1. Energy & Chemical Change
CHAPTER 7
Chemistry: The Molecular Nature of Matter, 6th edition
By Jesperson, Brady, & Hyslop
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

2. CHAPTER 7: Energy & Chemical Change
Learning Objectives
Potential vs Kinetic Energy
Internal Energy, Work, and Heat Calculations
System vs Surroundings
First Law of Thermodynamics
Units of Energy
Endothermic vs Exothermic
Heat of Temperature Changes (Specific Heat and Heat Capacity)
Calorimetry: Constant Volume vs Constant Pressure
Heat Stoichiometry (Thermochemical Equations)
Hess’s Law and Equation Summation
Heat of Reaction from Heats of Formation
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

3. Attractive forces that bind Atoms to each other in molecules, or
Energy
Chemical Bonds
Chemical bond
Attractive forces that bind
Atoms to each other in molecules, or
Ions to each other in ionic compounds
Give rise to compound’s potential energy
Chemical energy
Potential energy stored in chemical bonds
Chemical reactions
Generally involve both breaking and making chemical bonds
7.4 | Energy of Chemical Reactions
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

4. Atoms that are attracted to each other are moved closer together
Chemical Reactions
Energy
Forming Bonds
Atoms that are attracted to each other are moved closer together
Decrease the potential energy of reacting system
Releases energy
Breaking Bonds
Atoms that are attracted to each other are forced apart
Increase the potential energy of reacting system
Requires energy
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

5. Reaction where products have less chemical energy than reactants
Exothermic Reaction
Energy
Reaction where products have less chemical energy than reactants
Some chemical heat energy converted to kinetic energy
Reaction releases heat energy to surroundings
Heat leaves the system; q is negative ( – )
Heat energy is a product
Reaction gets warmer, temperature increases
Example.
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) + heat
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

6. Reaction where products have more chemical energy than reactants
Endothermic Reaction
Reaction where products have more chemical energy than reactants
Some kinetic energy converted to chemical energy
Reaction absorbs heat from surroundings
Heat added to system; q is positive (+)
Heat energy is a reactant
Reaction becomes colder,
temperature decreases
Example: Photosynthesis
6CO2(g) + 6H2O(g) + solar energy  C6H12O6(s) + 6O2(g)
© Michelle Molinari/Alamy
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

7. Larger amount of energy equals a stronger bond
Bond Strength
Energy
Measure of how much energy is needed to break bond or how much energy is released when bond is formed.
Larger amount of energy equals a stronger bond
Weak bonds require less energy to break than strong bonds
Key to understanding reaction energies
Example: If reaction has
Weak bonds in reactants and
Stronger bonds in products
Heat released
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

8. Methane and oxygen have weaker bonds
Energy
Why Fuels Release Heat
Fig 7.6
Methane and oxygen have weaker bonds
Water and carbon dioxide have stronger bonds
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

9. Amount of heat absorbed or released in chemical reaction
Heat of Reaction
Energy
Amount of heat absorbed or released in chemical reaction
Determined by measuring temperature change they cause in surroundings
Calorimeter
Instrument used to measure temperature changes
Container of known heat capacity
Use results to calculate heat of reaction
Calorimetry
Science of using calorimeter to determine heats of reaction
7.5 | Heat, Work, and the First Law of Thermodynamics
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

10. Calorimeter design not standard Depends on Type of reaction
Heats of Reaction
Energy
Calorimeter design not standard
Depends on
Type of reaction
Precision desired
Usually measure heat of reaction under one of two sets of conditions
Constant volume, qV
Closed, rigid container
Constant pressure, qP
Open to atmosphere
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

11. Amount of force acting on unit area
What is Pressure?
Energy
Amount of force acting on unit area
Atmospheric Pressure
Pressure exerted by Earth’s atmosphere by virtue of its weight.
~14.7 lb/in2
Container open to atmosphere
Under constant P conditions
P ~ 14.7 lb/in2 ~ 1 atm ~ 1 bar
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

