Thermodynamics Lecture Notes PDF

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THERMODYNAMICS Thermodynamics Lecture 1: Understand the theory of thermodynamics and its use in describing energy related changes What is thermodynamics?    

Energy flow in natural systems Allows us to understand biologically relevant processes e.g. transport of oxygen, dissolution ad distribution of drugs in the body Described properties of a system on a macroscopic scale Few variables are needed to fully describe a system  temperature, volume & pressure

Useful values: 

Temperature (K) = Temperature (°C) + 273.15



Boltzmann Constant (KB) - KB = 1.380658 x 1023 JK-1

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Molar Gas constant (R) - R = 8.3145101 J K-1mol-1 (DEFAULT) - R = 0.0820578 dm3 atm K-1mol-1 - R = 8.314 x 10-2 L bar K-1mol-1

Understand the Equations of State and Gas Law Equations of State:    

Useful for describing the properties of a material on a large scale Does not take into account intermolecular interactions e.g. dipole-dipole interactions Does not take into account the size of atoms Relate pressure, temperature and volume

9 9 9 9

(1) PV T T = temperature P = pressure V = volume α = constant (can be determined from (1))

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Since α is directly proportional to the mass of gas used:

9 9 

(2) R n = number of moles of gas R = constant (independent of the size of system)

Thus, the equation of state becomes:

9

(3) PV RT Ideal Gas Law

Example: An average sized person has about 5L of blood, which circulates the system in about a minute. Each litre of blood can carry 0.20 L of oxygen, measured ar 273 K and 1.0 atm pressure. How many moles of oxygen can the blood transport in a minute?

How many oxygen molecules are transported from the lungs per minute?

When there is a mixture of gases: 

The Ideal Gas Law becomes: PV = ΣnRT

9 9

Pi = partial pressure of each gas Xi = mole fraction of each gas

Example: A patient requires a mixture of 2L O2 supplied at 1.5 bar, 3L He supplied at 2.5 bar and 1L N2 supplied at 1.0 bar. Calculate the total pressure and the partial pressure of each gas in the mixture

The 1st Law of Thermodynamics Energy cannot be created or destroyed, rather transformed from one form to another   

The 1st law relates ΔU (internal energy) to heat (q) and work (w) It also introduces reversible and irreversible reactions it allows us to calculate enthalpy ΔH of a constant pressure system

What is ΔU?   

Change (Δ) in internal energy of a system It is a total of  kinetic energy, potential energy, internal energy of vibration & rotation and internal energy of chemical bond According to the 1st law, ΔU of an isolated system is constant

ΔUtotal = ΔUsystem + ΔUsurroundings = 0 ΔUsystem = - ΔUsurroundings  

Whatever happens to the system, the opposite will happen to surrounding If temperature of system decreases, temperature of surroundings must increase by exactly the same amount  achieved by work (w) and heat (q)

1st law: ΔU = q + w What is work?   

Any quantity of energy that is transferred across a boundary between system and surroundings, raising the potential energy of a mass in the surrounding ‘Work’ increases internal energy of a system ( ↑ internal energy of bonds = ↑ internal energy of molecule =↑ internal energy of system) It is calculated using the ‘path integral’

w = ∫F.dI 9 9 

w > 0 if ΔU > 0 If w > 0, work is done on the system by the surroundings

Some common types of work:

Example: An aqueous solution of glucose e is divided into 2 chambers by a membrane permeable to glucose. In one chamber, the concentration of glucose is 0.010 M and in the other, 0.050. Calculate the work required to transport 2g of glucose from the 0.010 M chamber to the 0.050 M chamber. (opposite of osmosis is happening; going from low high conc, hence work has to be done)

What is heat?    

Energy that flows (transfer of energy) between a system and its surrounding as a result of temperature difference If temperature of surrounding is lowered, q is positive (+) Temperature: a measure of the average kinetic energy of the molecules in a sample It is measured using a thermometer

How do we calculate ΔU? 

