O
2) NH3
3) AlF3
4) CCl4
5) CH2O


Which pair of molecules has the same molecular geometry?
1) CCl4 and NH3
2) H2O and SO2
3) NO3 and H2O
4) CO2 and CCl4
5) SO2 and NH3


According to the VSEPR theory, if the central atom has three bonded atoms and one lone pair,
the geometry (shape) at this atom will be ______________________.
1) linear
2) bent (angular)
3) trigonal planar
4) trigonal pyramidal
5) tetrahedral


Which of the following substance is/are planar?
(i) SO3 (ii) SO32- (iii) NO3- (iv) PF3 (v) BF3
1) only (i) and (ii)
2) only (i), (iii), and (v)
3) only (iv)
4) all are planar except (iv)
5) all are planar except (ii)


According to the VSEPR theory, the actual F-As-F bond angles in the AsF4- ion are predicted to be
1) 109.5°.
2) 90° and 120°.
3) 180°.
4) < 109.5°.
5) < 90° and < 120°.


Predict the geometry around the central atom in SO42-.
1) trigonal planar
2) trigonal pyramidal
3) tetrahedral
4) trigonal bipyramidal
5) octahedral


List the following molecules in order of increasing bond angle (smallest to largest):
NH3, BF3, ClNO, CF4
1) NH3, BF3, ClNO, CF4
2) ClNO, BF3, NH3, CF4
3) ClNO, CF4, NH3, BF3
4) CF4, NH3, BF3, ClNO
5) NH3, CF4, ClNO, BF3


Give the number of lone pairs around the central atom and the molecular geometry of SCl2.
1) 0 lone pairs, linear
2) 1 lone pair, bent
3) 2 lone pairs, bent
4) 3 lone pairs, bent
5) 3 lone pairs, linear


Give the number of lone pairs around the central atom and the geometry of the ion NO2–.
1) 0 lone pairs, linear
2) 1 lone pair, bent
3) 2 lone pair, bent
4) 3 lone pairs, bent
5) 3 lone pairs, linear


According to the VSEPR theory, which one of the following species should be linear?
1) H2S
2) HCN
3) BF3
4) H2CO
5) SO2


The C–N–O bond angle in nitromethane, CH3
NO2, is expected to be approximately
1) 60°
2) 90°
3) 109.5°
4) 120°
5) 180°


Which of the following molecules have the same geometries?
1) SF4 and CH4
2) CO2 and H2O
3) CO2 and BeH2
4) N2O and NO2 (NNO and ONO, respectively)


According to the VSEPR theory, the molecular shape of the carbonate ion, CO32-, is
1) square planar.
2) tetrahedral.
3) pyramidal.
4) trigonal planar.
5) octahedral.


According to VSEPR theory, which one of the following molecules should have a bent shape?
1. Cl2O
2. CO2
3. HCN
4. CCl4
5. none of them

10.2 Dipole Moments

Which one of the following molecules is nonpolar?

1) NH3

2) OF2

3) CH3Cl

4) H2O

5) BeCl2


Which of the following species has the largest dipole moment (i.e., is the most polar)?
1) CH4
2) CH3Br
3) CH2Br2
4) CHBr3
5) CBr4

10.4 Hybridization of Atomic Orbitals

Indicate the type of hybrid orbitals used by the C atom indicated by the arrow in the molecule below.

1) sp
2) sp2
3) sp3
4) sp3d
5) sp3d2



In which one of the following molecules is the central atom sp2 hybridized?
1. SO2
2. N2O
3. BeCl2
4. NF3
5. PF5


What is the hybridization of the central atom in ClO3-?
1. sp
2. sp2
3. sp3
4. sp3d
5. sp3d2

10.5  Hybridization in Molecules Containing Double and Triple Bonds

The number of pi bonds in this molecule is:

1) 2
2) 3
3) 4
4) 5
5) 6



The skeleton for methyl formate (C2H4O2) is given below.  Complete the following statement: Methyl formate has ______ pi bond(s); carbon #1 has _____ hybridization, while carbon #2 has _______ hybridization.
1) 1, sp2, sp3
2) 1, sp3, sp2
3) 3, sp2, sp3
4) 2, sp, sp3
5) 2, sp2, sp3



The number of pi bonds in the molecule below is

1) 2
2) 4
3) 6
4) 10
5) 15


Indicate the type of hybrid orbitals used by the central atom in CCl4.

