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Tuesday 11 June 2013

Chemical Thermodynamics

The scientific discipline that intersects the areas of chemistry and physic is commonly known as physical chemistry, and it is in that area that a thorough study of thermodynamics takes place. Physics concerns itself heavily with the mechanics of events in nature. Certainly changes in energy -- however measured, whether it be heat, light, work, etc. -- are clearly physical events that also have a chemical nature to them. Thermodynamics is the study of energy changes accompanying physical and chemical changes. The term itself clearly suggests what is happening -- "thermo", from temperature, meaning energy, and "dynamics", which means the change over time. Thermodynamics can be roughly encapsulated with these topics:



Heat and Work

Heat and work are both forms of energy. They are also related forms, in that one can be transformed into the other. Heat energy (such as steam engines) can be used to do work (such as pushing a train down the track). Work can be transformed into heat, such as might be experienced by rubbing your hands together to warm them up. Work and heat can both be described using the same unit of measure. Sometimes the calorie is the unit of measure, and refers to the amount of heat required to raise one (1) gram of water one (1) degree Celsius. Heat energy is measured in kilocalories, or 1000 calories. Typically, we use the SI units of Joules (J) and kilojoules (kJ). One calorie of heat is equivalent to 4.187 J. You will also encounter the term specific heat, the heat required to raise one (1) gram of a material one (1) degree Celsius. Specific heat, given by the symbol "C", is generally defined as:
C =
q

MΔT

Where:
C = specific heat in cal/g-°C
q = heat added in calories,
m = mass in grams
ΔT = rise in temperature of the material in °C.
The value of C for water is 1.00 cal/g-°C.
The values for specific heat that are reported in the literature are usually listed at a specific pressure and/or volume, and you need to pay attention to these settings when using values from textbooks in problems or computer models.
Example Problem: If a 2.34 g substance at 22°C with a specific heat of 3.88 cal/g-°C is heated with 124 cal of energy, what is the new temperature of the substance?

Answer:
  ΔT =
q

MC
  ΔT =
(124)

(2.34)(3.88)
= 13.7°C
  new T = 22 + 13.7 = 35.7°C
Two other common heat variables are the heat of fusion and the heat of vaporization. Heat of fusion is the heat required to melt a substance at its melting temperature, while the heat of vaporization is the heat required to evaporate the substance at its boiling point.
Chemical work is primarily related to that of expansion. In physics, work is defined as:
w = d × f
Where:
w = work, in joules (N×m) (or calories, but we are using primarily SI units)
d = distance in meters
f = opposing force in Newtons (kg*m/s2)
In chemical reactions, work is generally defined as :
w = distance × (area × pressure)
The value of distance times area is actually the volume. If we imagine a reaction taking place in a container of some volume, we measure work by pressure times the change in volume.
w = ΔV × P
Where:
ΔV is the change in volume, in liters
If ΔV=0, then no work is done.
Example Problem: Calculate the work that must be done at standard temperature and pressure (STP is 0°C and 1 atm) to make room for the products of the octane combustion:
2 C8H18 + 25 O2 --> 16 CO2 + 18 H2O

Answer:
  Knowing that 25 moles of gas are replaced by 34 moles of gas in this reaction, we can
  calculate a net increase of 9 moles of gas. Knowing the molar volume of an ideal gas at
  STP (22.4 L/mol), the change in volume and the work of expansion can be calculated
  dV = 9 moles ∗ 22.4 L/mol = 202 L
  The external pressure is 1.0 atm (standard pressure), so the work required is:
  w = dV ∗ P = 202 L ∗ 1.00 atm = 202 l-atm
  Using the conversion factor of 1 L-atm = 101 J, the amount of work in joules is:
  w = 202 L-atm ∗ 101 j/L-atm = 2000 J, or 2kJ of energy

