1. Chapter 16: Energy and Chemical Change
Table of Contents
Chapter 16: Energy and
Chemical Change
16.1: Energy
16.2: Heat in Chemical Reactions and Processes
16.3: Thermochemistry
16.3: Thermochemical Equations
16.4: Calculating Enthalpy Change

2. universe = system + surroundings
Energy and Chemical Change: Basic Concepts
Chemical Energy and the Universe
universe = system + surroundings
The Universe is everything around us
the System: the specific part of the universe that you are studying.
the Surroundings: everything else

3. Energy is the ability to do work or produce heat.
Energy and Chemical Change: Basic Concepts
The Nature of Energy
Energy is the ability to do work or produce heat.
Potential energy: stored energy.
Kinetic energy: the energy of movement.

4. Chemical potential energy: stored in bonds
Energy and Chemical Change: Basic Concepts
The Nature of Energy
Chemical potential energy:
stored in bonds
The energy stored in a substance because of its composition :
the type of atoms in the substance
the number and type of chemical bonds joining the atoms
the way atoms are arranged.

5. Molecular kinetic energy: Measure of molecular “speed”
Energy and Chemical Change: Basic Concepts
The Nature of Energy
Molecular kinetic energy: Measure of molecular “speed”
Measured by a thermometer: As temperature increases, the motion of submicroscopic particles increases. So a thermometer is like a molecular speedometer.

6. Chemical systems contain both kinetic energy and potential energy.
Energy and Chemical Change: Basic Concepts
The Nature of Energy
Chemical systems contain both kinetic energy and potential energy.
Law of conservation of energy: energy can be converted from one form to another, but it is neither created nor destroyed.
suppose you have money in two accounts at a bank and you transfer funds from one account to the other. Although the amount of money in each account has changed, the total amount of your money in the bank remains the same.
the first law of thermodynamics

7. Chemical potential energy
Energy and Chemical Change: Basic Concepts
Chemical potential energy
Heat (symbol=q), is the flow of kinetic energy due to temperature changes.
Exothermic = gives off heat (hot)
Forming bonds – releases energy
Endothermic = takes in heat (cold)
Breaking bonds – requires energy
Heat always travels from hot to cold.

8. 1,000 calories = 1 kilocal = 1 Calorie
Energy and Chemical Change: Basic Concepts
Measuring heat
UNITS
calorie (cal): metric unit of energy
the amount of energy required to raise the 1 gram of water 1° Celsius
1,000 calories = 1 kilocal = 1 Calorie
dietary Calories
chemistry calories

9. joule (J): The SI unit of heat and energy
Energy and Chemical Change: Basic Concepts
Measuring heat
UNITS
joule (J): The SI unit of heat and energy
4.184 joules = 1 calorie
1,000 J = 1 kJ
(Calories are bigger than joules)

10. Converting Energy Units Steps for converting units:
Energy and Chemical Change: Basic Concepts
Converting Energy Units
Steps for converting units:
1. Find your goal
2. Start with what you know
3. Use conversion factors to change units
4. Check that units cancel

11. Converting Energy Units
Energy and Chemical Change: Basic Concepts
Converting Energy Units
How much energy in joules is 230 nutritional Calories?
Goal = Joules Known = 230 Calories
J = 230 Cal
1000 cal
4.184 J
= J
962,320
1 Cal
1 cal
= 9.6 x 105 J
convert Cal to cal
convert cal to J

12. Practice converting energy units
Convert 142 Nutritional Calories into calories.
An exothermal reaction releases 86.5 kJ. How many kilocalories of energy are released?
If an endothermic process absorbs 256 J, how many kilocalories are absorbed?
142,000 cal
20.7 kcal
kcal

13. used to determine the specific heat of unknown metals
Energy and Chemical Change: Basic Concepts
Measuring Heat
Calorimeter: a device used to measure the amount of heat absorbed or released during a chemical or physical process.
used to determine the specific heat of unknown metals

14. Each substance has its own specific heat.
Energy and Chemical Change: Basic Concepts
Specific Heat
The specific heat (c): the amount of heat required to raise the temperature of 1 gram of any substance by 1° C.
Each substance has its own specific heat.
Why? Because different substances have different compositions.
specific heat (c) of water = J/(g∙°C)

15. ∆T = |Tfinal – Tinitial|
Energy and Chemical Change: Basic Concepts
Calculating heat evolved and absorbed
Heat absorbed or released by a substance during a change in temperature:
specific heat
(J/g ·°C)
mass (g)
Change in temperature (°C)
HEAT (J)
∆T = |Tfinal – Tinitial|