1) sp
2) sp2
3) sp3
4) sp3d
5) sp3d2


The number of pi bonds in the molecule below is

1. 1.
2. 2.
3. 3.
4. 5.
5. 9.



Key:
10.1: 1, 5, 4, 2, 4, 2, 5, 3, 5, 3, 2, 2, 4, 3, 4, 1
10.2: 5, 4
10.3:
10.4: 1, 1, 3
10.5: 5, 1, 2, 3, 3

Valence Bond Theory and Hybridization

Red = High Density

Valence bond theory

So, what is valence bond theory?  It's when bonds formed by sharing electrons form overlapping atomic orbitals.  As you bring bonded atoms closer and closer together you build up greater electron densities.

Hybridization - mixing of two or more atomic orbitals to form a new set of hybrid orbitals.  This is the bed-time level version, my professor says.  How does this work?

1.) Mix at least 2 nonequivalent atomic orbitals (e.g. s and p).  Hybrid orbitals have very different shape from original atomic orbitals.
2.) Number of hybrid orbitals is equal to number of pure atomic orbitals used in the hybridization process.
3.) Covalent bonds are formed by:
      a. Overlap of hybrid orbitals with other atomic orbitals
      b. Overlap of hybrid orbitals with other hybrid orbitals

Some of these bonds look different.  What are they each called?  A Sigma bond occurs between the two atoms.  A Pi bond appears above and below the plane of nuclei of the bonding atoms.  Single, double, and triple bonds have varying amounts of sigma and pi bonds:

- A single bond has 1 sigma bond.
- A double bond has 1 sigma bond and 1 pi bond.
- A triple bond has 1 sigma bond and 2 pi bonds.

As you can tell, each bond has 1 sigma bond and the pi bond is equal to the type of bond minus 1.  Ergo a triple bond (3 - 1) has 2 pi bonds.

So, how do I predict the hybridization of the central atom?

1.) Draw the Lewis structure of the molecule.
2.) Count the number of lone pairs AND the number of atoms bonded to the central atom.

Dipole Moments, Polar Molecules, Valence Bond Theory

Dipole Moments and Polar Molecules

---------------------->

Dipole moment = the product of the charge, Q, and the distance, r, between the charges:  Table 10.3 lists the dipole moments of several molecules.

Diatomic molecules containing atoms of different elements have dipole moments and are called polar molecules.  Examples include: HCl, CO, and NO.

Diatomic molecules containing atoms of the same element are examples of non-polar molecules because they do not have dipole moments.  For example: H2, O2, and F2.

For a molecule made up of three or more atoms both the polarity of the bonds and the molecular geometry determine whether there is a dipole moment.  Example is CO2 which can be either linear or bent.

Molecular Geometry

Valence shell electron pair repulsion (VSEPR) model:

We use this to predict the geometry of the molecule from the electrostatic repulsions between the electron (bonding and non-bonding) pairs.  The table below has molecules that have no non-bonding pairs:

In all of these examples, there are no lone pairs.  But it turns out that they can completely change the shape if present.  The repulsion between non-bonding electron pairs (lone pairs) is greater than the repulsion between a non-bonding electron pair and a bonding electron pair which is greater than the repulsion of two bonding electron pairs.

So let's look at molecular geometry when there are lone pairs present:

(Also, this is all covered in Table 10.2 in the textbook.  We're responsible for knowing this information.)

Predicting Molecular Geometry

1.) Draw Lewis structure for molecule
2.) Count number of lone pairs on the central atom and number of atoms bonded to the central atom
3.) Use VSEPR to predict the geometry of the molecule

And that's it!  Wear those flashcards out.

Types of Problems Likely to be on the Exam! - Chapter 9

9.4

Describe the bonding of C in the guanidinium cation, CN3H6+.  Carbon is the central atom.
1) Carbon forms two double bonds
2) Carbon forms four single bonds
3) Carbon forms one double bond and two single bonds
4) Carbon forms one triple bond and one single bond
5) Carbon forms one triple bond and two single bonds

The Lewis dot representation of formic acid, HCO2H, has
1) an incomplete octet for the C atom
2) four single bonds
3) three single bonds and one double bond
4) two double bonds
5) two equally plausible resonance forms

The skeleton for the, azide anion, N3-, is NNN. The Lewis dotrepresentation for the azide anion has
1) formal charges of zero on all three N atoms
2) two single bonds
3) one single bond and one double bond
4) two double bonds
5) two equally plausible resonance forms

