Physical Chemistry/Printable version

Physical Chemistry

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Introduction to Thermodynamics

Thermodynamics is a misnomer since we will only consider equilibrium phenomena, not time-dependence like in kinetics. Thermodynamics deals with things like energy, entropy, volume, heat, work, efficiency (ideal), free energy, chemical potential, pressure, temperature. It was developed to explain steam engines back in the 1800's. It has come a long way since then, able to explain a vast array of phenomena in chemistry, physics, and biology. It is an old, but beautiful theory. It is important in chemistry.

Sample problem

An ideal monoatomic gas at 1 bar is expanded in a reversible adiabatic process to a final pressure of ${\displaystyle {\begin{matrix}{\frac {1}{2}}\end{matrix}}}$  bar. Calculate ${\displaystyle q\;}$  per mole, ${\displaystyle w\;}$  per mole, and ${\displaystyle \Delta {\overline {U}}}$ .

To solve this, we need to apply the fact that ${\displaystyle {\frac {T_{2}}{T_{1}}}=\left({\frac {P_{2}}{P_{1}}}\right)^{\gamma -1/\gamma }}$ .

So, substitute and solve for ${\displaystyle T_{2}}$

${\displaystyle {\frac {T_{2}}{298.15\ K}}=\left({\frac {0.5\ bar}{1.0\ bar}}\right)^{\begin{matrix}{\frac {{\begin{matrix}{\frac {5}{3}}\end{matrix}}-1}{\begin{matrix}{\frac {5}{3}}\end{matrix}}}\end{matrix}}}$

${\displaystyle T_{2}={225.96\ K}}$

Now, we use

${\displaystyle w=\int _{T_{1}}^{T_{2}}{\overline {C}}_{v}\,dT}$

Which will give us ${\displaystyle w}$

${\displaystyle w=\int _{298.15\ K}^{225.96\ K}{\overline {C}}_{v}\,dT={\begin{matrix}{\frac {3}{2}}\end{matrix}}R\left(225.96\ K-298.15\ K\right)=-900.39\ J\ mol^{-1}}$

Hence, we have all of our answers...

Adiabatic means ${\displaystyle \Delta {\overline {U}}=w}$ , so ${\displaystyle q=0\ J\ mol^{-1}}$  and ${\displaystyle w=-900.39\ J\ mol^{-1}}$  while ${\displaystyle \Delta {\overline {U}}=-900.30\ J\ mol^{-1}}$

The Zeroth Law of Thermodynamics

This law of thermodynamics was developed after the establishment of the first and second laws of thermodynamics. The law is involved in defining temperature.

Let us consider three systems, A, B, and C. If systems A and C are in equilibrium, and B and C are in equilibrium, then A and B will be found to be in thermal equilibrium when connected by a heat conductor. This associative property of systems is the Zeroth Law of Thermodynamics.

The Zeroth Law of Thermodynamics states that if two systems are in thermodynamic equilibrium separately with a third system then both these systems are said to be in thermodynamic equilibrium with each other.

State Functions

The following demonstrates what's a state function and what's not a state function.

${\displaystyle q_{rev}\;}$  is not exact differential for a gas obeying van der Waals' equation, but ${\displaystyle {\frac {q_{rev}}{T}}}$  is as demonstrated below:

${\displaystyle dq_{rev}\;=\left({\frac {\partial U}{\partial T}}\right)_{v}dT+\left[P_{ext}+\left({\frac {\partial U}{\partial V}}\right)_{T}\right]dV}$

We assume quasistatic situation, so ${\displaystyle P_{ext}=P\;}$ .

${\displaystyle dq_{rev}\;=\left({\frac {\partial U}{\partial T}}\right)_{v}dT+\left[P+\left({\frac {\partial U}{\partial V}}\right)_{T}\right]dV}$

${\displaystyle =C_{v}dT+\left[P+\left({\frac {\partial U}{\partial V}}\right)_{T}\right]dV}$

${\displaystyle =C_{v}dT+\left[P+\left({\frac {a}{{\overline {V}}^{2}}}\right)\right]dV}$

${\displaystyle =C_{v}dT+\left({\frac {RT}{{\overline {V}}-b}}\right)dV}$

Now, you take the cross partial derivatives.

${\displaystyle \left({\frac {\partial C_{v}}{\partial V}}\right)_{T}=0}$

${\displaystyle \left({\frac {\partial \left({\frac {RT}{{\overline {V}}-b}}\right)}{\partial T}}\right)_{V}={\frac {R}{{\overline {V}}-b}}}$

They are not equal; hence, ${\displaystyle q_{rev}\;}$  is not exact differential (not a state function).

However, if we take ${\displaystyle {\frac {q_{rev}}{T}}}$  it will be exact differential (a state function).

