1. Energy Thermodynamics
Chapter 6
Energy
Thermodynamics

2. Energy is... …the ability to do work or produce heat. …conserved.
…defined as kinetic or potential.
…a state function (independent of the path, or how you get from point A to B...depends only on its present state).
Work is a force acting over a distance.
Heat is energy transferred between objects because of temperature difference.

3. The universe… …is divided into two halves.
…the system and the surroundings.
The system is the part you are concerned with (as a chemist).
The surroundings are the rest.
Exothermic reactions release energy to the surroundings.
Endothermic reactions absorb energy from the surroundings.

4. Heat
Potential energy

5. Heat
Potential energy

6. Direction Every energy measurement has three parts:
A unit ( Joules, kJ or Calories).
A number.
A sign to tell direction of energy flow.
negative - exothermic
Positive - endothermic

7. Surroundings
System
Energy
DE <0
Energy is leaving
the system!

8. Surroundings
System
Energy
DE >0
Energy is entering
the system!

9. Same rules for heat and work
Heat given off is negative.
Heat absorbed is positive.
Work done by system on surroundings is negative.
Work done on system by surroundings is positive.
Thermodynamics- The study of energy and the changes it undergoes.

10. First Law of Thermodynamics
The energy of the universe is constant.
Law of conservation of energy.
q = heat and w = work
DE = q + w (Internal Energy = DE)
DE is the sum of the KE and PE of all particles in a system…which can be changed by a flow of work and/or heat!
Take the systems point of view to decide signs.

11. What is work? Work is a force acting over a distance. w= F x Dd
P = F/ area
Typically, we will let the physicist worry about the work!

12. Work needs a sign…
If the volume of a gas increases, the system has done work on the surroundings (it pushed against the surroundings).
work is negative (or lost to the surr.)
w = - PDV
Expanding work is negative.
Contracting, surroundings do work on the system…w is positive.

13. Enthalpy (for nerds) abbreviated H H = E + PV (that’s the definition)
Calculated at a constant pressure.
DH = DE + PDV
the heat at constant pressure qp can be calculated from
DE = qp + w = qp - PDV
qp = DE + P DV = DH

14. Enthalpy for Chemists
Since qp = DE + P DV = DH…then we can say qp = DH
In a “nutshell”: at constant pressure, the change in enthalpy of the system is equal to the energy flow as heat.
Practical applications: the change in enthalpy is given by the equation DH = H products - H reactants

15. Calorimetry Science of measuring heat. Use a calorimeter: Two types…
#1: Constant pressure calorimeter (called a coffee cup calorimeter)
heat capacity for a material, C is calculated
C= heat absorbed/DT = DH/DT
specific heat capacity = C/mass (units J/ºC•g or J/K•g)

16. Calorimetry molar heat capacity = C/moles
heat = specific heat x m x DT
heat = molar heat x moles x DT
Make the units work and you’ve done the problem right.
A coffee cup calorimeter measures DH.
An insulated cup, full of water.
The specific heat of water is 1 cal/gºC
Heat of reaction= DH = Cp x mass x DT

17. Examples
The specific heat of graphite is 0.71 J/gºC. Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K.
A 46.2 g sample of copper is heated to 95.4ºC and then placed in a calorimeter containing 75.0 g of water at 19.6ºC. The final temperature of both the water and the copper is 21.8ºC. What is the specific heat of copper?

18. Calorimetry
#2 Constant volume calorimeter is called a bomb calorimeter.
Material is put in a container with pure oxygen. Wires are used to start the combustion. The container is put into a container of water.
The heat capacity of the calorimeter is known and tested.
Since DV = 0, PDV = 0, DE = q

19. Bomb Calorimeter thermometer stirrer full of water ignition wire
Steel bomb
sample

20. Properties
Intensive property - not related to the amount of substance.
density, specific heat, temperature.
Extensive property - does depend on the amount of substance.
Heat capacity, mass, heat from a reaction.

21. Hess’s Law Enthalpy is a state function (independent of the path).
We can “add equations” to come up with the desired final product, and then add the DH values to arrive at the final enthalpy change!
Two rules
#1 - If the reaction is reversed the sign of DH is changed
#2 - If the reaction is multiplied, so is DH

22. State Function One two-step pathway… N2 + O2  2NO ΔH = 180 kJ
2NO + O2  2NO2 ΔH = -112 kJ
A different single-step pathway…
N2 + 2O2  2NO2 ΔH = 68 kJ
Regardless of the pathway…68 kJ are absorbed in this endothermic reaction!

23. O2 +2NO -112 kJ H (kJ) 180 kJ 2NO2 68 kJ N2 +2O2 Two-step One-step N2

24. Standard Enthalpy
The enthalpy change for a reaction under standard conditions (25ºC, 1 atm, 1 M solutions)
Symbol DHº
When using Hess’s Law, work by adding the equations up to obtain the “target equation”.
The other species will cancel out…and you will sum the enthalpy values!

25. Procedure for solving…
Goal: Establish a pathway to the Target Equation by rearranging the “given equations”
Hints:
Work backwards from the target equation
Look for unique formulas (which only appear once)
Avoid formulas which appear multiple times until you get started

26. Hess’s Law Example Given DHº= -1300. kJ
calculate DHº for this reaction (target equation)
DHº= kJ
DHº= -394 kJ
DHº= -286 kJ

27. Standard Enthalpies of Formation
Hess’s Law is useful when you are given multiple sets of reactions….
However, the standard enthalpy of formation (DHfº) for a compound is defined as
the change in enthalpy that accompanies the formation of one mole of a compound from its elements in their standard states.

28. Standard Enthalpies of Formation
Standard states are 1 atm, 1M and 25ºC
See p. 246 of our text.
For an element it is 0 (zero)!
Why? It comes to us already formed!
There is a table of values listed in Appendix 4 (pg A19-22).

29. Standard Enthalpies of Formation
First, you need to be able to write the balanced equation from the elements.
What is the equation for the formation of NO2 ?
½N2 (g) + O2 (g) ® NO2 (g)
Have to make one mole to meet the definition.
Write the equation for the formation of methanol CH3OH.

30. Since we can manipulate the equations…
We can use heats of formation to figure out the heat of reaction.
Lets do it with this equation.
C2H5OH +3O2(g) ® 2CO2 + 3H2O
which leads us to this rule:

31. Since we can manipulate the equations…
We can use heats of formation to figure out the heat of reaction.
Lets do it with this equation.
C2H5OH +3O2(g) ® 2CO2 + 3H2O
which leads us to this rule.

32. C2H5OH(l) +3O2(g) ® 2CO2(g) + 3H2O(l)
Solve the problem:
C2H5OH(l) +3O2(g) ® 2CO2(g) + 3H2O(l)
ΔHf:
(kJ/mol)
ΔH = [(2•-393.5kJ)+(3•-286kJ)]-[-278kJ + 3•0kJ]
ΔH = -1367kJ of energy (released) per mole of ethanol burned.

33. Chapter 6 problems: