Week 6/Tu: Lecture Units ‘13 & 14’

© DJMorrissey, 2o12

Unit 12: Heat capacity -- heating curves, change of state -- “PV” work, enthalpy, state function Unit 13: Chemical Energy -- Hess’s Law (version of conservation of energy) -- bond energies Unit 14: Electronic structure -- light, electromagnetic energy -- Planck, Einstein & photons -- internal structure of atoms, Bohr Unit 15: Wave properties of matter -- particle waves ? -- electromagnetic transitions Issues: Temperature Scales Homework #4 due this Saturday, Oct. 6th 8am, back to usual

Germain Henri Hess 1802 - 1850

100 373.15 Boiling Point of Water at 1 atm 0 273.15 Freezing Point of Water -273.15 0 Absolute Zero

Week 6/Tu: Temperature Scales

© DJMorrissey, 2o12

What is “temperature”? Temperature is an intensive variable that represents the internal energy of a substance. (Remember that heat is simply a form of energy.) Celsius or Centigrade Kelvin Farhenheit ?

100 o F 0 o F

To change scales: K = oC + 273.15 oC = K – 273.15 Temperature Differences: Δ T (Kelvin) = Δ T (Centigrade)

Week 6/Tu: Hess’s Law (Sum of Heats) -1

© DJMorrissey, 2o12

Enthalpy H

Extent of Reaction

Enthalpy is a state function … thus, the energy change between reactants and products is a constant and independent of the path the reaction (actually) takes from beginning to end. Consider building up a reaction by a series of smaller reactions that already have been measured and tabulated in order to calc. overall ΔH. Data

C (s) + O2 (g) à CO (g) + ½ O2 (g) ΔH = -110 kJ

CO (g) + ½O2 (g) à CO2 (g) ΔH = ?

C(s) +O2(g)

CO2 (g)

CO(g) +½O2(g)

C (s) + O2 (g) à CO2 (g) ΔH = -394 kJ

Week 6/Tu: Hess’s Law (Sum of Heats) -2

© DJMorrissey, 2o12

Enthalpy is a state function … thus, the energy change between reactants and products is a constant and independent of the path the reaction (actually) takes from beginning to end. Consider building up a reaction by a series of smaller reactions that already have been measured and tabulated in order to calc. overall ΔH. Data:

C (s) + O2 (g) à CO (g) + ½ O2 (g) ΔH = -110 kJ

C (s) + O2 (g) à CO2 (g) ΔH = -394 kJ

CO (g) + ½O2 (g) à CO2 (g) ΔH = +110 + -394 = -284 kJ

CO (g) + ½ O2 (g) à C (s) + O2 (g) ΔH = +110 kJ

C (s) + O2 (g) à CO2 (g) ΔH = -394 kJ

Net:

Reverse 1:

Add 2:

Week 6/Tu: Enthalpy of Formation

© DJMorrissey, 2o12

0 Enthalpy H

Enthalpy is a state function … but there is no absolute scale. Choose a convenient zero to simplify calculations. Define the heat of formation of each element in its standard state to be zero, ΔHo

f The dimensions of ΔHof are kJ/mole.

This definition allows us to create tables of useful data.

C(s) + O2(g)

CO2 (g)

CO(g) +½O2(g) ΔHo

f #2 ΔHo

f #1

ΔH

Revised version of Hess’s Law for a chemical reaction: ΔH = Sum( ΔHo

f Products) - Sum( ΔHof Reactants)

Week 6/Tu: Reaction Enthalpy

© DJMorrissey, 2o12

Enthalpy H

Reactants

Products

Use the standard heats of formation to calculate the energy change: CH4 (g) + 2 O2 (g) à CO2 (g) + 2 H2O (l) + energy? ΔHo

f 1* -75 2 * 0 1* -394 2* -286

2 H2 (g) + O2 (g) à 2 H2O (g) -572 kJ

C (s) + O2 (g) à CO2 (g) -394 kJ

CH4 (g) à C (s) + 2 H2 (g) +75 kJ

ΔH = Sum( ΔHof Products) - Sum( ΔHo

f Reactants) = ( -394 + -572) – ( -75 + 0) kJ = -891 kJ as written

It should not be surprising that we get the same number if we go another route:

ΔH = 75 + -394 + -572 = -891 kJ

Week 6/Tu: Enthalpy of Bond Formation

© DJMorrissey, 2o12

The idea of using arbitrary paths to calculate reaction energies can be pushed further to include the calculation of bond enthalpies … imagine the reaction process by breaking all the molecules completely apart and then reforming the new molecules atom-by-atom. (Which would never happen but it be useful.)