The number of lone electron pairs in the N2 molecule is ___.
 1) 1
 2) 2
 3) 3
 4) 4
 5) 5

9.5

Which of the following pairs of elements would be most likely to form an ionic compound?
1) Cl and I
2) Al and K
3) Cl and Mg
4) C and S
5) Al and Mg

A polar covalent bond would form in which one of the following pairs of atoms?
 1) Cl and Cl
 2) Si and Si
 3) Ca and Cl
 4) Cs and Br
 5) P and Cl

9.6 

Give the number of bonding electron pairs around the central atom in a molecule of urea (CO(NH2)2)
1) 0
2) 1
3) 2
4) 3
5) 4

The total number of bonding electrons in a molecule of formaldehyde (H2CO) is
 1) 3
 2) 4
 3) 6
 4) 8
 5) 18

9.7

Assuming the octet rule is obeyed, how many covalent bonds will a nitrogen atom form to give a formal charge of zero?
 1) 1
 2) 2
 3) 3
 4) 4
 5) 5

In the best Lewis structure for the fulminate ion, CNO-, what is the formal charge on the central
nitrogen atom?
1. +2
2. +1
3. 0
4. -1
5. -2

9.9

Which of the following does not follow the octet rule?
1) CCl4
2) BF3
3) H2O
4) CO2
5) C2H4

Which response includes all the molecules below that do not follow the octet rule?
(1) H2S (2) BCl3 (3) PH3 (4) SF4
 1) (2) and (4)
 2) (2) and (3)
 3) (1) and (2)
 4) (3) and (4)
 5) (1) and (4)

Answer Key:
9.4: 3,3,4,2
9.5: ?, 5
9.6: 5,4
9.7: 3,2
9.9: 2,1

Miscellaneous Tips from Homework

Things to Know:

• Lattice Energy is indirectly related to the size of ions.  As ionic size increases from top to bottom in a group, lattice energy decreases.
• Electronegativity decreases from top to bottom in a group; increases from left to right in a period.
• Covalent bonds have a difference of 0; Ionic bonds have a difference that is greater than 2; Polar Covalent bonds are everything that fall in between.
• If two elements are in close proximity to each other on the periodic table, they're more likely to have similar electronegativities.
• Make sure all elements (except hydrogen) have a complete octet AND the valence electrons are all represented in the structure.

Writing Lewis Structures

This is how we do it:

• Draw skeletal structure of compound showing what atoms are bonded together.  Put the least electronegative element in the center.
• Count the total number of valence electrons.  Add 1 for each negative charge.  Subtract 1 for each positive charge.
• Draw single bonds between the central atom and the surrounding ones.  Complete the octet for all atoms bonded to the central atom except hydrogen.
• If central atom has less than eight electrons, form double and triple bonds on central atom as needed using lone pairs from neighboring atoms, to complete the octet on the central atom.

Example 1: Write the Lewis Structure of nitrogen trifluoride (NF3)

• N is less electronegative than F, so put N in the center.
• Count the valence electrons.  N has 5, and F has 7 each.  
• 5 + (7 x 3) = 26 valence electrons
• Draw single bonds between N and F atoms and complete octets on N and F atoms.
• Check, are the number of electrons in the structure equal to number of valence electrons?

Example 2: Write the Lewis structure of the carbonate ion (CO32-)

• C is less electronegative than O, put C in center.
• Count valence electrons.  C has 4 and O has 6.  Also, note the 2- charge.  4 + (6 x 3) + 2 = 24 valence electrons
• Draw single bonds between C and O atoms and complete octet on O atoms.  Which gives us this structure to the right.  But there's a problem.  Carbon, the central atom, hasn't formed a complete octet.  How do we correct this?
• In this situation we take one of the lone pairs of the oxygen atoms and make a double bond with the carbon atom.

So how about when there's more than one skeletal structure of a compound such as formaldehyfe (CH2O)?

An atom's formal charge is the difference between the number of valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis structure.  Formal charge on an atom in a Lewis structure = total number of valence electrons in the free atom - total number of nonbonding electrons - 1/2 (total number of bonding electrons).  The sum of the formal charges of the atoms in a molecule or ion must equal the charge on the molecule of ion.

So let's look at the above structure.  It appears to be pretty legit.  The hydrogen atoms both have two electrons.  The carbon atom has one non-bonding pair of electrons, a single bond, and a double bond.  That comes out to be a complete octet.  The oxygen atom appears to also be a complete octet.  It has one lone pair, a single bond, and a double bond.  In total there are 12 valence electrons.  Let's plug all of this into the formal charge recipe I have in rainbow colors in the paragraph above this one.

