1. Energy Thermodynamics pp
Chapter 6
Energy
Thermodynamics pp
Z5e 242 Section Nature of Energy

2. 6.1 The Nature of Energy Energy is . . . The ability to do work.
Conserved (It’s the Law).
Made of heat and work.
A state function (unlike work and heat).
I.e., it’s independent of the path, or how you get from point A to B.

3. Enthalpy is a STATE FUNCTION. State functions are PATH-INDEPENDENT

4. Work and Heat Work is a force acting over a distance.
Heat is energy transferred between objects because of temperature difference.

5. The universe Is divided into two halves.
The system and the surroundings.
The system is the part you are concerned with.
The surroundings are the rest.
Exothermic reactions release energy to the surroundings.
Endothermic reactions absorb energy from the surroundings.
Z5e 243

6. Heat
Z5e Rf. Fig 6.2 p. 244
Potential energy

7. Heat
Rf. Fig 6.3 p. 245
Potential energy

8. Direction Every energy measurement has three parts.
Unit ( Joules or calories).
Number - how many.
Sign - to tell direction.
Negative - exothermic
Positive - endothermic

9. Surroundings
System
Energy
Z5e 245
DE < 0

10. Surroundings
System
Energy
Z5e 245
DE >0

11. Same rules for heat and work pp
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.
Z5e 245

12. First Law of Thermodynamics
The energy of the universe is constant.
Law of conservation of energy.
q = heat
w = work
DE = q + w
Take the system’s point of view to decide signs.
Z5e 244

13. What is work? Work is a force acting over a distance. w= F x Dd
P = F/area
d = V/area
W = (P x area) x D (V/area) = PDV
Work can be calculated by multiplying pressure by the change in volume at constant pressure.
units of work: liter-atm or L-atm
d = distance

14. Work needs a sign
If the volume of a gas increases, the system has done work on the surroundings (memorize this!).
And work is negative: w = (-)
w = - PDV (don’t forget negative sign)
Expanding a gas: Work is negative.
Contracting: Surroundings do work on the system and w = (+).
1 L-atm = J (need for conversions)

15. Fig 6.4 p. 246, Volume of a Cylinder
The piston, moving a distance delta h against pressure P, does work on the surroundings.
b. Since V of cylinder = area of base x height, the delta V of gas = delta h x A
Z5e Fig. 6.4
The piston, moving a distance delta h against pressure P, does work on the surroundings.
Since V of cylinder = area of base x height, the delta V of gas = delta h x A

16. w = -p(chg V) = -24 l-atm Examples pp
What amount of work is done when L of gas is expanded to L at 2.40 atm pressure? Steps . . .
w = -P∆V = atm • 10.00L = -24 L-atm
If kJ of heat are absorbed what is the change in energy? Steps . . .
∆E = q + w = 2360 J + (-24 L-atm)( J/L-atm) = J
How much heat to change the gas without changing the internal energy of the gas?
Want ∆E = 0, so q + w = 0
since w = J; q added must = J
w = -p(chg V) = -24 l-atm
E = q + w = 2360 J + (-24 l-atm)( j/l-atm) = J
Want E = 0, so q + w = 0, so since w = J; q added must = J

17. 6.2 Enthalpy & Calorimetry
Enthalpy is abbreviated as H
H = E + PV (that’s the definition of heat)
At constant pressure DH = DE + PDV
The heat at constant pressure qp can be calculated from . . .
DE = qp + w = qp - PDV (w = - PDV)
qp = DE + P DV = DH
Z5e 248 Section 6.2 Enthalpy & Calorimetry

18. Calorimetry Measuring heat -- use a calorimeter.
Two kinds: Constant P or constant V
Constant pressure calorimeter (called a coffee cup calorimeter, open to outside)
Heat capacity for a material, Cp is calculated.
Cp= heat absorbed/DT = DH/DT
Specific heat capacity = Cp/mass
Z5e 250

19. Calorimetry Specific heat capacity = Cp/mass
Heat = specific heat x m x DT
q = m•∆T•Cp
Sometimes q is shown as ∆H ∆H = m•∆T•Cp (online HW)
Molar heat capacity = Cp/moles
Heat = molar heat x moles x DT
Make the units work and you’ve done the problem right (get dates in college).
Be careful to note if Cp uses grams or moles.