${\displaystyle {\frac {dq_{rev}}{T}}={\frac {C_{v}}{T}}dT+\left({\frac {R}{{\overline {V}}-b}}\right)dV}$

Take the cross partial derivatives.

${\displaystyle \left({\frac {\partial \left({\frac {C_{v}}{T}}\right)}{\partial V}}\right)_{T}=0}$

${\displaystyle \left({\frac {\partial \left({\frac {R}{{\overline {V}}-b}}\right)}{\partial T}}\right)_{V}=0}$

Both are equal making ${\displaystyle {\frac {q_{rev}}{T}}}$  exact differential (a state function).

Thermodynamic Processes for an Ideal Gas

Thermodynamic Processes for an Ideal Gas

Let us first consider the expansion and compression of an ideal gas from an initial volume V1 to a final volume V2 under constant-temperature (isothermal) conditions. From the definition of pressure-volume work, we have:
${\displaystyle w=-\int _{V_{1}}^{V_{2}}P_{ext}dV}$
If we assume the process occurs reversibly, then ${\displaystyle P_{ext}=P}$  where P is the pressure of the gas; therefore
${\displaystyle w=-\int _{V_{1}}^{V_{2}}PdV}$
Now let us use the equation of state for an ideal gas, which is ${\displaystyle PV=nRT}$  where n is the number of moles of gas, R is the ideal gas constant, and T is the absolute (Kelvin) temperature. Substituting for P, we get
${\displaystyle w=-\int _{V_{1}}^{V_{2}}{\frac {nRT}{V}}dV}$
If we have a closed system, then n is a constant; moreover, since we have stipulated that this process occurs under isothermal conditions, T is also a constant. Therefore, the integral simplifies to:
${\displaystyle w=-nRT\int _{V_{1}}^{V_{2}}{\frac {dV}{V}}}$
This integral may be solved to yield
${\displaystyle w=-nRT\ln \left({\frac {V_{2}}{V_{1}}}\right)\;}$

Next we consider the change in the internal energy of the ideal gas upon isothermal expansion/compression. It is helpful to review the two key assumptions that underlie the ideal gas model:

1. Ideal gas particles are point particles.
2. Ideal gas particles have no interparticular forces.

From these two statements, we now contemplate what would happen to the internal energy (U) of an ideal gas upon isothermal expansion or compression. Because ideal gas particles are point particles, if the gas sample is compressed, the particles will not experience any change in energy because they have no internal volume to get "squeezed". Similarly, because they have no interparticular forces, they do not experience any change in attraction or repulsion between different particles if they are squeezed closer together. Hence, we may (correctly) reason that for an isothermal process on an ideal gas, the change in internal energy is zero.
${\displaystyle \Delta U=0\;}$  (isothermal process, ideal gas)
(This same result may be rigorously proved using the thermodynamic master equations, or by using statistical thermodynamics.) Note that because U is a state function, ${\displaystyle \Delta U}$  will always be equal to zero for any isothermal process on an ideal gas, whether or not it occurs reversibly.
Because of the First Law of Thermodynamics,
${\displaystyle \Delta U=q+w\;}$
and because ${\displaystyle \Delta U=0}$ , the energy change associated with heat upon isothermal reversible expansion or compression is just the negative of the work, or
${\displaystyle q=-w=nRT\ln {\frac {V_{2}}{V_{1}}}}$
. In a similar fashion, other results may be derived for processes on ideal gases. A summary of results is given below.

Equation Table (${\displaystyle PV^{m}=constant}$ )

