1. Chapter 10
Energy Changes in Chemical Reactions 1

2. Energy Changes in Chemical Rxns
Most reactions give off or absorb energy
Energy is the capacity to do work or supply heat.
Heat: transfer of thermal (kinetic) energy between two systems at different temperatures (from hot to cold)
Metal bar in water
Metal bar drilled

3. Types of Energy
Work (w): energy transfer when forces are applied to a system
Heat (q): energy transferred from a hot object to a cold one
Radiant energy heat from the sun
Thermal energy  associated with motion of particles
Potential energy  energy associated with object’s position or substance’s chemical bonds
Kinetic energy  energy associated with object’s motion

4. Heat versus Temperature
Describe the difference between the two.
SI unit of energy: J
1 watt = 1 J/s, so a 100 Watt bulb uses 100 J each second
We often use the unit of kJ to refer to chemical heat exchanges in a reaction. 1 kJ = 1000 J
Energy is also reported in calories:
Amount of energy needed to raise 1 gram of water by 1oC
1 cal = J; 1 Cal = 4184 J
Cal (or kcal) is used on food labels
Molecular heat transfer

5. Energy and Energy Conservation
Heat: form of energy transferred from object at higher temperature to one at lower temperature (from hot object to cold object)
Thermochemistry: study of heat changes in chemical reactions, in part to predict whether or not a reaction will occur
Thermodynamics: study of heat and its transformations
First Law of Thermodynamics: Energy can be converted from one form to another but cannot be created or destroyed

6. System and Surroundings
System loses heat (negative); gains heat (positive)

7. Endothermic vs Exothermic
Endothermic reaction: q is positive (q > 0)
Reaction (system) absorbs heat
Surroundings feel cooler
Exothermic reaction: q is negative (q < 0)
Reaction releases heat
Surroundings feel warmer

8. Enthalpies of Reaction
Determine if the following processes are endothermic or exothermic…
Combustion of methane
Reacting Ba(OH)2 with NH4Cl
Neutralization of HCl
Melting
CaCO3 (s)  CaO (s) + CO2 (g) 8

9. Answers to Enthalpies of Reaction
Combustion of methane exothermic
Reacting Ba(OH)2 with NH4Cl endothermic
Neutralization of HCl exothermic
Melting endothermic
CaCO3 (s)  CaO (s) + CO2 (g) endothermic
Combustion, neutralization, and combination reactions tend to be exothermic
Decomposition reactions tend to be endothermic
Melting, boiling, and sublimation are endothermic 9

10. Applications of heat emission/absorption10

11. Specific Heat and Heat Capacity
Specific heat (sp. ht.): amount of heat required to raise 1 gram of substance by 1oC
Use mass, specific heat, and DT to calculate the amount of heat gained or lost:
q = msDT  ms = C  q = CDT
Heat capacity (C): amount of heat required to raise the temperature of a given quantity of a substance by 1oC; C = q / DT = J / oC
Molar heat capacity (Cm): amount of heat that can be absorbed by 1 mole of material when temperature increases 1oC; q = (Cm) x (moles of substance) x (DT) = J / mol • oC

12. Specific Heat and Heat Capacity

13. Practice Problem
Calculate the amount of heat transferred when 250 g of H2O (with a specific heat of J/g·oC) is heated from 22oC to 98oC.
q = msDT
Is heat being put into the system or given off by the system?
If a piece of hot metal is placed in cold water, what gains heat and what loses heat? Which one will have a positive q value and which will have a negative q value?

14. Practice Problem
34.8 g of an unknown metal at 25.2oC is mixed with 60.1 g of H2O at 96.2 oC (sp. ht. = J/g·oC). The final temperature of the system comes to 88.4oC. Identify the unknown metal.
Specific heats of metals:
Al J/g·oC
Fe J/g·oC
Cu J/g·oC
Sn 0.228J/g·oC 14

15. Calorimetry and Heat Capacity
Heat changes in a reaction can be determined by measuring the heat flow at constant pressure
Apparatus to do this is called a calorimeter.
Heat evolved by a reaction is absorbed by water; heat capacity of calorimeter is the heat capacity of water. 15

16. Example
A 28.2 gram sample of nickel is heated to 99.8oC and placed in a coffee cup calorimeter containing grams of water at 23.5oC. After the metal cools, the final temperature of the metal and water is 25.0oC.
qabsorbed + qreleased = 0
Which substance absorbed heat?
Which substance released heat?
Calculate the heat absorbed by the substance you indicated above.

17. Group Quiz #25
A hot piece of copper (at 98.7oC, specific heat = J/g•oC) weighs g. When placed in room temperature water, it is calculated that J of heat are released by the metal.
What gains heat?
What loses heat?
What is the final temperature of the metal?
Watch signs!!!!