The total number of valence electrons in the free atom (C) is 4.  Apparently this is the number of valence electrons an element, in this case Carbon, has.  The total number of nonbonding electrons is 2 aka that lone pair I mentioned.  There are 6 remaining electrons that are bonded, which we cut in half because that's what the formula says to do.  Formal charge on C = 4 - 2 - (1/2 x 6) = -1

We can do this again with O.  The total number of valence electrons in this case is 6.  It has the same number of non-bonding electrons as C, 2.  It also happens to have the same number of bonded electrons, 6.  That's pretty easy.  Formal charge on O = 6 - 2 - (1/2 x 6) = +1

Now let's do the same thing with this alternative structure and see if anything changes and if there's any significance to doing any of this at all.

Formal charge on C = 4 - 0 - (1/2 x 8) = 0  (It's different!!!!!!)

Formal charge on O = 6 - 4 - (1/2 x 4) = 0 

This second structure is favorable because it has a charge of 0 on the atoms.  

Guidelines for Formal Charge and Lewis Structures:

• For neutral molecules, a Lewis structure in which there are no formal charges is preferable to one in which formal charges are present.  Just like we saw above!
• Lewis structures with large formal charges are less plausible than those with small formal charges.
• Among Lewis structures having similar distributions of formal charges, the most plausible structure is the one in which negative formal charges are placed on the more electronegative atoms.

Resonance Structures

A resonance structure is one of two or more Lewis structures for a single molecule that cannot be represented accurately by only one Lewis structure.  These are the variations you can make by flipping around single and double bonds.  It's not that scary.

Exceptions to the Octet Rule

• The Incomplete Octet - when there are simply not enough electrons present to form an octet.
• Example: BeH2, BF3
• Odd-Electron Molecules - you just can't form an octet with an odd number =(
• Example: NO
• The Exapanded Octet (central atom with principal quantum number n > 2)
• Example: SF6

Electronegativity and Bonding

So what is electronegativity?
Besides being hard to spell, electronegativity is the ability of an atom to draw towards itself electron density in a chemical bond.  The scale used for electronegativity is unitless and all the numbers are relative.  However, it is related to electron affinity, somehow.  It doesn't say how in my notes.  Welp.  Also, Electronegativity increases left to right across a group, and increases from bottom to top in a period.  For example, F is the most electronegative while Fr is the least electronegative.

It turns out that you can classify bonds based on electronegativity difference of the bonded atoms!
If the difference is 0, then the bond is a covalent one.  If the difference is between 0 and 2, the bond is of the polar covalent variety.  A difference greater than 2 is an ionic bond.  So, in order of increasing difference of elctronegativity: Covalent (shares electrons) => Polar Covalent (partial transfer of electrons) => Ionic (tranfers electrons)


Now for practice, classify the following bonds as ionic, polar covalent, or covalent:

1. CsCl.
1. The electronegativity of Cs and Cl is 0.7 and 3.0 respectively.
2. 3.0 - 0.7 = 2.3
3. Ergo it is IONIC!
2. H2S
1. The electronegativity of H and S is 2.1 and 2.5 respectively.
2. 2.5 - 2.1 = 0.4
3. It is POLAR COVALENT!
3. NN
1. The electronegativity of N is 3.0
2. 3.0 - 3.0 = 0
3. SLYTHERIN!  COVALENT!

Basic Concepts of Chemical Bonding

Valence Electrons are the outer shell electrons of an atom.  They're the electrons that participate in chemical bonding.

Protip: A super easy way to check what an element's valence electrons are is to look at the group it belongs to.  Nitrogen, for example, belongs to group 5A and has five electrons.  

Types of Bonds:

The Ionic Bond is the electrostatic force that holds ions together in an ionic compound.

A Covalent Bond is a chemical bond in which two or more electrons are shared by two atoms.

Polar Covalent Bond or Polar Bond is a covalent bond with greater electron density around one of the two atoms.  The atom with a greater electron density is called an electron rich region.  Likewise, the atom with a lesser electron density is referred to as an electron poor region.

Bond Lengths:

1.) Single bond - two atoms share one pair of electrons
2.) Double bond - two atoms share two pairs of electrons
3.) Triple bond - two atoms share three pairs of electrons


Triple bond < Double bond < Single bond

• C-O is 143
• C=O is 121
• C-C is 154
• C=C is 133