Bond Enthalpy, Enthalpy of Reaction, and the Wave Nature of Light

Chapter 5: Thermochemistry

Bond Enthalpy

Bond enthalpy is a key concept in thermochemistry, representing the energy required to break one mole of a particular bond in a gaseous substance. Understanding bond enthalpy allows chemists to estimate the enthalpy changes in chemical reactions.

• Bond Enthalpy Definition: The enthalpy change associated with breaking one mole of a specific bond in the gas phase.

• Positive Value: Bond enthalpy is always positive because energy must be supplied to break chemical bonds.

• Bond Formation: Energy is released when new bonds form between gaseous fragments.

• Bond Strength: The greater the bond enthalpy, the stronger the bond.

Table: Average Bond Enthalpies (kJ/mol)

Bond	Bond Enthalpy (kJ/mol)
C—H	413
C—C	348
C=C	614
C≡C	839
O—H	463
O=O	495
N≡N	946
H—H	436
F—F	155
Cl—Cl	243
Br—Br	193
I—I	151

Additional info: Table values are averages and may vary slightly depending on molecular environment.

Bond Enthalpies and Enthalpy of Reaction

Bond enthalpies can be used to estimate the enthalpy change ($\Delta H$) of a chemical reaction by considering the bonds broken and formed during the process.

• Calculation Method:

  • Add bond energies for all bonds broken (reactants).

  • Subtract bond energies for all bonds formed (products).

• Formula: $\Delta H_{\text{rxn}} = \sum (\text{bond enthalpies of bonds broken}) - \sum (\text{bond enthalpies of bonds formed})$

• Predicting Reaction Type: If $\Delta H_{\text{rxn}}$ is negative, the reaction is exothermic; if positive, it is endothermic.

Example Calculation

Combustion of ethane:

$\mathrm{C_2H_6(g) + 7 O_2(g) \rightarrow 4 CO_2(g) + 6 H_2O(g)}$

• Identify all bonds broken and formed.

• Use average bond enthalpies from the table.

• Apply the formula to estimate $\Delta H_{\text{rxn}}$.

Additional info: This method provides an estimate; actual enthalpy may differ due to molecular environment.

Chapter 6: Electronic Structure of Atoms

The Wave Nature of Light

Understanding the electronic structure of atoms requires knowledge of electromagnetic radiation, which behaves as waves and particles.

• Electromagnetic Radiation: Moves as waves through space at the speed of light ($c = 3.00 \times 10^8$ m/s).

• Wavelength ($\lambda$): The distance between corresponding points on adjacent waves.

• Frequency ($\nu$): The number of complete waves passing a point per second (Hz).

• Relationship: $c = \lambda \nu$

• Amplitude: The height of the wave, related to intensity (brightness).

Table: Common Wavelength Units for Electromagnetic Radiation

Symbol	Length (m)	Type of Radiation
Å	$10^{-10}$	X-ray
nm	$10^{-9}$	Ultraviolet, visible
μm	$10^{-6}$	Infrared
mm	$10^{-3}$	Microwave
cm	$10^{-2}$	Microwave
m	$1$	Television, radio
km	$10^3$	Radio

Quantized Energy and Photons

Classical wave theory cannot explain certain phenomena, such as blackbody radiation and the photoelectric effect. Quantum theory introduces the concept of energy quantization.

• Quantum: The smallest packet of energy; plural: quanta.

• Planck's Explanation: Energy is emitted or absorbed in discrete packets (quanta).

• Einstein's Photoelectric Effect: Electrons are ejected from metal surfaces only if the incident light has sufficient energy (frequency).

• Energy of a Photon: $E = h\nu$ where $h = 6.626 \times 10^{-34}$ J·s (Planck's constant).

• Combined Formula: $E = \dfrac{hc}{\lambda}$

Example Calculation

• Given: Wavelength of red laser pointer $\lambda = 650$ nm.

• Convert to meters: $650$nm$= 6.50 \times 10^{-7}$ m.

• Calculate frequency: $\nu = \dfrac{c}{\lambda} = \dfrac{3.00 \times 10^8}{6.50 \times 10^{-7}} = 4.6 \times 10^{14}$ Hz

• Calculate photon energy: $E = h\nu = (6.626 \times 10^{-34})(4.6 \times 10^{14}) = 3.1 \times 10^{-19}$ J

Additional info: Higher frequency (shorter wavelength) light has higher energy photons.

Key Terms and Relationships

• Wavelength ($\lambda$): Distance per cycle of a wave.

• Frequency ($\nu$): Cycles per second (Hz).

• Amplitude: Intensity or brightness of a wave.

• Inverse Relationship: As wavelength decreases, frequency increases: $c = \lambda \nu$

• Energy and Frequency: $E = h\nu$

• Energy and Wavelength: $E = \dfrac{hc}{\lambda}$