1. Thermodynamics I

2. Energy Energy is the capacity to do work
The forms of energy most important in chemistry are:
thermal energy (kinetic energy) associated with the random motion of atoms and molecules
chemical energy (potential energy) stored in the structural units of chemical substances, i.e. , covalent bonds, ionic bonds, hydrogen bonds, etc.
Other forms exist, but they are of less interest to a chemist

3. Energy Energy can not be lost, nor created
The first law of thermodynamics says that the total amount of energy in the universe remains constant
The principle of conservation of energy
Energy is transferred from one place to another, but it is never lost nor created

4. Energy Changes in Chemical Reactions
Heat is a transfer of thermal energy from one object to a colder object
The system is the part of the universe under observation (in chemistry, it is often just the reactants and the products of a reaction)
The environment is the rest of the universe not included within the system

5. Energy Changes in Chemical Reactions
(a) An open system allows for the exchange of matter and energy with the environment
(b) A closed system allows for the exchange of energy with the environment, but not matter
(c) An isolated system does not allow the exchange of energy or matter with the environment

6. Energy Changes in Chemical Reactions
Any process that generates heat is called exothermic
For example, the combustion of hydrogen is exothermic, as the heat leaves the system and enters the environment

7. Energy Changes in Chemical Reactions
Any process that requires heat from the environment to be brought into ​​to the system is called endothermic
e.g., the melting of ice is endothermic, because you have to provide heat to melt the ice

8. Enthalpy
Most of the physical and chemical transformations of interest occur at a constant pressure
Enthalpy, H, is defined as H = E + PV
During a reaction at constant pressure, the enthalpy of a reaction, DH = H(products) - H(reactants), is equivalent to the heat released or absorbed
If DH > 0, the reaction is endothermic
If DH < 0, the reaction is exothermic

9. Thermochemical Equations
e.g.; CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) DH = kJ is a thermochemical equation
The above thermochemical equation says that the combustion of one mole of CH4(g) produces one mole of CO2(g) and two moles of H2O(l), and releases kJ of heat
For the inverse reaction CO2(g) + 2 H2O(l) CH4(g) + 2 O2(g) DH = kJ We must provide kJ of heat to make it happen

10. Thermochemical Equations
If we double the amount of reactants and products, the value of DH doubles as well CH4(g) + 4 O2(g) CO2(g) + 4 H2O(l) DH = kJ
It is important to note the phase of each reactant and product ex.; CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) DH = kJ but CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) DH = kJ because H2O(g) is higher in enthalpy H2O(l) H2O(g) DH = kJ

11. Specific Heat and Heat Capacity
The specific heat (s) of a substance is the amount of heat required to raise the temperature of a one gram substance by one Kelvin
The heat capacity (C) of a substance is the amount of heat required to raise the temperature of a given mass of this substance by one Kelvin
An object with a high heat capacity requires a lot of heat to warm up and releases a lot of heat when cooling down

12. Specific Heat and Heat Capacity
The specific heat is an intensive quantity and heat capacity is an extensive quantity C = m s where m is the mass of the substance (in grams)
The heat capacity relies on changes in temperature, DT, with the quantity of heat (Q) absorbed or released during the process Q = C DT = m s DT

13. Specific Heat and Heat Capacity
Example: The specific heat of iron is J/(g K). A 869 g bar of iron goes from 94oC to 5oC. Calculate the amount of heat released by the metal.
Solution: Exothermic reaction, thus 34.3 kJ of heat is released.

14. Constant Volume Calorimetry
The heat of combustion is measured in a bomb calorimeter which has a fixed volume
The bomb is insulated so that there is no heat transfer with the environment Qsystem = Qwater + Qbomb + Qreaction = Qreaction = - (Qwater + Qbomb)

15. Constant Volume Calorimetry
The heat released by the reaction contributes to the warming of the water and the bomb calorimeter Qwater = mwater swater DT Qbomb = Cbomb DT where Cbomb was determined in advance by the combustion of a standard within the bomb calorimeter in question
N.B. Qreaction is not exactly equal to DH because it is not operating at constant P (It is a constant V)

16. Constant Volume Calorimetry
Example: g of methanol (CH3OH) is burned in a constant-volume calorimeter. The temperature of the water and the bomb calorimeter is increased 4.20 oC. If the quantity of water was exactly 2000 g and the heat capacity of the calorimeter is 2.02 kJ/K, calculate the molar heat of combustion of methanol. The specific heat of water is J/(g K).

