TTU Chem. 1307: Exam 3

Thermodynamics

The study of energy and its transformations.

Thermochemistry

The branch of thermodynamics that deals with heat in chemical and physical change

Kinetic Energy

The energy possessed by a particle due to its motion.

K.E. = 1⁄2 mass x (velocity)2

Potential Energy

The energy stored, or possessed by a
particle due to its position or condition.

P.E. = m x g x h
m = mass; h = height g = gravitational
const (9.8 m s-2)

Energy can be exchanged

for example, as an object falls, potential energy is converted to kinetic energy.

Thermal energy

associated with the temperature of a particle, or system.

Mechanical energy

energy that enables a force to be applied and cause displacement.

Chemical energy

the energy stored in chemical bonds.

System

part of the universe of interest (i.e., reactants and products in a beaker)

Surroundings

everything else

E = the total internal energy of a system

ΔE = Efinal - Einitial = Eproducts - Ereactants

Work, w

work is done on a particle when a force
causes a displacement (change in position).

Heat, q

thermal energy in the process of transfer or conversion. The transfer is the result of a temperature difference.

E = the total internal energy of a system (in terms of Work and Heat) - Equation

ΔE = q + w

Energy flow out of the system

- sign

Energy flow into the system

+ sign

Heat transfer: system to surroundings
ΔE is (negative or positive?)
q is (negative or positive?)

- (negative)
- (negative)

Ice melts

ΔE is + (heat transfer: surroundings to system)
q is +

Gasoline burns in a car engine. The heat released expands the CO2 and H2O vapor, moving the pistons. Excess heat released is removed by the radiator. If the expanding gases do 451 J of work on the pistons and the system releases 325 J to the surroundings as heat, what are the values for q, w and ΔE in Joule (J) units?

ΔE = q + w w = -451 J
q = -325 J
ΔE = (-325 J) + (-451 J) = -776 J

What is the SI unit of energy?

The SI unit of energy is the Joule (J): Joule = N m = kg m2 s-2

1 calorie (cal) =

4.184 J

1 (Cal), or Kilocalorie (kcal) =

= 1000 * 4.184 = 4184 J

1 kilowatt-hour (kWh)

= 3.60 * 10^6 J

Pressure-Volume Work

Joule = N m

P x V = atm L

GasConstants: 0.08206atm L/(K mol) 8.314 J / (K mol)

So, 1 atm L = 8.314 / .08206 = 101.3 J 0.08206

Work

work is done on a particle when a force causes a displacement (change in position).

Pressure-Volume ("PV") work

Pressure-Volume ("PV") work

Energy flow

system to surroundings. w is - (negative)

w = - PΔV (expansion against an external pressure)

Total internal energy of a system

E is the sum of the kinetic and the potential energy of all particles in the system

ΔE = Efinal - Einitial = Eproducts - Ereactants

Is ΔE a state function?

Yes

ΔE = Efinal - Einitial

A state function depends only upon the initial and
final states, not the path taken!

State Functions

ΔE total internal energy of a system ΔP pressure change of a system
ΔV volume change of a system
ΔT temperature change of a system ΔH (enthalpy) of a system

Not state functions

q = heat of a system w = work

First Law of Thermodynamics

the total energy of the universe is conserved (energy is neither created nor destroyed).

ΔE Summary

E = the total internal energy of a system

E is the sum of the kinetic and the potential energy of all particles in the system

Typically refer to ΔE, which is a state function

ΔE=q+w

Enthalpy

The heat released (at constant pressure) is called enthalpy and is given the symbol H.

ΔE = ΔH - PΔV

PΔV = Pressure-volume Work
ΔH = q at constant P

ΔHreaction is negative

reaction is exothermic

ΔHreaction is positive

reaction is endothermic

Heat Out

Exothermic

Heat In

Endothermic

Wood burns when ignited

Negative, exothermic

Water freezes on a day when the temp is < 0 °C

Negative, exothermic

Exercising supplies heat, resulting in the evaporation
of sweat and cooling of the skin

positive, endothermic
Sweat + heat -> dry, cooler skin

One means to quantify heat: ΔT (temp change)

Heat capacity (C ) = q/ΔT

Specific Heat Capacity

(Cs) = q/(mass x ΔT)

C(s)=J/g*Degree Celcius

mass = grams

ΔTemp.= (Temp. Final -Temp. Initial) in degrees Celsius

Q = Hear Joules

The specific heat capacity

is the amount of heat energy required to raise the temperature of one gram of a substance 1 °C.
- Cs
- Units J/(g ∙ °C)

The molar heat capacity

The molar heat capacity is the amount of heat energy required to raise the temperature of one mole of a substance 1 °C.