20. Calorimetry A coffee cup calorimeter measures DH.
It’s an insulated cup, full of water.

21. Figure 6.5 A Coffee-Cup Calorimeter Made of Two Styrofoam Cups
Z5e 251

22. Calorimetry Specific heat of H2O = 1 cal/gºC = 4.184 J/goC
Heat of reaction= DH = sh x mass x DT (see this form on online HW)
Also seen as: q = mTCp
See, Table 6.1 p. 251 (next page) for Cp of some common substances. Note the units! (J/ºC•g)

23. Z5e 251 Table 6.1

24. Convert Kg to g; q = mTCp . Ans . . . = 2 880 000 J = 2880 kJ
Examples
Cp of graphite is J/gºC. Calculate q needed to raise the temperature of 75.0 kg of it from 294 K to 348 K. . .
Convert Kg to g; q = mTCp . Ans . . .
= J = 2880 kJ
Convert Kg to g; q = mTCp = J = 2880 kJ
q gained = q lost; set both equations equal to each other. Ans J/goC
Be sure to discuss that we are looking at change in T, so since C and K degree increments are equal, do NOT convert change in T value to Kelvin. E.g., 10 Kelvin degree change should not be converted to a minus 263 Celsius change.

25. Examples pp
46.2 g of copper heated to 95.4ºC & placed in calorimeter with 75.0 g water at 19.6ºC. Final temp. of both is 21.8ºC. What is Cp of Cu? . . .
q gained by Cu = q lost by H2O; set both equations equal to each other, solve . . .
qCu = mcu∆TCuCpCu = mwat∆TwatCpwat = qwat Cpwat = 4.18 J/ºC•g (memorize). Use absolute value for ∆T; plug-and-chug to get answer of . . .
0.203 J/gºC for Cu
Don’t need to convert ºC to K since both have the same degree increments & we are only interested in the change.
Need to know this for online HW05 Q8
Convert Kg to g; q = mTCp = J = 2880 kJ
q gained = q lost; set both equations equal to each other. Ans J/goC
Be sure to discuss that we are looking at change in T, so since C and K degree increments are equal, do NOT convert change in T value to Kelvin. E.g., 10 Kelvin degree change should not be converted to a minus 263 Celsius change.

26. Calorimetry 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
Z5e 252

27. Bomb Calorimeter thermometer stirrer full of water ignition wire
Steel bomb
sample
Z5e 253 and see Fig. 6.6 p. 254

28. Bomb Calorimetry Example pp
T increases by 3.2ºC when g of quinone (C6H4O2) is burned in a bomb calorimeter with heat capacity of 1.56 kJ/ºC. What is the combustion energy of quinone per gram and also per mole?
Heat gained by calorimeter = 1.56kJ/ºC x 3.2 ºC = 5.0 kJ = heat loss by quinone
∆Ecomb = -5.0kJ/ g = -25kJ/g
∆Ecomb = -25kJ/g x g/mol = kJ/mol
Text problem #49 is like this (HW)
(q = m∆TCp + heat gained by calorimeter)
Z5e 284 q. 50

29. Properties
Intensive properties: Not related to the amount of substance.
E.g., density, specific heat, temperature.
Extensive property - does depend on the amount of stuff.
E.g., heat capacity, mass, heat from a reaction.

30. Note on Hydrates pp
A hydrate means a salt (formula unit) that has some water hooked onto it.
Copper(II) chloride dihydrate is . . .
CuCl2•2H2O
The M of the hydrate is g/mol
The molar mass of the anhydride is g/mol ( less 2 waters)
You need to know this for the online HW

31. More online HW notes pp
HW05 Q4 - answer in kJ/mol of Z (not J/g). Find the q, then ÷ by # of moles.
HW05 Q2 & Q8 - use the examples just done today. Use internet for Cp of Pb.
HW05 Q11 - look at the example problems in your assigned reading.