Energy

You might remember the first law of thermodynamics: energy cannot be created or destroyed. Energy can only change form. Chemically, that usually means energy is converted to work, energy in the form of heat moves from one place to another, or energy is stored up in the constituent chemicals. You have seen how to calculate work. Heat is defined as that energy that is transferred as a result of a temperature difference between a system and its surroundings. Mathematically, we can look at the change in energy of a system as being a function of both heat and work:
ΔE = q - w
Where:
ΔE is the change in internal energy of a system
q is the heat flowing into the system
w is the work being done by the system
If q is positive, we say that the reaction is endothermic, that is, heat flows into the reaction from the outside surroundings. If q is negative, then the reaction is exothermic, that is, heat is given off to the external surroundings.
You might also remember the terms kinetic energy and potential energy. Kinetic energy is the energy of motion -- the amount of energy in an object that is moving. Potential energy is stationary, stored energy. If you think of a ball sitting on the edge of a table, it has potential energy in the energy possible if it falls off the table. Potential energy can be transformed into kinetic energy if and when the ball actually rolls off the table and is in motion. The total energy of the system is defined as the sum of kinetic and potential energies.
In descriptions of the energy of a system, you will also see the phrase "state properties". A state property is a quantity whose value is independent of the past history of the substance. Typical state properties are altitude, pressure, volume, temperature, and internal energy.

Enthalpy

Enthalpy is an interesting concept: it is defined by its change rather than a single entity. A state property, the word enthalpy comes from the Greek "heat inside". If you have a chemical system that undergoes some kind of change but has a fixed volume, the heat output is equal to the change in internal energy (q = ΔE). We will define the enthalpy change, ΔH, of a system as being equal to its heat output at constant pressure:
dH = q at constant pressure
Where:
ΔH = change in enthalpy
We define enthalpy itself as:
H = E + PV
Where:
H = enthalpy
E = energy of the system
PV = pressure in atm times volume in liters
You will not need to be able to calculate the enthalpy directly; in chemistry, we are only interested in the change in enthalpy, or ΔH.
ΔH = Hfinal - Hinitial or ΔH = H(products) - H(reactants)
Tables of enthalpies are generally given as ΔH values. Example Problem: Calculate the ΔH value of the reaction:

HCl + NH3 → NH4Cl
(ΔH values for HCl is -92.30; NH3 is -80.29; NH4Cl is -314.4)
Answer:
  ΔH = ΔHproducts - ΔHreactants
  ΔHproducts = -314.4
  ΔHreactants = -92.30 + (-80.29) = -172.59
  ΔH = -314.4 - 172.59 = 141.8
We can also represent enthalpy change with the equation:
ΔH = ΔE + P ΔV
Where:
ΔV is the change in volume, in liters
P is the constant pressure
If you recall, work is defined as P ΔV, so enthalpy changes are simply a reflection of the amount of energy change (energy going in or out, endothermic or exothermic), and the amount of work being done by the reaction. For example, if ΔE = -100 kJ in a certain combustion reaction, but 10 kJ of work needs to be done to make room for the products, the change in enthalpy is:
ΔH = -100 kJ + 10 kJ = -90 kJ
This is an exothermic reaction (which is expected with combustion), and 90 kJ of energy is released to the environment. Basically, you get warmer. Notice the convention used here -- a negative value represents energy coming out of the system.
You can also determine ΔH for a reaction based on bond dissociation energies. Breaking bonds requires energy while forming bonds releases energy. In a given equation, you must determine what kinds of bonds are broken and what kind of bonds are formed. Use this information to calculate the amount of energy used to break bonds and the amount used to form bonds. If you subtract the amount to break bonds from the amount to form bonds, you will have the ΔH for the reaction.
Example Problem: Calculate ΔH for the reaction:

N2 + 3H2 → 2NH3
(The bond dissociation energy for N-N is 163 kJ/mol; H-H is 436 kJ/mol; N-H is 391 kJ/mol)
Answer:
  ΔH = ΔHproducts - ΔHreactants
  To use the bond dissociation energies, we must determine how many bonds
  are in the products and the reactants. In NH3 there are 3 N-H bonds so in 2 NH3
  there are 6 N-H bonds. In N2 there is 1 N-N bond and in 3H2 there are 3 H-H bonds.
  ΔHproducts = 6(391) = 2346
  ΔHreactants = 163 + 3(436) = 1471
  ΔH = 2346 - 1471 = 875

Entropy

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