If a 32.5 g cube of Al at 45.8 °C is placed into 105.3 g of water at 15.4 °C, what is the expected temperature of the metal and water at thermal equilibrium? (Cs values: Al = 0.903 J/g °C), water = 4.18 J/g °C)

q metal = -q water

Mass of metal Specific heat capacity of water Change in Temperature

=

-mass of water Specific heat capacity of water (4.184) Change in temperature


Al cube:
32.5 g
Ti =45.8°C 0.903 J/g °C

Water:
105.3 g
Ti =15.4°C 4.18 J/g °C

Remember ΔT = (Tfinal - Tinitial); Plug values into eqn and solve for Tfinal (Tf). thermal equilibrium => Tf metal = Tf water

By convention, we calculate the enthalpy change for the number of moles of reactants in the reaction as written.

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) ΔH = −2044 kJ

1 mol C3H8(g) = -2044 kJ
5 mol O2(g) = -2044 kJ

Coffee cup calorimeter:

-Constant Pressure
-Aqueous solution

The qreaction measured = Enthalpy (ΔH)

qreaction = -qsolution
= -(masssolution × Cs, solution × ΔT)

To obtain ΔHreaction per mol, divide by the number of moles.

The bomb calorimeter

is usually made of a sealed, insulated container filled with water. The heat change in the reaction is measured as the temperature change in the surrounding water bath:

qsurroundings = qcalorimeter = -qsystem

If a chemical reaction is reversed

ΔHrxn changes sign.

In performing addition or subtraction with reactions

- if the equation is multiplied by a factor, then ΔHrxn should be multiplied by the same factor.

ΔHrxn= -393kJ

2C(s)+2O2 (g)→2CO2 (g) ΔHrxn =2(-393kJ)=-786kJ

Hess' Law
- When a chemical reaction can be expressed as the sum of a series of reaction steps

ΔHrxn is the sum of the heats of reaction of the steps.

A + 2B →2D ΔHrxn =ΔH1 +ΔH2

When 155 mL of water at 26 °C is mixed with 75 mL of water at 85 °C, what is the final temperature? (H2O density = 1.00 g/mL; Cs H2O = 4.18 J/g °C)

155 g×4.18 J/g °C ×(Tf - 26 °C)
=
- 75 g× 4.18 J/g °C×(Tf - 85 °C)

Tf =45°C

S8(s)+8H2(g)→8H2S(g) ΔH= −162kJ

a) Is the reaction endothermic or exothermic?

b) What is ΔH for the reverse reaction?

c) What is ΔH when 2.6 mol of S8 reacts?

a) exothermic

b)+162kJ

c) 2.6 mol S8 x (-162 kJ/mol S8) = -4.2 x 102 kJ

Standard State

• For a gas Pure gas at 1 atm (exact) pressure.
• For a liquid or solid Pure substance in its most stable form at 1 atm (exact) and a specified temperature (usually 25 °C)
• For a substance dissolved in solution 1 M (exact) concentration.

Standard enthalpy change (ΔHo)

ΔH for a reaction when all reactants and products are in their standard states.

Standard enthalpy of formation (ΔHfo)

• For a pure element in its standard state, ΔHfo = 0 (exact).
• For a pure compound -> ΔH when 1 mol of the compound forms from its elements in their standard states.
• ΔHfo is also called the "standard heat of formation"

Calculating ΔHrxno From ΔHfo

ΔH°reaction = Σ n ΔHf°(products) − Σ n ΔHf°(reactants)

ex.

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

ΔHf°:
CH4 (g) = -74.6 kJ/mol
CO2 (g) = -393.5 kJ/mol
H2O (g) = -214.8 kJ/mol
Answer:

ΔHrxn = [(-393.5) + 2(-241.8)] - [(-74.6) + 0] = -802.5 kJ

When does ΔHfo = ΔH?

Which of the following gives the standard heat of formation (ΔHfo) for isopropyl alcohol (C3H8O (l))?