17. Constant Pressure Calorimetry
For reactions other than combustion (i.e., endothermic or less exothermic reactions)
Qreaction = -(Qwater + Qcalorimeter) where Qcalorimeter was determined in advance
Because this is occurring at constant pressure Qreaction = DH

18. Constant Pressure Calorimetry
Example: We mix 100 mL of a mol/L HCl solution with 100 mL of a mol/L NaOH solution in a constant-pressure calorimeter whose heat capacity is 335 J/K. The initial temperature of the solutions is oC. The final temperature is oC. Calculate the heat of the neutralization reaction :
NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)
Assume that the densities and heat capacities of the solutions are the same as those of pure water (1.00 g/mL and J/(g K), respectively)

19. Constant Pressure Calorimetry
Solution:
N.B. This heat was released by the reaction of (0.500mol/L)(0.100L) = moles of acid and base, so if 1.00 mole of HCl reacts with 1.00 mole of NaOH heat of neutralization = -2.81kJ / mol = kJ/mol

20. The Standard Enthalpy of Formation
We are only able to measure changes in enthalpy, DH, and not absolute enthalpies (i.e., H)
Because we are only interested in changes in enthalpies, we are free to choose any reference point (i.e., our “zero”)
Arbitrarily, we chose to say that the standard enthalpy of formation, DHfo, of any element in its most stable allotropic form (i.e., their standard state) is zero
Then, DHfo for any other substance is the amount of heat produced or absorbed when one mole of the substance is formed from these elements in their standard states at a pressure of 1 atm

21. The Standard Enthalpy of Formation
A few examples:
ΔHfo (H2, g) = 0, by definition
ΔHfo (O2, g) = 0, by definition
ΔHfo (O3, g) > 0, since O2 is more stable
ΔHfo (C, graphite) = 0, by definition
ΔHfo (C, diamond) > 0, since graphite is more stable
ΔHfo (H2O, l) = kJ, because the reaction H2(g) + ½ O2(g)  H2O(l) releases kJ of heat

22. The Standard Enthalpy of Reaction
Using the standard enthalpy of formation, DHfo, we can calculate the standard enthalpy of reaction, DHoreaction, i.e., the enthalpy of a reaction that occurs under a pressure of 1 atm
ex.; For the reaction
The general formula is where m et n are the stoichiometric coefficients of the reactants and products

23. The Standard Enthalpy of Reaction
Sometimes, we can determine DHfo of a compound by directly reacting its elements in their standard states
e.g.; For the reaction C(graphite) + O2(g) CO2(g) The reaction is complete and kJ of heat is released, thus by definition, DHfo(CO2, g) = kJ
However, for other compounds, it is more difficult to find DHfo because the direct synthesis of the compound from its elements in their standard states is impossible e.g.; The reaction: C(graphite) + 2 H2(g) CH4(g) does not happen no matter what experimental conditions are used

24. Hess’ Law
Hess’ Law: when the reactants are converted to products, the enthalpy is the same whether the reaction takes place in one step or in several steps
In other words, as long as the starting point (reactants) and end point (products) are the same, the DH value does not change with the chosen trajectory to get reactants to products

25. Hess’ Law
Example: Using the following thermochemical equations, calculate the enthalpy of formation of methane
C(graphite) + O2(g) CO2(g) DHoreaction = kJ H2(g) + O2(g) H2O(l) DHoreaction = kJ CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) DHoreaction = kJ
Solution: Add the following three thermochemical equations: C(graphite) + O2(g) CO2(g) DHoreaction = kJ H2(g) + O2(g) H2O(l) DHoreaction = kJ CO2(g) + 2 H2O(l) CH4(g) + 2 O2(g) DHoreaction = kJ C(graphite) + 2 H2(g) CH4(g) DHoreaction = kJ Thus DHfo(CH4, g) = kJ

26. Thermodynamics
Thermochemistry: The study of heat transfer in chemical reactions
Thermochemistry is a branch of thermodynamics
Thermodynamics: The study of transformations from the point of view of the energy exchanges that accompany these transformations
We also consider work (not just heat) in thermodynamics