12. Difference between qV and qP can be significant
Energy
Comparing qV and qP
Difference between qV and qP can be significant
Reactions involving large volume changes,
Consumption or production of gas
Consider gas phase reaction in cylinder immersed in bucket of water
Reaction vessel is cylinder topped by piston
Piston can be locked in place with pin
Cylinder immersed in insulated bucket containing weighed amount of water
Calorimeter consists of piston, cylinder, bucket, and water
(Fig 7.7)
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

13. Heat capacity of calorimeter = 8.101 kJ/°C
Energy
Comparing qV and qP
Heat capacity of calorimeter = kJ/°C
Reaction run twice, identical amounts of reactants
Run 1: qV - Constant Volume
Same reaction run once at constant
volume and once at constant pressure
Pin locked;
ti = C; tf = C
qCal = Ct
= J/C  (28.91 – 24.00)C = 39.8 kJ
qV = – qCal = –39.8 kJ
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

14. Run at atmospheric pressure Pin unlocked ti = 27.32 C; tf = 31.54 C
Comparing qV and qP
Energy
Run 2: qP
Run at atmospheric pressure
Pin unlocked
ti = C; tf = C
Heat absorbed by calorimeter is
qCal = Ct
= J/C  (31.54  27.32)C
= 34.2 kJ
qP = – qCal = –34.2 kJ
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

15. Uses up some energy that would otherwise be heat
Comparing qV and qP
Energy
qV = kJ
qP = kJ
System (reacting mixture) expands, pushes against atmosphere, does work
Uses up some energy that would otherwise be heat
Work = (–39.8 kJ) – (–34.2 kJ) = –5.6 kJ
Expansion work or pressure volume work
Minus sign means energy leaving system
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

16. P = opposing pressure against which piston pushes
Work Convention
Energy
Work = –P × V
P = opposing pressure against which piston pushes
V = change in volume of gas during expansion
V = Vfinal – Vinitial
For Expansion
Since Vfinal > Vinitial
V must be positive
So expansion work is negative
Work done by system
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

17. First Law of Thermodynamics
Energy
First Law of Thermodynamics
In an isolated system, the change in internal energy (E) is constant:
E = Ef – Ei = 0
Can’t measure internal energy of anything
Can measure changes in energy
E is state function
E = heat + work
E = q + w
E = heat input + work input
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

18. First Law of Thermodynamics
Energy
First Law of Thermodynamics
Energy of system may be transferred as heat or work, but not lost or gained.
If we monitor heat transfers (q) of all materials involved and all work processes, can predict that their sum will be zero
Some energy transfers will be positive, gain in energy
Some energy transfers will be negative, a loss in energy
By monitoring surroundings, we can predict what is happening to system
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

19. First Law of Thermodynamics
Energy
E = q + w
q is (+)
Heat absorbed by system (IN)
q is (–)
Heat released by system (OUT)
w is (+)
Work done on system (IN)
w is (–)
Work done by system (OUT)
Endothermic reaction
E = +
Exothermic reaction
E = –

20. E is Independent of Path
Energy
q and w
NOT path independent
NOT state functions
Depend on how change takes place
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

21. Discharge of Car Battery
Energy
Discharge of Car Battery
Path a
Short out with wrench
All energy converted to heat, no work
E = q (w = 0)
Path b
Run motor
Energy converted to work and little heat
E = w + q (w >> q)
E is same for each path
Partitioning between two paths differs
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

22. Bomb Calorimeter (Constant V)
Energy
Bomb Calorimeter (Constant V)
Apparatus for measuring E in reactions at constant volume
Vessel in center with rigid walls
Heavily insulated vat
Water bath
No heat escapes
E = qv
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