For constant volume conditions:

w = - ∫Pexternal dV = 0 

Thus:

ΔU = qv (constant volume)

Understand the concept of enthalpy ΔH  

Since P, V and U are all state functions, U + PV is also a state function This new state function is (ENTHALPY  heat of formation):

ΔH ≡ ΔU +PV 9 9



Total internal energy + any contribution that we get from change in P or V Important: chemical reactions are carried out generally at constant pressure, not constant volume  it is the volume change that sometimes has greatest impact

As such,

ΔH = qp (constant pressure) 9

ΔH can be used as a measure of heat flow in a chemical reaction

Everything in thermodynamics is a result of change in internal energy

QUICK QUESTIONS 1) What is the equation of state for an ideal gas? PV = nRT 2) Why are there different values of R and what are they used for? R has different values depending on the different units given in the question e.g. pressure 3) What units are always used for temperature? Kelvin (K) 4) Does the ideal gas law describe properties of a material on a large or small scale? Large scale (macroscopic) 5) What is the first law of thermodynamics? Energy cannot be created or destroyed, rather transformed from one form to another 6) What is U and what are the 4 sources it comes from? U is internal energy of a system. The 4 sources it comes from are: Kinetic energy, Potential energy, Internal energy of vibration and rotation, Internal energy of chemical bond 7) How does ΔU relate to system and surroundings? ΔU = 0  If heat is lost from the surrounding, the equivalent amount of heat must be taken in by the system (balance) 8) Define the two terms that contribute to ΔU Work (w) = energy that is transferred across a boundary between system and surroundings Heat (q) = energy that is transferred (between system & surrounding or vice versa) as a result of a temperature difference between surrounding and system 9) When is heat positive? When energy is absorbed by the system and the temperature of surrounding is lowered 10) Define enthalpy and explain how it relates to heat Enthalpy is a measure of heat transfer, defned by; H ≡ U + PV. Since we do reactions at constant volume & pressure, ΔH = qp

Thermodynamics Lecture 2: What is thermochemistry?  

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It is the study of heat flow into and out of a reaction system and deduces energy stores in chemical bonds (e.g. how much energy do I need to break a chemical bond?) For a reaction carried out at constant temperature & pressure: - The heat that flows out of or into a system = ΔU for that reaction - The heat flow into or out of a system = ΔH of the reaction In this case, ΔH = ΔU - Enthalpy can be used as a measure of internal energy for a molecule

The standard enthalpy of formation: 

It is a measure of energy released/consumed when 1 mole of a substance is created under standard conditions from its pure elements

ΔH°f 9 9 9 9

Δ = a change in enthalpy ° = signifes that it’s a standard enthalpy change f = indicates that the substance is formed from its elements ΔH°f = 0 for an element in its standard state

State function: 

Since H is a state function, the reaction enthalpy can be written as the enthalpies of formation of products – enthalpies of the formation of reactants

ΔH = Hp - Hr 

A state function is: - A function whose values depend only upon the initial and final states of the system - E.g. internal energy (U), enthalpy (H), entropy (S)

Chemical bonds, energy and reactions:    

All chemical reactions are unimolecular or bimolecular Energy stored in chemical bonds is taken up or released in chemical reactions A chemical reaction (bond) is the sharing or electrons  this is done to get the electrons in a lower energy state Dipoles are formed as a result of differing electronegativities in an atom

Keesom forces 

 

INTERMOLECULAR FORCES Debye forces

Permanent dipole-dipole  interactions between molecules Dipoles aligned = repulsive force Dipoles not aligned = attractive force 

When molecule with a permanent dipole meets a non-dipole molecule, the electron clouds around non-dipole molecule get polarized  temp dipole induced Permanent dipole induces a dipole in a molecule

London forces 



Both molecule approach each other and induce dipoles in each other from the electron clouds around the molecule Temporary imbalance of electrons of a non-polar molecule induce temporary dipole in another molecule

Thermochemistry cont.      

½ N2 (g) + 3/2 H2 (g)  NH3 (g) To determine H and U in the equation above, use a thermometer If the reaction is carried out in a constant pressure reaction vessel, the temperature of the water bath will either increase or decrease If water temperature increases, heat flows from system  surrounding = - ΔH (exo) If water temperature decreases, heat flows from surrounding  system = + ΔH (endo) The total measured ΔU (or ΔH) will include a contribution from both process: - ΔU from constant T and P and - ΔU from the change in T and P

Enthalpy: 

Enthalpy change:

ΔH = ΣΔHp - ΣΔHr 9

H = enthalpy per mole

Example: Using ΔH°f (C12H22O11, s) = -2226.1 kJ mol-1 , ΔH°f (CO2, aq) = -412.9 kJ mol-1, ΔH°f (H2O, l) = -285.8 kJ mol-1, calculate the heat evolved in oxidation of 1 mol sucrose at 1 bar pressure

   

It is not possible to measure the ΔH°f for every compound  due to incomplete side reactions In 1840, Hess showed that ΔH°f only depends on the initial and final states of a system Hess proposed the Law of Constant Heat Summation Summation of heats or reactants and heats of products

