2. Thermochemistry
Chapter 6

3. Energy Capacity to do work or to produce heat.
Law of conservation of energy – energy can be converted from one form to another but can be neither created nor destroyed.
The total energy content of the universe is constant.
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4. Energy Potential energy – energy due to position or composition.
Kinetic energy – energy due to motion of the object and depends on the mass of the object and its velocity.
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5. Initial Position
In the initial position, ball A has a higher potential energy than ball B.
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6. Final Position
After A has rolled down the hill, the potential energy lost by A (changed to kinetic energy) has been converted to random motions of the components of the hill (frictional heating) and to the increase in the potential energy of B (which is less than the P.E of A ) The temp. of the hill slightly increases.
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8. Transfer of Energy
System – part of the universe on which we wish to focus attention.
Surroundings – include everything else in the universe.
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9. System and Surrounding

10. Chemical Energy Endothermic Reaction: Heat flow is into a system.
Absorb energy from the surroundings.
Exothermic Reaction:
Energy flows out of the system.
Energy gained by the surroundings must be equal to the energy lost by the system.
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11. - remove energy in order to slow the molecules down to form a solid.
CONCEPT CHECK!
Is the freezing of water an endothermic or exothermic process? Explain.
- remove energy in order to slow the molecules down to form a solid.
Exothermic process because you must remove energy in order to slow the molecules down to form a solid.
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12. (system is underlined)
CONCEPT CHECK!
Classify each process as exothermic or endothermic. Explain. The system is underlined in each example.
(system is underlined)
Your hand gets cold when you touch ice.
The ice gets warmer when you touch it.
Water boils in a kettle being heated on a stove.
Water vapor condenses on a cold pipe.
Ice cream melts.
Exo
Endo
Exothermic (heat energy leaves your hand and moves to the ice)
Endothermic (heat energy flows into the ice)
Endothermic (heat energy flows into the water to boil it)
Exothermic (heat energy leaves to condense the water from a gas to a liquid)
Endothermic (heat energy flows into the ice cream to melt it)
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13. Methane is burning in a Bunsen burner in a laboratory.
CONCEPT CHECK!
For each of the following, define a system and its surroundings and give the direction of energy transfer.
Methane is burning in a Bunsen burner in a laboratory.
Water drops, sitting on your skin after swimming, evaporate.
System – methane and oxygen to produce carbon dioxide and water; Surroundings – everything else around it; Direction of energy transfer – energy transfers from the system to the surroundings (exothermic)
System – water drops; Surroundings – everything else around it; Direction of energy transfer – energy transfers from the surroundings (your body) to the system (water drops) (endothermic)
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14. Hydrogen gas and oxygen gas react violently to form water. Explain.
CONCEPT CHECK!
Hydrogen gas and oxygen gas react violently to form water. Explain.
Which is lower in energy: a mixture of hydrogen and oxygen gases, or water?
Water is lower in energy because a lot of energy was released in the process when hydrogen and oxygen gases reacted.
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15. Thermodynamics
The study of energy and its interconversions is called thermodynamics.
Law of conservation of energy is often called the first law of thermodynamics.
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16. Internal Energy
Internal energy E of a system is the sum of the kinetic and potential energies of all the “particles” in the system.
To change the internal energy of a system: ΔE = q + w
q represents heat
w represents work
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17. Work vs Energy Flow To play movie you must be in Slide Show Mode
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18. Internal Energy (magnitude and direction)
Thermodynamic quantities consist of two parts:
Number gives the magnitude of the change.
Sign indicates the direction of the flow.
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19. Internal Energy Sign reflects the system’s point of view.
Endothermic Process:
q is positive
Exothermic Process:
q is negative
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20. Internal Energy Sign reflects the system’s point of view.
System does work on surroundings:
w is negative
Surroundings do work on the system:
w is positive
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21. Work Work = P × A × Δh = PΔV P is pressure. A is area.
Δh is the piston moving a distance.
ΔV is the change in volume.