Constant Pressure Constant Volume Isothermal Adiabatic
Variable ${\displaystyle \Delta P=0\;}$  ${\displaystyle \Delta V=0\;}$  ${\displaystyle \Delta T=0\;}$  ${\displaystyle q=0\;}$
${\displaystyle m\;}$  ${\displaystyle 0\;}$  ${\displaystyle \infty \;}$  ${\displaystyle 1\;}$  ${\displaystyle \gamma ={\frac {C_{p}}{C_{v}}}={\frac {5}{3}}\;}$
[for a monoatomic ideal gas]
Work
${\displaystyle {\begin{matrix}w=-\int _{V_{1}}^{V_{2}}PdV\end{matrix}}}$
${\displaystyle -P\left(V_{2}-V_{1}\right)\;}$  ${\displaystyle 0\;}$  ${\displaystyle -nRT\ln \left({\frac {V_{2}}{V_{1}}}\right)\;}$  ${\displaystyle C_{v}\left(T_{2}-T_{1}\right)\;}$
Heat Capacity, ${\displaystyle C\;}$  ${\displaystyle C_{p}=(5/2)nR\;}$  ${\displaystyle C_{v}=(3/2)nR\;}$  ${\displaystyle \infty \;}$  ${\displaystyle C_{p}\;}$  or ${\displaystyle C_{v}\;}$
Internal Energy, ${\displaystyle \Delta U=q+w\;}$  ${\displaystyle q+w\;}$
${\displaystyle q_{p}+P\Delta V\;}$
${\displaystyle q\;}$
${\displaystyle w=0\;}$
${\displaystyle C_{v}\left(T_{2}-T_{1}\right)\;}$
${\displaystyle 0\;}$
${\displaystyle q=-w\;}$
${\displaystyle w\;}$
${\displaystyle q=0\;}$
${\displaystyle C_{v}\left(T_{2}-T_{1}\right)\;}$
Enthalpy, ${\displaystyle \Delta H\;}$
${\displaystyle H=U+PV\;}$
${\displaystyle C_{p}\left(T_{2}-T_{1}\right)\;}$  ${\displaystyle q_{v}+V\Delta P\;}$  ${\displaystyle 0\;}$  ${\displaystyle C_{p}\left(T_{2}-T_{1}\right)\;}$
Entropy
${\displaystyle {\begin{matrix}\Delta S=-\int _{T_{1}}^{T_{2}}{\frac {C}{T}}dT\end{matrix}}}$
${\displaystyle C_{p}\ln \left({\frac {T_{2}}{T_{1}}}\right)\;}$  ${\displaystyle C_{v}\ln \left({\frac {T_{2}}{T_{1}}}\right)\;}$  ${\displaystyle nR\ln \left({\frac {V_{2}}{V_{1}}}\right)\;}$  ${\displaystyle 0\;}$

Molar Thermodynamic Properties for an Ideal Gas

We are given that the molar Gibbs free energy of an ideal is ${\displaystyle {\bar {G}}={\bar {G}}^{\circ }+RT\ln {\frac {P}{P^{\circ }}}\;}$ . Let's try and derive ${\displaystyle {\bar {U}}={\bar {U}}^{\circ }={\bar {H}}^{\circ }-RT\;}$ ; noting that ${\displaystyle {\bar {U}}^{\circ }={\bar {G}}^{\circ }+T{\bar {S}}^{\circ }-RT\;}$  and assuming that the internal energy ${\displaystyle U\;}$  and enthalpy ${\displaystyle H\;}$  of an ideal gas are independent of pressure and volume.

Introduction to QM

Quantum Mechanics

Quantum mechanics is a branch of physics developed in the first part of the 20th century that was highly successful in explaining the behavior of atoms, molecules and nuclei. Developed between 1900 and 1930 and later combined with the general special theory of relativity, it revolutionized the field of physics. The new concepts, which were the particle properties of radiation, the wave properties of matter, quantization of physical properties and the idea that one can no longer know exactly where a single particle such as an electron is at any one time were necessary to explain all of the new experimental evidence that was available at the time.

At the turn of the 20th century, classical theories were well-established for thermodynamics, optics, electromagnetism, and motion of objects. Classical predictions were very accurate for nearly every system that had been studied, and there was a common belief that all of the conceptual problems in chemistry and physics were essentially solved (although solving the equations for a large complex system could be computationally challenging).

A series of experimental results that could not be explained using classical mechanics ultimately led to the development of quantum mechanics. As a result, much of modern chemistry has been explained using the principles of quantum mechanics: for example, atomic structure, molecular bonding, and all types of spectroscopy are directly related to quantum chemistry. Quantum mechanics has also been applied to develop devices we use every day including semiconductors in the circuitry of our computers and lasers in our CD players.

From Classical to Quantum Mechanics

Classical View of Light

The classical view of light is simply a wave. This wave has two orthogonal (perpendicular) components. These are an electric field and a magnetic field. In the diagram shown the light is traveling to the right, the electric field is in the same plane as the page, and the magnetic field is traveling in and out of the page. All three of these are perpendicular to each other. The electric and magnetic component share anti-nodes. Anti-nodes are the points where the wave passes through the zero and changes sign. Since both waves are similar it is possible to refer to the wavelength of light, and not just the wavelength of the electric component of light.

Photoelectric Effect

The Photoelectric effect is the emission of electrons from metal when it is shined on by light. The amount of electrons released depends on the type of metal and the frequency of the light. For some metals, if the light is below a certain frequency, it will not emit any electrons. Light was originally thought as a form of energy, but for this to be true, electrons would have been emitted, regardless of the frequency.

Lattice Enthalpy

Lattice enthalpy is the enthalpy change when one mole of an ionic compound is formed from its gaseous ions under standard conditions.

M+ (g) + X- (g) ----> MX (s)


It is a measure of how strong the ionic lattice is within the compound. Lattice energies are exothermic. The more negative the value, the stronger the bonding.