ΔH°f = ΔH°fp - ΔH°fr 

“The enthalpy change for any sequence of reactions that sum to the same overall reaction, is the same”

Example: Consider urea H2H(CO)NH2 C(s) + N2 (g) + 2H2 (g) + ½O2(g)  H2H(CO)NH2 (s) 9 9

It is highly unlikely that this reaction would work and only yield urea So, we start with urea and we fnd the heat evolved when urea is oxidised - This will give us a theoretical idea of the formation heat of urea

H2H(CO)NH2 (s) + 7/2O2(g)  CO2(g) + 2NO2 (g) + 2H2O (l) C(s) + O2(g)  CO2(g)

ΔH°I

ΔH°II

H2(g) + ½O2(g)  H2O(l) ΔH°III ½N2 (g) + O2 (g)  NO2 (g) ΔH°IV Combine all of these: CO2(g) + 2NO2 (g) + 2H2O (l)  H2H(CO)NH2 (s) + 7/2O2(g) - ΔH°I (- because it is the opposite) C(s) + O2(g)  CO2(g)

ΔH°II

2 x [H2(g) + ½O2(g)  H2O(l)] 2 x ΔH°III 2 x [½N2 (g) + O2 (g)  NO2 (g)] 2 x ΔH°IV Therefore, ΔH°f = - ΔH°I + ΔH°II + 2ΔH°III + 2ΔH°IV *Remember: it isn’t necessary for the reaction to proceed at 298.15K. ΔH°f can be determined this way as long as the fnal temperature returns to 298.15K.

Example: The average bond enthalpy of the O-H bond is defined as ½ the enthalpy change for the reaction H2O(l)  2H(g) + O(g). The formation enthalpies for H(g) and O(g) are 218.0 and 249.2 kJmol-1 respectively, at 298.15K. The formation enthalpy for H2O(l) is -241.8kJmol-1 at the same temperature. Calculate the average bond enthalpy of the O-H bond in water at 298.15K. H2O(l)  H2(g) + ½O2(g) ΔH°= - 241.8 kJmol-1 H2(g)  2H(g) ½O2(g)  O(g)

ΔH°= 2 x 218.0 kJmol-1 ΔH°= 249.2 kJmol-1

[(2 x 218.0) + (249.2)] – [-241.8] = 927.0 H2O(l)  2H(g) + O(g)

ΔH°= 927.0 kJmol-1

Therefore, the average bond enthalpy of the O-H bond = ½ x 927.0 kJmol-1 = 461.0 kJmol-1 - We frst found the ΔH° of the whole molecule and then halved it Determine the average bond energy, ΔU, of the O-H bond in water at 298.15K. Assume ideal gas behaviour. ΔH° ≡ ΔU° +PV ΔU° = ΔH° - PV Since PV = nRT: ΔU° = ΔH° - nRT = 927.0 kJmol-1 – 2 x 8.314 Jmol-1K-1 x 298.15K = 922.0 kJmol-1 Therefore, ΔU° for O-H bond = ½ x 922.0 kJmol-1 = 461.0 kJmol-1 Example: Is Hell exothermic (gives off heat) or endothermic (absorbs heat)? Boyle's Law states that in order for the temperature and pressure in Hell to stay the same, the volume of Hell has to expand proportionately as souls are added. Boyles law: assuming constant mass and temperature, the pressure of a gas tends to increase as volume of the container decreases (P1V1 = P2V2)

The 2nd Law of Thermodynamics In a system, a process that occurs will tend to increase the total entropy of the universe Other versions:   

“Heat generally cannot spontaneously flow from a material at a lower temperature to a material at a higher temperature” “It is impossible to convey heat completely into work in a cyclic process” There are 3 things that link the 1st law to the 2nd law: - Spontaneous vs. non-spontaneous reactions - Reversible vs. irreversible reactions - Entropy

Spontaneous & non spontaneous reactions:   

  

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 

Occur with no outside interference (no work needs to be done) Not putting energy in, not taking energy out E.g. if we open the valve, the gas distributes itself evenly. The reverse will not happen spontaneously Processes that are spontaneous in one direction are not spontaneous in the reverse direction e.g rusting nail This is temperature dependent E.g: - Above 0°C  spontaneous for ice to melt - Below 0°C  spontaneous for ice to freeze Reactions are favourable when they result in a decrease in enthalpy and an increase in entropy of the system E.g. bonfire  exothermic (decrease in the energy of the system as energy is released to surroundings as heat). Entropy of the system increases during a combustion reaction. A non-spontaneous reaction does not favour the formation of products at the given set of conditions.