22. Work
For an expanding gas, ΔV is a positive quantity because the volume is increasing. Thus ΔV and w must have opposite signs:
w = –PΔV
To convert between L·atm and Joules, use 1 L·atm = J.
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23. They perform the same amount of work.
EXERCISE!
Which of the following performs more work?
a) A gas expanding against a pressure of 2 atm from 1.0 L to 4.0 L.
A gas expanding against a pressure of 3 atm from 1.0 L to 3.0 L.
They perform the same amount of work.
They both perform the same amount of work. w = -PΔV
a) w = -(2 atm)( ) = -6 L·atm
b) w = -(3 atm)( ) = -6 L·atm
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24. Change in Enthalpy State function ΔH = q at constant pressure
ΔH = Hproducts – Hreactants
Energy is a state function; work and heat are not
State Function – property that does not depend in any way on the system’s past or future (only depends on present state).
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25. –252 kJ Consider the combustion of propane: EXERCISE! ΔH = –2221 kJ
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
ΔH = –2221 kJ
Assume that all of the heat comes from the combustion of propane. Calculate ΔH in which 5.00 g of propane is burned in excess oxygen at constant pressure.
–252 kJ
(5.00 g C3H8)(1 mol / g C3H8)(-2221 kJ / mol C3H8)
ΔH = -252 kJ
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26. Calorimetry Science of measuring heat Specific heat capacity:
The energy required to raise the temperature of one gram of a substance by one degree Celsius.
Molar heat capacity:
The energy required to raise the temperature of one mole of substance by one degree Celsius.
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27. HEAT CAPACITY The heat required to raise an object’s T by 1 ˚C.
Which has the larger heat capacity?

28. specific heat(s) and heat capacity (C)
The specific heat(s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius.
C = m x s
Heat (q) absorbed or released
q = m x s x Dt
q = C x Dt
Dt = tfinal - tinitial

29. specific heat(s) and heat capacity (C)
This Table is not from Zumdahl’s book

30. Calorimetry
If two reactants at the same temperature are mixed and the resulting solution gets warmer, this means the reaction taking place is exothermic.
An endothermic reaction cools the solution.
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31. A Coffee–Cup Calorimeter Made of Two Styrofoam Cups
RReview Expt.11RReview Expt.11
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32. Constant-Volume Calorimetry
q qrxn = - (qwater + qbomb) rxn = - (qwater + qbomb)
Reaqwater = m x s x Dt
ction at Constant V
qwater = m x s x Dt
qbomb = Cbomb x Dt
Reaction at Constant V
DH ~ q rxn
DH ~ rxn
No heat enters or leaves!

33. Calorimetry Energy released (heat) = s × m × ΔT
s = specific heat capacity (J/°C·g)
m = mass of solution (g)
ΔT = change in temperature (°C)
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34. ENTHALPY ∆H = Hfinal - Hinitial
Most chemical reactions occur at constant P, so
Heat transferred at constant P = qp
qp = ∆H where H = enthalpy
and so ∆E = ∆H + w (and w is usually small)
∆H = heat transferred at constant P ≈ ∆E
∆H = change in heat content of the system
∆H = Hfinal - Hinitial

35. A 466-g sample of water is heated from 8.50°C to 74.60°C.
Calculate the amount of heat absorbed (in kilojoules) by the water.

36. Strategy We know the quantity of water and the specific heat of water
Strategy We know the quantity of water and the specific heat of water. With this information and the temperature rise, we can calculate the amount of heat absorbed (q).
Solution Using Equation (6.12), we write
Check The units g and °C cancel, and we are left with the desired unit kJ. Because heat is absorbed by the water from the surroundings, it has a positive sign.

37. q= q= ms∆t = 1000g x 4.184 j/ g oC x 6 oC =25104J
17. How many kJ of heat must be removed from 1000 g of water (specific heat = J /g oC) to lower the temperature from 18.0°C to 12.0°C?
A x 10–2 kJ B kJ C kJ D. 25 kJ
q= ms∆t = 1000g x j/ g oC x 6 oC =25104J
= 25 kJ
q= q= ms∆t = 1000g x j/ g oC x 6 oC =25104J
ms∆t = 1000g x j/ g oC x 6 oC =25104J