A) C3H8O(l) + 9/2 O2(g) → 3 CO2(g) + 4 H2O(g)
B) 3 C(s, graphite) + 4 H2(g) + 1⁄2 O2(g) → C3H8O(l)
C) 3 CO(g) + 4 H2O(g) → C3H8O(l) + 3 O2(g)
D) C3H8O(l) → CO(g) + 2 C (s, diamond) + 4 H2(g)

Answer:

B) ΔHfo = ΔH when 1 mol of the compound forms from its elements in their standard states

Wave Properties

c = λ×ν (classical)

Particle properties

E = h×ν (quantum)

Particles:

can have wave properties (λ = h (mv)-1)

Planck's constant

Symbol = h

(6.626 x 10-34 J s)

Schrödinger Eqn

HΨ=EΨ

1. wavefunctions orbitals

2. Energy levels (transitions between levels give characteristic light emission)

Frequency

v = frequency (Hz = s-1 units)

Speed of light in vacuum

c = const. (speed of light in vacuum) 3.00 x 108 m/s

c=ν×λ

Relationships between energy, wavelength and frequency

i.e., high energy radiation has short λ and high ν.

c = ν × λ E = h ν = (h c) / λ

Radio-Microwave-Infrared-Ultraviolet-X-ray-Gamma Ray

Long, Wavelength, Low Energy, Low Frequency

--------- >

Short Wave Length, High Energy ,High Frequency

Particle Properties of EM Radiation

Particle Energy = Plank's Constant * Frequency

E = h*v

E =h*c/λ
E = Energy per photon
λ = Wavelength
c = speed of light (2.998*10^8 m/s)

Calculate the energy (in J) per photon of red light with wavelength 700 nm? What is this energy per mol of photons (c = 3.00 x 108 m/s)?

(6.626 x 10-34 J-s) x (3.00 x 108 m/s)
------------------------------------------- =
((700 nm) x (10-9 m/nm))

= 2.84 x 10-19 J. <---- Energy per photon

(2.84 x 10-19 J/photon)*(6.022 x 10^23 photons/mol) =

= 1.71 x 105 J/mol

Spectra

• When atoms or molecules absorb energy, that energy is
often released as light energy, - Fireworks, neon lights, etc.
• When that emitted light is passed through a prism, a pattern of particular wavelengths of light is seen that is unique to that type of atom or molecule - the pattern is called an emission spectrum.
- Noncontinuous
- Can be used to identify the material
(Flame tests)

Color

• The color of light is determined by its wavelength or frequency.
• White light is a mixture of all the colors of visible light.
- A spectrum
- Red Orange Yellow Green Blue Indigo
Violet
• When an object absorbs some of the wavelengths of white light and
reflects others, it appears colored; the observed color is predominantly the colors reflected.

Discrete lines ("line spectra")

- arise from transitions between specific energy states

ex.
n = 1 , n = 2

Spectra

are a "fingerprint" for the compound

Which of the following explains why the absorbance spectrum for chlorophyll has peaks near 450 nm and 550 nm, but no peaks in between.

A) The green and yellow colors interfere
B) The light cannot transmit through the leaf
C) The pigment does not have the appropriate energy levels to enable light absorption between 450 nm and 550 nm
D) Only light reflected from the pigment is detected in an absorbance spectrum

C)

As an electron absorbs energy

it is excited to an unstable energy level

ex. n=1 to n=2

As an electron emits energy

light is emitted as electron falls back to lower energy level.

ex. n=2 to n=1

Rydberg Equation:

1 1 1
--- = R ( ------------ - ------------)
λ (n-final)^2 (n-initial)^2

R= 1.096776 x 10^7 m^-1

or 1/(1.096776x10^7) m

n-initial = initial energy level

n-final = final energy level

If solving for the Energy of a hydrogen atom's energy level use ( R = 2.18x10^-18 J )

R
E= - ------
n^2

ΔE = Efinal - Einitial

ΔE = -2.18 x 10-18 J [(1/nf2) - (1/ni2)]

n= energy level of hydrogen atom

What is the Ground State of an H electron

When the H electron is in the first orbit, the atom is in its lowest energy state, called the ground state.

n=1

The higher the orbital

the more unstable the atom

Each state is associated with a fixed circular orbit of the electron around the nucleus.
The higher the energy level, the farther the orbit is from the nucleus.

Excited state

When the electron is in any orbit higher than n = 1, the atom is in an excited state

Bohr's Model

- The atom does not radiate energy while in one of its stationary states.

- The atom changes to another stationary state only by absorbing or emitting a photon.

-The energy of the photon (hν) equals the difference between the energies of the two energy states.