Reversible reactions: 

In a reversible process, the system changes in such a way that the system and surroundings can be put back in their original states by exactly revering the process

Irreversible reactions:   

Cannot be undone by exactly reversing the change to the system All spontaneous and real processes are irreversible Non spontaneous reactions can also be irreversible & can be reversible if work is applied

QUICK QUESTIONS

1) Define exothermic When heat goes from system  surrounding 2) Define endothermic When heat goes from surrounding  system 3) Which one is numerically positive? Why? Endothermic. This is because when heat (q) goes into the system, and the temperature of surroundings is lowered, q will be positive. 4) What is ΔH°f? When is it 0? It is the standard enthalpy of formation. It is 0 when an element is in its standard state (e.g. C in solid form, O in gas form) 5) What is Hess’ Law? Do I understand the process of combining reactions to produce an overall reaction (i.e. formation of urea)? The Law of Constant Heat Summation: The enthalpy change for any sequence of reactions that sum to the same overall reaction, is the same. 6) What is the 2nd law of thermodynamics? Entropy of the universe is always increasing 7) What is entropy? Degree of disorder/randomness in a system 8) What is the difference between spontaneous and non-spontaneous reactions? A spontaneous reaction is one that does not require any work to be done and favours formation of products under conditions in which reaction is occurring. A non-spontaneous reaction is one that does not favour the formation of products at the given set of conditions. 9) Explain temperature dependence of spontaneous reactions Whatever is spontaneous at one temperature is will be non-spontaneous at a different temperature e.g. melting ice is spontaneous above 0°C but not below 0°C, as it will freeze. Thermodynamics Lecture 3:

Understand the concept of entropy ΔS      

Entropy (s) came about in 19th century by Rudolph Clausius He was convinced of the signifcance of the ratio of heat (q) delivered and the temperature (T) at which it is delivered (heat had temperature dependence) It is a measure of disorder/randomness in a system It is related to the various modes of motion in molecules It is a state function It is temperature dependent

ΔS = Sfinal - Sinitial 

For a process occurring at constant temperature (isothermal):

ΔS = qrev 9 9

qrev = heat that is transferred when the process is carried out reversibly at a constant temperature T = temperature in Kelvin

Entropy and the second law: The 2nd law of thermodynamics: the entropy of the universe does not change for reversible processes and increase for spontaneous processes 

For reversible (ideal) systems:

ΔSuniverse = Ssystem + Ssurrounding = 0 9



If we lose entropy of system, we will gain entropy of surrounding by exactly the same amount and vice versa

For irreversible (real, spontaneous systems):

ΔSuniverse = Ssystem + Surrounding > 0 

The entropy of the universe is increasing (real, spontaneous process) - BUT entropy can decrease for individual systems Entropy on the molecular scale:   

Ludwig Boltzmann described the concept of entropy on the molecular level Temperature is a measure of the average kinetic energy of the molecules in a sample Molecules exhibit several types of motion:

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  

Translational: movement of entire molecule from one place to another Vibrational: periodic motion of atoms within a molecule Rotational: rotation of the molecule about an x axis or rotation about σ bonds

Each thermodynamic state has a specifc number of microstates (W) associated with it More microstates = greater entropy Entropy is:

S = KB lnW 9

KB = 1.380658 x 1023 JK-1

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What is a microstate? - If molecule is rotating, every degree it rotates is microstate - If molecule is stretching, every cm it stretches is a microstate

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What can increase the number of microstates? - more particles  more states  higher entropy - higher temp (greater degree of freedom)  more energy states  higher entropy - less structure (gas vs. solid)  more states  higher entropy

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Hence, number of microstates, and therefore, entropy, increases with: - increase in temperature - greater volume (gases)  more freedom for molecule to move around - increasing number of independently moving molecules  greater number of molecules (moles)  greater number of bonds

Entropy and physical states:      

Entropy increases with the freedom of motion of molecules S (g) > S (l) > S (s) Dissolution of a solid: ions have more entropy (more states) BUT, some water molecules have less entropy (they are grouped around ions) 99.99% of the time, there is an overall increase in S during dissolution EXECEPTION: highly charged ions that make a lot of water molecules align around them - In the image, we have an ionic solid  there are delta - oxygens around the + charge - Similarly, the delta + hydrogens have coordinated around the – charge - Let’s say, we have liberated 1 + and 1 – charge  increased entropy of the system by a value of 2 (because 1 + and 1 - charge has moved out of the solid)

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They had no entropy when bonded in solid state, but after rem...