23. Calorimeter Problem: Ex 4
Energy
When g of olive oil is completely burned in pure oxygen in a bomb calorimeter, the temperature of the water bath increases from ˚C to ˚C.
a) How many Calories are in olive oil, per gram? The heat capacity of the calorimeter is kJ/˚C.
t = ˚ C – ˚C = ˚C
qabsorbed by calorimeter = Ct = kJ/°C × ˚C
Calculate heat absorbed by calorimeter
Convert this to heat released by combustion of olive oil
Convert heat to Cal/gram
= kJ
qreleased by oil = – qcalorimeter = – kJ
–8.740 Cal/g oil

24. Calorimeter Problem: Ex 4
Energy
Calorimeter Problem: Ex 4
Olive oil is almost pure glyceryl trioleate, C57H104O6. The equation for its combustion is
C57H104O6(l) + 80O2(g)  57CO2(g) + 52H2O
What is E for the combustion of one mole of glyceryl trioleate (MM = g/mol)? Assume the olive oil burned in part a) was pure glyceryl trioleate.
E = qV = –3.238 × 104 kJ/mol oil
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

25. Heat of reaction at constant Pressure (qP) H = E + PV
Enthalpy (H)
Energy
Heat of reaction at constant Pressure (qP)
H = E + PV
Similar to E, but for systems at constant P
Now have PV work + heat transfer
H = state function
At constant pressure
H = E + PV = (qP + w) + PV
If only work is P–V work, w = – P V
H = (qP + w) – w = qP
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

26. Enthalpy Change (H)
Energy
H is a state function
H = Hfinal – Hinitial
H = Hproducts – Hreactants
Significance of sign of H
Endothermic reaction
System absorbs energy from surroundings
H positive
Exothermic reaction
System loses energy to surroundings
H negative

27. Coffee Cup Calorimeter
Energy
Coffee Cup Calorimeter
Simple
Measures qP
Open to atmosphere
Constant P
Let heat be exchanged between reaction and water, and measure change in temperature
Very little heat lost
Calculate heat of reaction
qP = Ct
Ignore the tiny amount of heat that the cup and the thermometer absorb.
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

28. Coffee Cup Calorimetry: Ex 5
Energy
Coffee Cup Calorimetry: Ex 5
NaOH and HCl undergo rapid and exothermic reaction when you mix 50.0 mL of 1.00 M HCl and 50.0 mL of 1.00 M NaOH. The initial t = 25.5 °C and final t = 32.2 °C. What is H in kJ/mole of HCl? Assume for these solutions s = J g–1°C–1. Density: 1.00 M HCl = 1.02 g mL–1; 1.00 M NaOH = 1.04 g mL–1
NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(aq)
qabsorbed by solution = mass  s  t
massHCl = 50.0 mL  1.02 g/mL = 51.0 g
massNaOH = 50.0 mL  1.04 g/mL = 52.0 g
massfinal solution = 51.0 g g = g
t = (32.2 – 25.5) °C = 6.7 °C
1. Determine how much heat is absorbed by the calorimeter.
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

29. Coffee Cup Calorimetry: Ex 5
Energy
Coffee Cup Calorimetry: Ex 5
qcal = g  J g–1 °C–1  6.7 °C = 2890 J
Rounds to qcal = 2.9  103 J = 2.9 kJ
qrxn = –qcalorimeter = –2.9 kJ
= mol HCl
Heat evolved per mol HCl =
2. Heat evolved by reaction
= -58 kJ/mol
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

30. Enthalpy Changes in Chemical Reactions
Energy
Enthalpy Changes in Chemical Reactions
Focus on systems
Endothermic
Reactants + heat  products
Exothermic
Reactants  products + heat
Want convenient way to use enthalpies to calculate reaction enthalpies
Need way to tabulate enthalpies of reactions
7.7 | Thermochemical Equations
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