32. Note on online HW pp
When a question asks, “how much heat is liberated,” your answer will be positive because there is no “negative” heat.
When a question asks, “what is the change in heat” then you have to indicate the change by a (+) or (-) sign.
When you use energy or heat in a mathematical equation (e.g., q = m∆TCp then you also have to show the sign.
(2 * * -286) - (1 * * 0) = kJ/mol = exothermic
Oxygen = O since element; liquid water value is not same as gaseous water

33. Z5e 256 6.3 Hess’ Law 6.3 Hess’s Law Enthalpy is a state function.
So, it is independent of the path.
We can add the pathway equations to come up with the desired final product, and add the pathway DH’s.
Two rules:
(1) If the reaction is reversed the sign of DH is changed.
(2) If the reaction is multiplied, so is DH
Z5e Hess’ Law

34. O2
2NO
-112 kJ
H (kJ)
180 kJ
2NO2
Z5e 257 Fig same change occurs whether in 1 or 2 steps.
N2 + 2O2 --> 2NO2 ;  H1 = 68kJ
Or 2 step:
N2 + O2 --> 2NO ;  H2 = 180kJ
2NO + O2 --> 2NO2l;  H3 = -112kJ
68 kJ N2
2O2

35. Hints for using Hess’ Law pp
When using Hess’s Law, work backwards from the required reaction to decide how to manipulate each reaction step so you can add them up to make it look like the answer.
Reverse any reactions as needed (change DH sign) & multiply the coefficients so the extra parts cancel.
Note: H2O(l) will not cancel H2O(g)!
Z5e 259

36. Example of Hess’ Law pp
Calculate ∆Hºf for C(s) + 2H2(g) ® CH4(g) given: C(s) + O2(g) ® CO2(g) ∆Hºc = kJ/mol H2(g) + 1/2 O2(g) ® H2O(l) ∆Hºc = kJ/mol CH4(g) + 2O2(g) ® CO2(g) + 2H2O(l) ∆Hºc = kJ/mol
Need CH4 as a product, so reverse the 3rd equation & change its ∆H sign.
Hrw 520

37. Example of Hess’ Law pp
Calculate ∆Hºf for C(s) + 2H2(g) ® CH4(g) given: C(s) + O2(g) ® CO2(g) ∆Hºc = kJ/mol H2(g) + 1/2 O2(g) ® H2O(l) ∆Hºc = kJ/mol CO2(g) + 2H2O(l) ® CH4(g) + 2O2(g) ∆Hºf = kJ/mol
H2O is not in the original equation so need to cancel it out.
But, Eq. 2 has only 1 water as a product while Eq. 3 has 2 waters as a reactant.
Multiply Eq. 2 by 2 (including ∆Hºc)
Hrw 520

38. Example of Hess’ Law pp
Calculate ∆Hºf for C(s) + 2H2(g) ® CH4(g) given: C(s) + O2(g) ® CO2(g) ∆Hºc = kJ/mol 2H2(g) + O2(g) ® 2H2O(l) ∆Hºc = (2)( kJ/mol) CO2(g) + 2H2O(l) ® CH4(g) + 2O2(g) ∆Hºf = kJ/mol
Add the 3 equations and cancel like terms on both sides of the reaction.
Hrw 520

39. Example of Hess’ Law pp
Calculate ∆Hºf for C(s) + 2H2(g) ® CH4(g) given : C(s) + O2(g) ® CO2(g) ∆Hºc = kJ/mol 2H2(g) + O2(g) ® 2H2O(l) ∆Hºc = (2)( kJ/mol) CO2(g) + 2H2O(l) ® CH4(g) + 2O2(g) ∆Hºf = kJ/mol
C(s) + 2H2(g) ® CH4(g) ∆Hºf = kJ
Since ∆H is (-) the reaction is exothermic.
Hrw 520

40. Standard Enthalpy
The enthalpy change for a reaction at standard conditions (25ºC, 1 atm , 1 M solutions). (Std T = 0 ºC when working with the gas laws!)
Symbol DHº