Limitation of the Bohr Atomic Model

Applies to the H atom

Does not predict properties of atoms with more than one electron well.

Does not account for repulsive forces between electrons.

Bohr Atomic Model leads to what concepts?

• Principal quantum number "n"
• Ground state
• Excited states
• Light Absorption
• Light Emission

Schrödinger Equation

Expresses the total energy of an atom in terms of the kinetic and potential energy of the electrons and nucleus.

Is a partial differential equation - Solving the Schrödinger equation for an atomic system gives the wavefunction associated with a particular energy state.

HΨ=EΨ

E = energy
Ψ = wavefunction
H = Hamiltonian operator
(differential eqn.)

Wavefunction - a mathematical function that describes the wave- like nature of the electron.

Orbital - given by the wavefunction, but usually depicted as a plot of the wavefunction squared (Ψ2), which gives a probability distribution map that shows where the electron is likely to be found relative to the nucleus of the atom.

Solutions to the Schrödinger Equation are in

Quantum Numbers
n = principal quantum number
n = 1, 2, 3...(integers >= 1)

Examples of Quantum Number, n

Of Hydrogen

n = 4 E4 = -1.36*10^-19 J

n = 3 E3 = -2.42*10^-19 J

n = 2 E2 = -5.45*10^-19 J

n = 1 E1 = -2.18*10^-18 J

Quantum Number, l

l = Angular Momentum

The angular momentum quantum number determine the shape of the orbital.

Possible values from 0,1....(n-1)

Ex. n=3, the possible values of
l are 0, 1 and 2.

The l Quantum Number is related to orbital shapes:

l = 0 ---> s-orbital

l = 1 ---> p-orbital

l = 2 ---> d-orbital

l = 3 ---> f-orbital

Each Value of l is called by a particle letter that designates the shape of the orbital.

- s orbitals are spherical.
- p orbitals are like two balloons tied at the knots.
- d orbitals are mainly like four balloons tied at the knots.
- f orbitals are mainly like eight balloons tied at the knots.

Magnetic Quantum Number

M (l) = Magnetic Quantum Number
indicates orbital orientation values range from
+1, 0, -1

ex. l = 1, the possible values of m are -1, 0, 1

Spin Quantum Number

M(s) = Spin Quantum Number

electron spin up => +1/2

electron spin down => -1/2

What values of the angular momentum quantum number, l, are possible for the principal level n = 3?
A) only l = 0 is possible
B) 0,1and2
C) only 3
D) 1,2and3

B)

a) Identify the orbitals associated with l = 0, 1 and 2.

b) For each sub-level above, identify the m(l) quantum numbers.

a)

l = 0 ---> s-orbital

l = 1 ---> p-orbital

l = 2 ---> d-orbital

b)

l=0(s);ml = 0

= 1(p);ml = -1,0,+1

= 2(d);ml = -2,-1,0,+1,+2

An s-orbital in the n = 2
level is designated as

A 2s orbital

An f-orbital in the n = 6 level is designated as

A 6f orbital

Probability is

proportional to Ψ^2

no mater what the n value in an orbit???

List the possible ml values for each l value above and calculate the total number of orbitals in the n = 4 level.

l || Possible m(l) Values || Orbital Name
0 || 0 || 4s (1 orbital)
1 || -1, 0, +1 || 4p (3 orbitals)
2 || -2, -1, 0, 1, 2. || 4d (5 orbitals)
3 || -3, -2, -1, 0, 1, 2, 3. || 4f (7 orbitals)

total number of orbitals n=4 = 4^2 = 16 total orbitals

The Wave Nature of Matter:

de Broglie relation

λ = h
----
m*v

v = particle velocity
h = plank's constant = 6.626*10^-34

What is the de Broglie wavelength for a baseball (150 g)
traveling 90 mph (40 m/s)?

(6.626 x 10-34 J s) / (0.150 kg) x (40 m/s)

= 1.10 x 10-34 m

Which set of quantum numbers below is not allowed?
A) n = 5, l = 3, ml = 0, ms = +1/2
B) n = 5, l = 2, ml = -1, ms = -1/2
C) n = 3, l = 2, ml = 3, ms = +1/2
D) n = 4, l = 3, ml = 0, ms = -1/2

C)

What is the orbital assigned to the quantum numbers n = 5, l = 4 ?
A) 5p orbital
B) 4f orbital
C) 4d orbital
D) 5g orbital

D)

g-orbitals are possible, we just do not need them to account for electron structure of currently known elements.