16. Calculating Specific Heat
Energy and Chemical Change: Basic Concepts
Calculating Specific Heat
The temperature of a sample of iron with a mass of 10.0 g changed from 50.4°C to 25.0°C with the release of 114 J heat. What is the specific heat of iron? q
114 J
c = = =
m · ΔT
(10.0 g)
|25.0°C-50.4°C|
114 J = =
.449 J/(g·°C)
(10.0 g)
(25.4°C)

17. Using Data from Calorimetry
Energy and Chemical Change: Basic Concepts
Using Data from Calorimetry
A piece of metal with a mass of 4.68 g absorbs 256 J of heat when its temperature increases by 182°C. What is the specific heat of the metal?
Known
Unknown
mass of metal = 4.68 g metal
specific heat
c = ? J/(g·°C)
heat absorbed, q = 256 J
∆T = 182°C
c = .301 J/(g·°C)

18. Practice specific heat equation
If the temperature of 34.4 g of ethanol increases from 25.0°C to 78.8°C, how much heat has been absorbed?
4.50 g of gold [c = J/(g·°C)] absorbed J of heat. If the initial temperature was 25.0°C, what was the final temperature?
4.52 x 103 J
5.00 x 102 °C

19. End day 1

20. Chapter 16: Energy and Chemical Change
Table of Contents
Chapter 16: Energy and
Chemical Change
16.2-3: Thermochemistry

21. Practice specific heat equation
If the temperature of 34.4 g of ethanol increases from 25.0°C to 78.8°C, how much heat has been absorbed?
4.50 g of gold [c = J/(g·°C)] absorbed J of heat. If the initial temperature was 25.0°C, what was the final temperature?
4.52 x 103 J
5.00 x 102 °C

22. Practice specific heat equation
Endothermic or exothermic?
Melting ice
Boiling water
Freezing water
Making water from steam
endo
endo
exo
exo

23. Endo- or exothermic?

24. Chemical Energy and the Universe
Energy and Chemical Change: Basic Concepts
Chemical Energy and the Universe
Thermochemistry: the study of heat changes that accompany chemical reactions and phase changes.
Enthalpy (H): the heat content of a system at constant pressure.
Enthalpy (heat) of reaction (∆Hrxn) = the change in enthalpy for a reaction.
always measured as change.

25. Thermochemical Equations
Exothermic
4Fe(s) + 3O2(g) → 2Fe2O3(s) kJ
4Fe(s) + 3O2 → 2Fe2O3(s); ∆H = kJ
Endothermic
27 kJ + NH4NO3(s) → NH4+(aq) + NO3-(aq)
NH4NO3(s) → NH4+(aq) + NO3-(aq); ∆H = 27 kJ

26. Standard enthalpy changes have the symbol ∆H°.
Energy and Chemical Change: Basic Concepts
Enthalpy changes
Standard enthalpy changes have the symbol ∆H°.
Standard conditions:
1 atm pressure
298 K (25°C)
Units of enthalpy:
kJ released
1 mol substance

27. ∆Hcomb is exothermic (-)
Energy and Chemical Change: Basic Concepts
Enthalpy changes
Types of enthalpy
Enthalpy (heat) of combustion (∆Hcomb) = enthalpy change for the complete burning of one mole of the substance.
∆Hcomb is exothermic (-)

28. PHASE CHANGES
TEMPERATURE
boiling
evaporating
gas BP
condensing
liquid
melting MP
freezing
solid
ENERGY

29. ∆Hvap and ∆Hfus are endothermic (+)
Energy and Chemical Change: Basic Concepts
Changes of state
Molar enthalpy (heat) of vaporization (∆Hvap) = heat required to evaporate one mole of a liquid (liquid → gas)
Molar enthalpy (heat) of fusion (∆Hfus) = heat required to melt one mole of a solid (solid → liquid)
∆Hvap and ∆Hfus are endothermic (+)

30. ∆Hcondensation = -∆Hvap ∆Hsolidification = -∆Hfus
Energy and Chemical Change: Basic Concepts
Changes of state
Molar enthalpy (heat) of condensation (∆Hcond) = heat required to condense one mole of a liquid (gas → liquid)
Molar enthalpy (heat) of solidification (∆Hsolid) = heat required to solidify one mole of a solid (liquid → solid)
∆Hcondensation = -∆Hvap
∆Hsolidification = -∆Hfus

31. PHASE CHANGES
TEMPERATURE
ΔHchange
boiling
evaporating
gas BP
condensing
liquid
melting MP
freezing
q=c∙m∙ΔT
solid
ENERGY

32. Converting Energy Units
Energy and Chemical Change: Basic Concepts
Converting Energy Units
How much heat is released when 82.1 g of methanol is burned?
Goal = heat released Known = 82.1 g methanol
82.1 g CH3OH
1 mol CH3OH
- 726 kJ
32.05 g CH3OH
1 mol CH3OH
Use ENTHALPY to convert mol to kJ
convert
g to mol
MOLAR MASS
q = kJ
1860