31. Standard state in thermochemistry
The Standard State
Energy
A standard state specifies all the necessary parameters to describe a system. Generally this includes the pressure, temperature, and amount and state of the substances involved.
Standard state in thermochemistry
Pressure = 1 atmosphere
Temperature = 25 °C = 298 K
Amount of substance = 1 mol (for formation reactions and phase transitions)
Amount of substance = moles in an equation (balanced with the smallest whole number coefficients)
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

32. Thermodynamic Quantities
Energy
Thermodynamic Quantities
E and H are state functions and are also extensive properties
E and H are measurable changes but still extensive properties.
Often used where n is not standard, or specified
E ° and H ° are standard changes and intensive properties
Units of kJ /mol for formation reactions and phase changes (e.g. H °f or H °vap)
Units of kJ for balanced chemical equations (H °reaction)
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

33. H in Chemical Reactions
Energy
H in Chemical Reactions
Standard Conditions for H 's
25 °C and 1 atm and 1 mole
Standard Heat of Reaction (H ° )
Enthalpy change for reaction at 1 atm and 25 °C
Example:
N2(g) H2(g)  2 NH3(g)
1.000 mol mol mol
When N2 and H2 react to form NH3 at 25 °C and 1 atm kJ released
H= –92.38 kJ
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

34. Thermochemical Equation
Energy
Thermochemical Equation
Write H immediately after equation
N2(g) + 3H2(g)  2NH3(g) H = –92.38 kJ
Must give physical states of products and reactants
H different for different states
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l ) H ° rxn = –890.5 kJ
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H ° rxn = –802.3 kJ
Difference is equal to the energy to vaporize water
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

35. Thermochemical Equation
Energy
Thermochemical Equation
Write H immediately after equation
N2(g) + 3H2(g)  2NH3(g) H= –92.38 kJ
Assumes coefficients is the number of moles
92.38 kJ released when 2 moles of NH3 formed
If 10 mole of NH3 formed
5N2(g) + 15H2(g)  10NH3(g) H= –461.9 kJ
H° = (5 × –92.38 kJ) = – kJ
Can have fractional coefficients
Fraction of mole, NOT fraction of molecule
½N2(g) + 3/2H2(g)  NH3(g) H°rxn = –46.19 kJ
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

36. C3H8(g) + 5O2(g) → 3 CO2(g) + 4 H2O(g) ΔH °rxn= –2043 kJ
Energy
State Matters!
C3H8(g) + 5O2(g) → 3 CO2(g) + 4 H2O(g)
ΔH °rxn= –2043 kJ
C3H8(g) + 5O2(g) → 3 CO2(g) + 4 H2O(l )
ΔH °rxn = –2219 kJ
Note: there is difference in energy because states do not match
If H2O(l ) → H2O(g) ΔH °vap = 44 kJ/mol
4H2O(l ) → 4H2O(g) ΔH °vap = 176 kJ/mol
Or –2219 kJ kJ = –2043 kJ
If we burn propane, we produce 2043 kJ of energy and make gas phase products. But the standard enthalpy of combustion for propane is –2219 kJ. Why?
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

37. Thermochemical Equations in Reverse
Energy
Thermochemical Equations in Reverse
To get energy out when form products, must put energy in to go back to reactants
Consequence of Law of Conservation of Energy
If you know H ° for reaction, you also know H ° for the reverse
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
H°reaction = – kJ
Reverse thermochemical equation
Must change sign of H
CO2(g) + 2H2O(g)  CH4(g) + 2O2(g)
H°reaction = kJ
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

38. Multiple Paths; Same H °
Energy
Multiple Paths; Same H °
Can often get from reactants to products by several different paths
Intermediate A
Products
Reactants
Intermediate B
7.8 | Hess’s Law
Should get same H °
Enthalpy is state function and path independent
Let’s see if this is true
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