41. Given calculate DHº for this reaction DHº= -1300. kJ
Example
Given calculate DHº for this reaction
DHº= kJ
DHº= -394 kJ
DHº= -286 kJ
Calculate DHº for this reaction Ans. . .
Reverse equation 1; multiply eq. 2 by 2 (include delta H)
Ans. 226 kJ, so endothermic
Reverse equation 1; multiply eq. 2 by 2 (include delta H)
Ans. 226 kJ, so endothermic

42. Example Given DHº= +77.9kJ DHº= +495 kJ DHº= +435.9kJ
Calculate DHº for this reaction. Ans. . .
Cut all rxns in 1/2
Reverse #s 2 & 3
Add up to = -426 kJ = exothermic
Cut all reactions in 1/2
Reverse #s 2 & 3
Add up to = -426 kJ = exothermic (not d/t SF)

43. Standard Enthalpies of Formation
Hess’s Law is much more useful if you know lots of reactions.
The amount of heat needed to form 1 mole of a compound from its elements in their standard states; e.g., H2O(l) v. H2O(g)
Standard states are 1 atm, 1M and 25ºC
(Std T = 0ºC for gas laws like PV = nRT)
For an element DHfº = 0 (don’t forget!!)
See Table 6.2 p. 262; also Appendix 4 (pg A21)
Z5e Section 6.4 Standard Enthalpies of Formation

44. 6.4 Standard Enthalpies of Formation
Need to be able to write the equations.
What is equation for formation of NO2 ?
½N2 (g) + O2 (g) ® NO2 (g) because . . .
Have to make one mole to meet the definition. So, what coeff. for N2?
Use 1/2 for coefficient of N2
1/2N2 (g) + O2 (g) ® NO2 (g)
Why not N2 (g) + 2O2 (g) ® 2NO2 (g) ?
Because by definition only looking at one mole of NO2
Z53 261
a. Use 1/2 for coefficient of N2
b. C(s) + 2H2(g) + 1/2 O2(g) = CH3OH

45. Standard Enthalpies of Formation cont.
Write equation for formation of methanol, CH3OH from its elements. Answer . . .
C(s) + 2H2(g) + 1/2 O2(g) ® CH3OH
Note: heat of combustion is the reverse of heat of formation (sign also changes). This will be on the test!!
The above reaction showing heat of combustion instead of formation is . . .
CH3OH ® C(s) + 2H2(g) + 1/2 O2(g)
Z53 261
a. Use 1/2 for coefficient of N2
b. C(s) + 2H2(g) + 1/2 O2(g) = CH3OH

46. Since we can manipulate the equations
We can use heats of formation to figure out the heat of reaction.
Let’s look at the schematic (see fig. 6.9, p. 263, next slide) . . .

47. Figure A Schematic Diagram of the Energy Changes for the Reaction CH4(g) + 2O2(g) ® CO2(g) + 2H2O(l)
Z5e 263

48. 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(l) +3O2(g) ® 2CO2(g) + 3H2O(l) (2 • • -286) - (1• • 0) = kJ = exothermic
Oxygen = O since element; liquid water value is not same as gaseous water
Which leads us to this rule:
(2 * * -286) - (1 * * 0) = kJ/mol = exothermic
Oxygen = O since element; liquid water value is not same as gaseous water

49. Example 1 pp
Using the standard enthalpies of formation listed in Table 6.2, p. 262, calculate the enthalpy change for the following reaction Steps
4NH3(g) + 7O2(g) ® 4NO2(g) + 6H2O(l)
Z5e 264 SE 6.9

50. Example 1 pp 4NH3(g) + 7O2(g) ® 4NO2(g) + 6H2O(l)
Step 1: Decompose NH3(g) into its elements. 4NH3(g) ® 2N2(g) + 6H2(g)
The reverse of this is the formation reaction for NH3(g) 1/2N2(g) + 3/2H2(g) ® NH3(g) DHfº = -46 kJ/mol
We need 4 times the reverse of the formation reaction for our equation above.
Z5e 264 SE 6.9