33. Practice finding heat ∆H = enthalpy = (+) endothermic = (-) exothermic
WHY? It tells us if it’s endo- or exothermic
q = heat
= always (+)
WHY? Because it tells us HOW MUCH energy there is, NOT if it’s endo- or exothermic

34. Practice finding heat 2.58 kJ 377 kJ 232 g
Calculate the heat required to melt 25.7 g of solid methanol.
How much heat is evolved when 275 g of ammonia gas condenses to a liquid?
What mass of methane must be burned in order to release 12,880 kJ of heat?
2.58 kJ
377 kJ
232 g

35. End Day 2

36. WebAssign #14 q=c∙m∙ΔT ΔHchange 693 kJ
How much heat is required to change 213 g of ice from -46.5ᵒC to 0.0ᵒC, melt the ice, warm the water from 0.0ᵒC to 100ᵒC, boil the water, and heat the steam to 173.0ᵒC?
100ᵒ to 173ᵒC
q=c∙m∙ΔT
boil
ΔHchange
melting
0.0ᵒC to 100ᵒC
-46.5ᵒC to 0.0ᵒC
693 kJ

37. Hess’ Law (adding equations) Comparing products to reactants
Energy and Chemical Change: Basic Concepts
There are two ways that we will calculate the standard enthalpy of reaction:
Hess’ Law (adding equations)
Comparing products to reactants
- both use enthalpies of formation
The standard enthalpy (heat) of formation (∆H°f) of a compound is the change in enthalpy for the formation of one mole of a compound

38. Calculating Enthalpy Change
Energy and Chemical Change: Basic Concepts
Calculating Enthalpy Change
Hess’s law: If you can add two equations, you can add their enthalpies
Used to find the enthalpy of the overall reaction.
Think of it like you’re doing a puzzle

39. Some have 3 equations – follow the same principles
Applying Hess’ Law
Use the following chemical equations to determine ΔH for this equation: 2H2O2(l) → 2H2O(l) +O2(g)
2H2(g) +O2(g) → 2H2O(l) ΔH = -572 kJ
H2(g) +O2(g) → H2O2(l) ΔH = -188 kJ
Identify products and reactants
Match coefficients to overall equation
Add equations (cancel spectators)
Add enthalpies
ΔH = -196 kJ
Some have 3 equations – follow the same principles

40. Determine ΔH for each reaction
2CO(g) + 2NO(g) → 2CO2(g) + N2(g) ∆H = ?
2CO(g) + O2(g) → 2CO2(g) ∆H = kJ
N2(g) + O2(g) → 2NO(g) ∆H = kJ
4Al(s) + 3MnO2(s) → 2Al2O3(s) + 3Mn(s) ∆H = ?
4Al(s) + 3O2(g) → 2Al2O3(s) ∆H = kJ
Mn(s) + O2(g) → MnO2(s) ∆H = -521kJ kJ
-1789 kJ

41. Uses the standard enthalpies of formation for each individual molecule
Energy and Chemical Change: Basic Concepts
A simpler way:
Uses the standard enthalpies of formation for each individual molecule
f reactants
f products
The standard heat of formation of an element in its standard state is zero. Ex) Au, O2; ΔH°f = 0 kJ

42. Standard Enthalpies of formation
Use standard enthalpies of formation to calculate ∆H°rxn for CH4(g) + 2O2(g) → CO2(g) +2H2O(l)
1. Find ∆H°f for known elements
∆H°f (CH4(g)) = -75 kJ
∆H°f (O2(g)) = 0.0 kJ
∆H°f (CO2(g)) = -394 kJ
∆H°f (H2O(g)) = -286 kJ

43. Standard Enthalpies of formation
Use standard enthalpies of formation to calculate ∆H°rxn for CH4(g) + 2O2(g) → CO2(g) +2H2O(l)
2. Use formula ∆H°rxn = ∆H°f(prod) - ∆H°f(react)
∆H°rxn = (∆H°f(CO2) + ∆H°f(H2O)) – (∆H°f(CH4) + ∆H°f(O2))
∆H°rxn = [(-394 kJ + 2(-286 kJ)] – [(-75 kJ + 2(0.0 kJ)]
= [-966 kJ] – [-75 kJ]
= -966 kJ + 75 kJ
= -891 kJ
The combustion of
1 mol CH4 releases 891 kJ

44. Practice ∆H°rxn ∆H°rxn = 178.3 kJ ∆H°rxn = 66.36 kJ
Use standard enthalpies of formation to calculate ∆H°rxn for each of the following reactions.
CaCO3(s) → CaO(s) + CO2(g)
N2(g) + 2O2(g) → 2NO2(g)
4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(l)
∆H°rxn = kJ
∆H°rxn = kJ
∆H°rxn = kJ

45. End Ch.16