Molecular Orbital Theory

Introduction

The valence electrons are considered to be associated with all the nuclei in the molecule. Thus, the atomic orbitals from different atoms must be combined to product molecular orbitals. Developed by Friedrich Hund and Robert S. Millikan in 1932, molecular orbital theory takes the view that a molecule is similar to an atom in one important respect. Both have energy levels that correspond to various orbitals that can be populated by electrons. In atoms, these orbitals are known as atomic orbitals; in molecules however, they are called molecular orbitals (MOs).

Assumptions to be considered

1. When two atoms approach each other, their atomic orbitals lose their identity and mutually overlap to form new orbitals called molecular orbitals (MOs).
2. The MOs are polycentric and are associated with the nuclei of all the atoms constituting the molecule. The electron probability distribution around the group of nuclei constituting the molecule is given by the molecular orbitals just as atomic orbitals give probability of finding an electron around the nucleus in an atom.
3. Only atomic orbitals of about the same energy and same symmetry interact significantly.
4. The total number of MOs produced is always equal to the total number of atomic orbitals contributed by the atoms that have combined.
5. When two atomic orbital overlap in-phase, it leads to an increase in the intensity of negative charge in region of overlap; the MO formed has lower potential energy than the separate atomic orbitals and is called a bonding molecular orbital.
6. When two atomic orbital overlap out-of-phase, it leads to an decrease in the intensity of negative charge between the nuclei; the MO formed has higher potential energy than the separate atomic orbitals and is called an anti-bonding molecular orbital. Electrons in this type of MO destabilize the bond between atoms.
7. Amount of stabilization of bonding orbitals equals the amount of destabilization of the anti-bonding orbitals.
8. Each MO can accommodate electrons according to Pauli's exclusion principle and Hund's rule of maximum multiplicity.

Formation of Molecular Orbitals by Linear Combination of Atomic Orbitals (LCAO)

Consider two atoms ${\displaystyle A}$  and ${\displaystyle B}$  which have atomic orbitals described by the wave functions ${\displaystyle \Psi _{a}}$  and ${\displaystyle \Psi _{b}}$ . When the electron clouds of these two atoms overlap as the atoms approach, the wave function for the molecule can be obtained by linear combination of the atomic orbitals ${\displaystyle \Psi _{a}}$  and ${\displaystyle \Psi _{b}}$ .

The atomic orbitals ${\displaystyle \Psi _{a}}$  and ${\displaystyle \Psi _{b}}$  combine to give rise to a pair of molecular orbitals ${\displaystyle \Psi _{g}}$  and ${\displaystyle \Psi _{u}}$ . The function ${\displaystyle \Psi _{g}}$  corresponds to the increased electron density in between the nuclei due to in-phase overlap and is, therefore, a bonding molecular orbital ${\displaystyle [\Psi _{g}=\Psi _{a}+\Psi _{b}]}$ . It is thus lower in energy than the original atomic orbitals. Similarly, the function ${\displaystyle \Psi _{u}}$  corresponds to the minimized electron density in between the nuclei. It is called the anti-bonding orbital which is higher in energy ${\displaystyle [\Psi _{g}=\Psi _{a}-\Psi _{b}]}$ .

Valence Bond Theory

Introduction

The Valence Bond Theory was proposed by Heitler and London (1927) and further explained by Pauling et. al..

According to this theory, atoms with unpaired electrons tend to combine with other atoms which also have unpaired electrons. In this way, the unpaired electrons are paired up and the atoms involved attain a stable electronic arrangement. The spins on two electrons however, must be opposite (antiparallel) because of the Pauli Exclusion Principle.

The basic idea of the theory is that a bond between two atoms is formed when two electrons with their spins paired are shared by two overlapping atomic orbitals, one orbital from each of the atoms joined by the bond. This electron pair becomes concentrated in the region of overlap and help the nucleus bind together. The amount of decrease in potential energy when the bond is formed is determined in part by the extent to which the orbitals overlap.

Orbital Overlap Concept

According to the above theory, a covalent bond is formed between two atoms in a molecule, when a half-filled valence atomic orbitals (AOs) of the two atoms containing unpaired electron overlap with one another and the electrons pair up in the overlapping region. Due to this overlapping process, electrons are localized between the atoms in the bond region. When the attractive forces (electron-nucleus of two different atoms) are stronger than the repulsive forces (electron-electron or nucleus-nucleus), energy is released, lowering the energy of the bonded atoms and increasing the molecule's stability.

Types of Overlapping and Nature of Covalent Bonds

The following bonds are observed based on type of overlap. Covalent bonds are classified into two main types: sigma (${\displaystyle \sigma }$ ) bonds and pi (${\displaystyle \pi }$ ) bonds.