27. State Functions
The state of a system is defined by the values ​​of all macroscopic variables such as composition, temperature, pressure, and volume
A state function is a property of the system which is determined by the state of the system, regardless of the way the system has reached this state

28. State Functions
e.g.; If 1 L of water is cooled from 30 oC to 20 oC at 1 atm or if 1 L of water is heated from 10 oC to 20 oC at 1 atm, the two samples of water will have the same pressure, volume, temperature, mass, energy, enthalpy, specific heat, etc., as these properties are state functions
A state equation is an equation that relates state functions eg.; PV = nRT is a state equation because it sets the value of the fourth function, if three of the four (P, V, n, T) are known

29. The First Law of Thermodynamics
Energy can be converted from one form to another, or transferred from one place to another, but it can neither be created nor destroyed.
The First Law says that the change in the internal energy of a system (DU) is offset by a variation of the energy of the environment so that the energy of the universe does not change

30. The First Law of Thermodynamics
We can express the First Law mathematically as:
DU = Q + W
where Q is the heat transferred to the system and W is the work done on the system
If Q > 0, we have an endothermic transformation
If Q < 0, we have an exothermic transformation
If W > 0, work is done on the system and the energy of the system increases
If W < 0, work is done by the system and the energy of the system decreases

31. PV Work
We will consider electrical work when we study electrochemistry
For mechanical work, the expansion work done by a gas (PV work) will be especially significant W = -P DV
For a compression (we require calculus to get this formula) W = - nRT ln (Vf/Vi)

32. PV Work
Example: The volume of gas passes from 264 mL to 971 mL at a constant temperature. Calculate the work done (in joules) by a gas if it expands (a) in a vacuum, and (b) opposed to a constant pressure of 4.00 atm.
Solution: We have to work with SI units: atm = Pa and mL = 1 x 10-6 m (a) P = 0 so W = (b)

33. Work and Heat
In this last example, we saw that the value of W depends on the path taken and is therefore not a state function
Because DU = Q + W, and we also know that U is a state function and W is not, we know that Q, like W, is not a state function
the value of W depends on the path chosen, but the value of DU (Q + W) is independent of the path, so the value of Q must also depend on the selected path

34. Work and Heat
Example: A gas expands and performs PV work equal to 279 J on the external environment. At the same time, it absorbs 216 J of heat. What is the variation in energy of the system?
Solution: W = -279 J Q = +216 J DU = Q + W = +216 J + (-279 J) = -63 J

35. Enthalpy and the First Law of Thermodynamics
At constant volume, the PV work is equal to zero because DV = 0, and thus DU = Q + W = Q
For all reactions at constant volume, Q = DU
However, constant-pressure reactions are of greater interest and if a gas is produced or consumed, DV is not negligible e.g.; During the reaction, Na(s) + 2 H2O(l) NaOH(aq) + H2(g) The H2(g) produced must do PV work when it has to push back the air so as to enter the atmosphere

36. Enthalpy and the First Law of Thermodynamics
For any reaction, DU = Q + W
For a reaction at constant pressure, DH = Q
So for a reaction at constant pressure, DU = DH + W
if one considers only the PV work at constant pressure, W = -P DV DU = DH - P DV or DH = DU + P DV
Because U, P, and V are state functions, H is also a state function

37. Enthalpy and the First Law of Thermodynamics
If we say that DV only depends on changes in the number of moles of gas produced or consumed (i.e., the volumes of solids and liquids are negligible in comparison) DH = DU + PDV = DU + D(PV)
Making the approximation that the gases behave as ideal gases DH = DU + D(nRT)
Since the temperature is constant : DH = DU + RTDn Where Dn = moles of gas products – moles of gas reactants and we use R = J/(K mol)

38. We want to heat a sample of water via the combustion of ethane (C2H6) at a temperature of 25.0 ◦C and a pressure of 1.00 atm (N.B. combustion is the reaction of a substance with O2(g) to produce CO2(g) and H2O(l)). All of the heat released in this reaction enters kg of water so as to increase the temperature from 25.0 ◦C to 60.0 ◦C. What volume of ethane (C2H6) must be used?
∆H◦f (C2H6, g) = −103.9 kJ mol−1
∆H◦f (CO2, g) = −393.5 kJ mol−1
∆H◦f(H2O, l) = −285.8 kJ mol−1
s (H2O, l) = J K−1 g−1