39. Multiple Paths; Same H °: Ex 6
Energy
Multiple Paths; Same H °: Ex 6
Path a: Single step
C(s) + O2(g)  CO2(g) H°rxn = –393.5 kJ
Path b: Two step
Step 1: C(s) + ½O2(g)  CO(g) H °rxn = –110.5 kJ
Step 2: CO(g) + ½O2(g)  CO2(g) H °rxn = –283.0 kJ
Net Rxn: C(s) + O2(g)  CO2(g) H °rxn = –393.5 kJ
Chemically and thermochemically, identical results
True for exothermic reaction or for endothermic reaction
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

40. Multiple Paths; Same H °rxn: Ex 7
Energy
Multiple Paths; Same H °rxn: Ex 7
Path a: N2(g) + 2O2(g)  2NO2(g) H °rxn = 68 kJ
Path b:
Step 1: N2(g) + O2(g)  2NO(g) H °rxn = kJ
Step 2: 2NO(g) + O2(g)  2NO2(g) H °rxn = –112 kJ
Net rxn: N2(g) + 2O2(g)  2NO2(g) H °rxn = kJ
Hess’s Law of Heat Summation
For any reaction that can be written into steps, value of H °rxn for reactions = sum of H °rxn values of each individual step
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

41. Graphical description of Hess’ Law: Vertical axis = enthalpy scale
Energy
Enthalpy Diagrams
Graphical description of Hess’ Law:
Vertical axis = enthalpy scale
Horizontal line =various states of reactions
Higher up = larger enthalpy
Lower down = smaller enthalpy
Gives energy description of alternative pathways
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

42. Arrow down Hrxn = negative Arrow up Hrxn = positive Calculate cycle
Energy
Enthalpy Diagrams
Use to measure Hrxn
Arrow down Hrxn = negative
Arrow up Hrxn = positive
Calculate cycle
One step process = sum of two step process
Example: H2O2(l )  H2O(l ) + ½O2(g)
–286 kJ = –188 kJ + Hrxn
Hrxn = –286 kJ – (–188 kJ )
Hrxn = –98 kJ
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

43. Hess’s Law of Heat Summation Going from reactants to products
Energy
Hess’s Law of Heat Summation
Going from reactants to products
Enthalpy change is same whether reaction takes place in one step or many
Chief Use
Calculation of H °rxn for reaction that can’t be measured directly
Thermochemical equations for individual steps of reaction sequence may be combined to obtain thermochemical equation of overall reaction
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

44. Rules for Manipulating Thermochemical Equations
Energy
When equation is reversed, sign of H°rxn must also be reversed.
If all coefficients of equation are multiplied or divided by same factor, value of H°rxn must likewise be multiplied or divided by that factor
Formulas canceled from both sides of equation must be for substance in same physical states
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

45. Strategy for Adding Reactions Together:
Energy
Strategy for Adding Reactions Together:
Choose most complex compound in equation for one-step path
Choose equation in multi-step path that contains that compound
Write equation down so that compound
is on appropriate side of equation
has appropriate coefficient for our reaction
Repeat steps 1 – 3 for next most complex compound, etc.
Choose equation that allows you to
cancel intermediates
multiply by appropriate coefficient
Add reactions together and cancel like terms
Add energies together, modifying enthalpy values in same way equation modified
If reversed equation, change sign on enthalpy
If doubled equation, double energy
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

46. ] –1[ Energy Calculate H°rxn: Ex 8 C (s, graphite)  C (s, diamond)
Given C (s, gr) + O2(g)  CO2(g) H°rxn = –394 kJ
C (s, dia) + O2(g)  CO2(g) H°rxn = –396 kJ
To get desired equation, must reverse second equation and add resulting equations
C(s, gr) + O2(g)  CO2(g) H°rxn = –394 kJ
CO2(g)  C(s, dia) + O2(g) H°rxn = –(–396 kJ)
C(s, gr) + O2(g) + CO2(g)  C(s, dia) + O2(g) + CO2(g)
H° = –394 kJ kJ = + 2 kJ
–1[ ]
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