18. Enthalpies of Physical/Chemical Changes
Enthalpy (H) describes heat flow into and out of a system under constant pressure
Enthalpy (a measure of energy) is heat transferred per mole of substance.
At constant pressure,
qp = DH = Hproducts – Hreactants
DH > 0  endothermic (net absorption of energy from environment; products have more internal energy)
DH < 0  exothermic (net loss of energy to environment; reactants have more internal energy)

19. Heating a Pure Substance (Water)
Why does T become constant during melting and evaporating?
Melting, vaporization, and sublimation are endothermic
We can calculate total heat needed to convert a 15 gram piece of ice at -20oC to steam at 120oC.
2.09 J/goC
4.184 J/goC
2.080 J/goC
334 J/g
2250 J/g 19

20. Enthalpies of Phase Changes
Heat of fusion (DHfus): Amount of heat required to melt (solid  liquid)
Heat of vaporization (DHvap): Amount of heat required to evaporate (liquid  gas)
Heat of sublimation (DHsub): Amount of heat required to sublime (solid  gas)
Why are there no values for DHfreezing, DHcondendsation, or DHdeposition?

21. Thermochemical Equations
Shows both mass and enthalpy relationships
2Al (s) + Fe2O3 (s)  2Fe (s) + Al2O3 (s) DHo = -852 kJ
Amount of heat given off depends on amount of material:
852 kJ of heat are released for every 2 mol Al, 1 mol Fe2O3, 2 mol Fe, and 1 mol Al2O3

22. Thermochemical Equations
2Al (s) + Fe2O3 (s)  2Fe (s) + Al2O3 (s) DHo = -852 kJ
How much heat is released if 10.0 grams of Fe2O3 reacts with excess Al?
What if we reversed the reaction?
Heat would have to be put in to make the reaction proceed:
2Fe (s) + Al2O3 (s)  2Al (s) + Fe2O3 (s) DHo = +852 kJ

23. Hess’s Law
If a compound cannot be directly synthesized from its elements, we can add the enthalpies of multiple reactions to calculate the enthalpy of reaction in question.
Hess’s Law: change in enthalpy is the same whether the reaction occurs in one step or in a series of steps
Look at direction of reaction and amount of reactants/products

24. Hess’s Law
Value changes sign with direction
Figure 8.5 24

25. Hess’s Law Values of enthalpy change DHT = DH1 + DH2 + DH3 + ….
For a reaction in the reverse direction, enthalpy is numerically equal but opposite in sign
Reverse direction, heat flow changes; endothermic becomes exothermic (and vice versa); sign of DH changes
Proportional to the amount of reactant consumed
Twice as many moles = twice as much heat; half as many moles = half as much heat
DHT = DH1 + DH2 + DH3 + …. 25

26. Enthalpy of Chemical Reaction
Thermochemical equation:
H2(g) + I2(s)  2HI(g) DH = kJ
Two possible changes:
Reverse the equation:
2HI(g)  H2(g) + I2(s) DH = kJ
Double the amount of material:
2H2(g) + 2I2(s)  4HI(g) DH = kJ 26

27. Hess’s Law Calculate DHo for 2NO (g) + O2 (g)  N2O4 (g) DHo = ?
N2O4 (g)  2NO2 (g) DHo = kJ
NO (g) + ½ O2 (g)  NO2 (g) DHo = kJ

28. Hess’s Law
We can use known values of DHo to calculate unknown values for other reactions
P4 (s) + 3 O2 (g)  P4O6 (s) DH = kJ
P4 (s) + 5 O2 (g)  P4O10 (s) DH = kJ
What is DHo for the following reaction? P4O6 (s) + 2 O2 (g)  P4O10 (s) DH = ? 28

29. Hess’s Law

30. Hess’s Law Given: 2NH3(g)  N2H4(l) + H2(g) DH = 54 kJ
N2(g) + H2(g)  NH3(g) DH = -69 kJ
CH4O(l)  CH2O(g) + H2(g) DH = -195 kJ
Find the enthalpy for the following reaction:
N2H4(l) + CH4O(l)  CH2O (g) + N2(g) + 3H2(g)
DH = ? kJ

31. Group Quiz #26 Given the following equations:
2CO2 (g)  O2 (g) + 2CO (g) DH = kJ
½ N2 (g) + ½ O2 (g)  NO (g) DH = 90.3 kJ
Calculate the enthalpy change for:
2CO (g) + 2NO (g)  2CO2 (g) + N2 (g) DH = ? 31

32. Standard Heats of Formation
Standard heat of formation (DHof): heat needed to make 1 mole of a substance from its stable elements in their standard states
DHof = 0 for a stable (naturally occurring) element
Which of these have DHof = 0?
CO(g), Cu(s), Br2(l), Cl(g), O2(g), O3(g), O2(s), P4(s)
Do the following equations represent standard enthalpies of formation? Why or why not?
2Ag (l) + Cl2 (g)  2AgCl (s)
Ca (s) + F2 (g)  CaF2 (s) 32