51. Example 1 pp 4NH3(g) + 7O2(g) ® 4NO2(g) + 6H2O(l)
Step 2: Since O2(g) is in standard state its DHfº = 0.
We now have the elements N2(g), H2(g), and O2(g), which can be combined to form the products of the overall reaction as follows: . . .
Z5e 264 SE 6.9

52. Figure 6.10 A Pathway for the Combustion of Ammonia
Z5e 264 Fig used with SE 6.9.

53. Example 1 pp 4NH3(g) + 7O2(g) ® 4NO2(g) + 6H2O(l)
So far, we’ve figured DHfº for the reactants.
Now we do the same for the products.
Step 3: Synthesis of NO2. We need 4 times the formation reaction (from Table 6.2).
Step 4: Need 6 times formation reaction for H2O(l) (not H2O(g)).
Z5e 264 SE 6.9

54. Example 1 cont. pp 4NH3(g) + 7O2(g) ® 4NO2(g) + 6H2O(l) Use:
[4x(34kJ) + 6x(-286 kJ)] - [4x(-46kJ)] = kJ.
(2 * * -286) - (1 * * 0) = kJ/mol = exothermic
Oxygen = O since element; liquid water value is not same as gaseous water

55. Calculating Horeaction using standard enthalpies of formation -
Values for Hof are looked up in tables such as Appendix 4 in your book
Hrw 523

56. Calculate Horxn for CaCO3(s)  CaO(s) + CO2(g) compound Hof (kJ/mol) CaCO3(s) CaO(s) CO2(g)
[ ] - [ ] = = endothermic 57

57. DHo = S DHfoproducts - S DHforeactants
Calculation of DHo
DHo = S DHfoproducts - S DHforeactants

58. Use above plus Appendix 4, p. A21 to get . . .
Example What is the value of DHrx for the reaction: 2 C6H6(l) O2(g) ® 12 CO2(g) H2O(g)
C6H6(l) DHfo = kJ/mol
Use above plus Appendix 4, p. A21 to get . . .
O2(g) DHfo = 0 kJ/mol
CO2(g) DHfo = kJ/mol
H2O(g) DHfo = kJ/mol
D Hrx = [S D Hfo]product - [S D Hfo]reactants

59. Example What is the value of DHrx for the reaction: 2 C6H6(l) O2(g) ® 12 CO2(g) H2O(g) from Appendix 4 p. A21ff C6H6(l) DHfo = kJ/mol; O2(g) DHfo = 0 CO2(g) DHfo = ; H2O(g) DHfo = D Hrx = [S D Hfo]product - [S D Hfo]reactants
D Hrx = [12( ) + 6( )]product
- [2( ) + 15(0)]reactants kJ
= x 103 kJ

60. Horxn and Stoichiometry
Nitroglycerine is a powerful explosive, giving four different gases when detonated C3H5(NO3)3(l)  3 N2(g) + O2(g) CO2(g) + 5 H2O(g) Given that Hof for nitroglycerine is -364 kJ/mol, and consulting a table for the enthalpies of formation of the other compounds, calculate the enthalpy change when 10.0 g of nitroglycerine is detonated.
Find enthalpy for 1 mole of NTG (use previous method)
Then convert 10.0 g NTG to moles and use mole ratio between that and 1 mole. 10g ÷ 227 g/mol = mol 61

61. Horxn and Stoichiometry
C3H5(NO3)3(l)  3 N2(g) + O2(g) CO2(g) + 5 H2O(g)
calculate ∆Hrxn when 10.0 g of NTG is detonated.
[6• •-242] - [-364] = x 103 kJ/mol
This is for 1 mol. Have 10.0 g, convert to moles
10g ÷ 227 g/mol = mol
Then mol • x 103 kJ = -141 kJ exothermic 62

62. Note on online HW pp
When a question asks, “how much heat is liberated,” your answer will be positive because there is no “negative” heat.
When a question asks, “what is the change in heat” then you have to indicate the change by a (+) or (-) sign.
When you use energy or heat in a mathematical equation (e.g., q = m∆TCp then you also have to show the sign.
(2 * * -286) - (1 * * 0) = kJ/mol = exothermic
Oxygen = O since element; liquid water value is not same as gaseous water