47. Need to Tabulate H° values Major problem is vast number of reactions
Energy
Tabulating H° values
Need to Tabulate H° values
Major problem is vast number of reactions
Define standard reaction and tabulate these
Use Hess’s Law to calculate H° for any other reaction
Standard Enthalpy of Formation, Hf°
Amount of heat absorbed or evolved when one mole of substance is formed at 1 atm (1 bar) and 25 °C (298 K) from elements in their standard states
Standard Heat of Formation
7.9 | Standard Heats of Reaction
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

48. Element Standard state
Energy
Most stable form and physical state of element at 1 atm (1 bar) and 25 °C (298 K)
Element
Standard state O
O2(g) C
C (s, gr) H
H2(g) Al
Al(s) Ne
Ne(g)
Note: All Hf° of elements in their standard states = 0
Forming element from itself.
See Appendix C in back of textbook and Table 7.2
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

49. Uses of Standard Enthalpy (Heat) of Formation, Hf°
Energy
Uses of Standard Enthalpy (Heat) of Formation, Hf°
From definition of Hf°, can write balanced equations directly
Hf° of C2H5OH(l )
2C(s, gr) + 3H2(g) + ½O2(g)  C2H5OH(l ) Hf° = – kJ/mol
Hf° of Fe2O3(s)
2Fe(s) + 3/2O2(g)  Fe2O3(s) Hf° = –822.2 kJ/mol
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

50. H°rxn has units of kJ H°f has units of kJ/mol
Energy
Using Hf°
2. Way to apply Hess’s Law wighout needing to manipulate thermochemical equations
Consider the reaction:
aA + bB  cC + dD
H°reaction = c × H°f(C) + d × H°f(D) – {a×H°f(A) + b×H°f(B)}
H°rxn has units of kJ because
Coefficients  heats of formation have units of mol  kJ/mol
H°reaction = –
Sum of all H°f of all of the products
Sum of all H°f of all of the reactants
H°rxn has units of kJ H°f has units of kJ/mol
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

51. Calculate H°rxn Using Hf°: Ex 9
Energy
Calculate H°rxn Using Hf°: Ex 9
Calculate H°rxn using Hf° data for the reaction
SO3(g)  SO2(g) + ½O2(g)
Multiply each Hf° (in kJ/mol) by the number of moles in the equation
Add the Hf° (in kJ/mol) multiplied by the number of moles in the equation of each product
Subtract the Hf° (in kJ/mol) multiplied by the number of moles in the equation of each reactant
H°rxn has units of kJ H°f has units of kJ/mol
H°rxn = 99 kJ
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

52. Don’t always want to know H°rxn
Energy
Other Calculations
Don’t always want to know H°rxn
Can use Hess’s Law and H°rxn to calculate Hf° for compound where not known
Example 10: Given the following data, what is the value of Hf°(C2H3O2–, aq)?
Na+(aq) + C2H3O2–(aq) + 3H2O(l )  NaC2H3O2·3H2O(s)
H°rxn = –19.7 kJ/mol
Naaq)
H f= –239.7 kJ/mol
NaC2H3O2•3H2O(s)
H f= kJ/mol
H2O(l)
H f= kJ/mol
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

53. Other Calculations: Ex 10 Continued
Energy
Other Calculations: Ex 10 Continued
H°rxn = Hf° (NaC2H3O2·3H2O, s) – Hf° (Na+, aq) – Hf° (C2H3O2–, aq) – 3Hf° (H2O, l )
Rearranging
Hf°(C2H3O2–, aq) = Hf°(NaC2H3O2·3H2O, s) – Hf°(Na+, aq) – H°rxn – 3Hf° (H2O, l)
Hf°(C2H3O2–, aq) = –710.4 kJ/mol – (–239.7kJ/mol) – (–19.7 kJ/mol) – 3(– kJ/mol)
= kJ/mol
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E