33. Standard Enthalpies of Formation
Can use measured enthalpies of formation to determine the enthalpy of a reaction (use Appendix B in back of book)
DHorxn = SnDHof (products) – SnDHof (reactants)
S = sum; n = number of moles (coefficients)
Direct calculation of enthalpy of reaction if the reactants are all in elemental form
Sr (s) + Cl2 (g)  SrCl2 (g)
DHorxn = [DHof (SrCl2)] – [DHof (Sr) + DHof (Cl2)] = kJ/mol

34. Standard Enthalpies of Formation
Some Common Substances (25oC)

35. Heats of Formation DHorxn = S DHof,products - S DHof,reactants
Calculate values of DHo for the following rxns:
1) CaCO3 (s)  CaO (s) + CO2 (g)
2) 2C6H6 (l) + 15O2 (g)  12CO2 (g) + 6H2O (l)
DHof values:
CaCO3: kJ/mol; CaO: kJ/mol; CO2: kJ/mol; C6H6: 49.0 kJ/mol; H2O(l): kJ/mol 35

36. Group Quiz #27
Use Standard Heat of Formation values to calculate the enthalpy of reaction for:
C6H12O6(s)  C2H5OH(l) + CO2(g)
Hint: Is the equation balanced?
DHof (C6H12O6(s)) = kJ/mol
DHof (C2H5OH(l)) = kJ/mol
DHof (CO2(g)) = kJ/mol

37. Bond Dissociation Energies
Bond Dissociation Energy (or Bond Energy, BE): energy required to break a bond in 1 mole of a gaseous molecule
Reactions generally proceed to form compounds with more stable (stronger) bonds (greater bond energy)
H2 Bond Energy 37

38. Bond Dissociation Energies
Bond energies vary somewhat from one mole- cule to another so we use average bond dissociation energy (D)
H-OH 502 kJ/mol Avg O-H = 453
H-O 427 kJ/mol kJ/mol
H-OOH 431 kJ/mol 38

39. Bond Dissociation Energies39

40. Bond Dissociation Energies
DHorxn = SBE (reactants) SBE (products) endothermic exothermic
energy input energy released
SBE(react) > SBE(prod)  endothermic
SBE(react) < SBE(prod)  exothermic
Use only when heats of formation are not available, since bond energies are average values for gaseous molecules 40

41. Heats of Reaction
Use bond energies to calculate the enthalpy change for the following reaction: N2(g) + 3H2(g)  2NH3(g)
DHrxn = [BEN  N + 3BEH-H] + [-6BEN-H]
DHrxn = [ (436)] – [6(390)] = -87 kJ
measured value = kJ
Why are the calculated and measured values different? 41

42. Heats of Reaction
Use bond energies to calculate the enthalpy change for the decomposition of nitrogen trichloride: NCl3 (g)  N2 (g) Cl2 (g)
How many distinct bond types are there in each molecule?
How many of each bond type do we need to calculate DHrxn?
BE(N-Cl) = 200 kJ/mol
BE(N≡N) = 945 kJ/mol
BE(Cl-Cl) = 243 kJ/mol 42

43. Answer 6(N-Cl) + -1(N N) + -3(Cl-Cl)
6(200) + -(945) + -3(243) = -474 kJ ≡

44. Thermochemistry Calculation Summary
Use q = msDT (s = J/g·oC)
If given mass of reactant, convert to moles and multiply by enthalpy to find total heat transferred
If given multiple equations with enthalpies, use Hess’s Law
If given DHof values: products – reactants
If given bond energy (BE) values:
+reactants + -products

45. Practice Problems Identify how to set up the following problems:
Calculate the DHo of reaction for:
C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (l)
DHof C3H8(g): kJ/mol; DHof CO2(g): kJ/mol; DHof H2O(l): kJ/mol
8750 J of heat are applied to a 170 g sample of metal, causing a 56oC increase in its temperature.  What is the specific heat of the metal? Which metal is it?

46. Practice Problems C2H4(g ) + 6F2(g)  2CF4(g) + 4HF(g) DHo = ?
H2 (g) + F2 (g)  2HF (g) DHo = -537 kJ
C (s) + 2F2 (g)  CF4 (g) DHo = -680 kJ
2C (s) + 2H2 (g)  C2H4 (g) DHo = 52.3 kJ
Use average bond energies to determine the enthalpy of the following reaction
CH4 (g) + Cl2 (g)  CH3Cl (g) + HCl (g)
(BEC-Cl = 328 kJ/mol)

47. The End