N2 (g) + 2 H2 (g) à N2 H4 (g) + energy?

2 N (g) + 4 H (g)

2H + H2

Break N2 molecule 946 kJ/mol N=N Make 4 x N-H bonds 389 kJ/mol each Break two H2 molecules 436 kJ/mol H-H Make 1 N-N bond 159 kJ/mol

ΔH = Energy of Bonds Broken – Energy of Bonds Made = (1* 946 + 2* 436) – (4* 389 + 1*159) kJ = + 103 kJ

Week 6/Tu: Example of Bond Enthalpy

© DJMorrissey, 2o12

Another example of bond enthalpies and state functions: What is the bond enthalpy for a C-H bond using the std. heat of formation of methane?

C (s) + 2 H2 (g) à CH4 (g) ΔHof = -75 kJ/mol

C (g) + 2 H2 (g)

C (g) + 4H (g)

ΔH(sublim.) =+717 kJ/mol

Break two H2 molecules 436 kJ/mol Make four C-H bonds Bond Energy = x ?

ΔH path 1 (direct) = ΔH path 2 (long way) 1mol* -75 kJ/mol = (1* 717) + (2* 436) + (4 x) -75 = 1589 + 4x -1664 /4 = x -416 = x

Week 6/Tu: Fundamentals of Energy -1-

© DJMorrissey, 2o12

Following the development of thermodynamics and atomic chemistry, scientists turned their attention towards understanding the fundamental basis of energy and atomic structure. First of all “energy” was identified as electromagnetic radiation that travels at constant speed and comes in an infinite range of wavelengths. For any travelling wave, the speed is the product of the frequency times the wavelength. For light: c = λ ν

CARTOON à

EXPANDED à

Week 6/Tu: Fundamentals of Energy -2-

© DJMorrissey, 2o12

Following the development of thermodynamics and atomic chemistry, scientists turned their attention towards understanding the fundamental basis of energy and atomic structure. One of the early discoveries was that electromagnetic energy seemed to be emitted in two forms by natural objects. 1)  Continuous spectrum from hot objects (e.g., kiln) 2)  Line spectrum from (pure) atoms or molecules

View inside a hot kiln

Top shelf, cold

Week 6/Tu: Fundamentals of Energy -3-

© DJMorrissey, 2o12

Following the development of thermodynamics and atomic chemistry, scientists turned their attention towards understanding the fundamental basis of energy and atomic structure. One of the early discoveries was that electromagnetic energy seemed to be emitted in two forms by natural objects. 1)  Continuous spectrum from hot objects – explained by Max Planck (1900) as the

emission spectrum created by a huge number of tiny oscillators, each oscillator has a fixed frequency … we recognize these oscillators today as vibrations of atoms in molecules, or electrons bound in atoms. Chemical bonds have different strengths, the vibrations have different but fixed energies. Atoms have different numbers of protons, so the binding of electrons have different energies.

2)  Line spectrum from pure chemicals – these are individual vibrations from specific molecules, or from the transitions of electrons in individual atoms.

Week 6/Tu: Fundamentals of Energy -4-

© DJMorrissey, 2o12

Einstein contributed an important part to the modern understanding of electromagnetic radiation by explaining the photoelectric effect.

Facts: 1)  No electric current (no electrons

emitted) for long wavelength light – a threshold

2)  Once threshold is crossed, the current does not depend on wavelength

3)  Once threshold is crossed, the current depends on the intensity of light

His theory was that light was quantized into photons and that each one had a unique energy: Ephoton = h ν = h c / λ Planck’s Constant h = 6.625 x10-34 J-s

Experiment: shine light onto a clean metal surface, observe e-’s