38. CONCEPT CHECK!
A g sample of water at 90°C is added to a g sample of water at 10°C. The final temperature of the water is: a) Between 50°C and 90°C b) 50°C c) Between 10°C and 50°C
The correct answer is b).
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39. CONCEPT CHECK!
A g sample of water at 90.°C is added to a g sample of water at 10.°C. The final temperature of the water is: a) Between 50°C and 90°C b) 50°C c) Between 10°C and 50°C Calculate the final temperature of the water. 23°C
The correct answer is c). The final temperature of the water is 23°C.
- (100.0 g)(4.18 J/°C·g)(Tf – 90.) = (500.0 g)(4.18 J/°C·g)(Tf – 10.)
Tf = 23°C
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40. CONCEPT CHECK!
You have a Styrofoam cup with 50.0 g of water at 10.°C. You add a 50.0 g iron ball at 90. °C to the water. (sH2O = 4.18 J/°C·g and sFe = 0.45 J/°C·g) The final temperature of the water is: a) Between 50°C and 90°C b) 50°C c) Between 10°C and 50°C Calculate the final temperature of the water. 18°C
The correct answer is c). The final temperature of the water is 18°C.
- (50.0 g)(0.45 J/°C·g)(Tf – 90.) = (50.0 g)(4.18 J/°C·g)(Tf – 10.)
Tf = 18°C
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41. Qreaction for the combustion of the one-gram sample.
Problem
A g sample of rocket fuel hydrazine, N2 H4 , is burned in a bomb calorimeter containing 1200 g of water. The temperature rises from to ⁰C. Taking C for the bomb to be 840 J/ ⁰C , calculate
Qreaction for the combustion of the one-gram sample.
Qreaction for the combustion of the one mole of hydrazine in the bomb calorimeter.

42. QAAaction = - (qwater + q calorimeter)
qQreaction = - (qwater + q calorimeter)
qwater = g x J/g ⁰C x (28.16 – 24.62) ⁰C
= J
qcalorimeter = 840J/ ⁰C x ( ) = J
(a) Qreaction for 1 g N2H4 = J J = J = kJ
(b) Qreaction for 1 mole of N2H4 = 20.7 kJ /gx 32 g/mole = kJ
water – 1200 g x J/g ⁰C x (28.16 – 24.62) ⁰C
Qreaction for 1 g N2H4 = J J = J = kJ
Qreaction for 1 mole of N2H4 = 20.7 kJ /gx 32 g/mole =

43. In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.
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44. N2(g) + 2O2(g) → 2NO2(g) ΔH1 = 68 kJ
This reaction also can be carried out in two distinct steps, with enthalpy changes designated by ΔH2 and ΔH3.
N2(g) + O2(g) → 2NO(g) ΔH2 = 180 kJ
2NO(g) + O2(g) → 2NO2(g) ΔH3 = – 112 kJ
N2(g) + 2O2(g) → 2NO2(g) ΔH2 + ΔH3 = 68 kJ
ΔH1 = ΔH2 + ΔH3 = 68 kJ
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45. The Principle of Hess’s Law
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47. Characteristics of Enthalpy Changes
If a reaction is reversed, the sign of ΔH is also reversed.
The magnitude of ΔH is directly proportional to the quantities of reactants and products in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of ΔH is multiplied by the same integer.
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48. Example Consider the following data: Calculate ΔH for the reaction
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49. Problem-Solving Strategy
Work backward from the required reaction, using the reactants and products to decide how to manipulate the other given reactions at your disposal.
Reverse any reactions as needed to give the required reactants and products.
Multiply reactions to give the correct numbers of reactants and products.
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50. Example Reverse the two reactions: Desired reaction:
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51. Example
Multiply reactions to give the correct numbers of reactants and products:
4( ) 4( )
3( ) 3( )
Desired reaction:

52. Example Final reactions: Desired reaction: ΔH = +1268 kJ
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53. Hess’s Law
FoFor example, suppose you are given the following data:
r example, suppose you are given the following data:
Could you use these data to obtain the enthalpy change for the following reaction?
you use these data to obtain the ealpy change for the following reaction? 3

54. Hess’s Law
If we multiply the first equation by 2 and reverse the second equation, they will sum together to become the third. 3

55. Standard Enthalpy of Formation (ΔHf°)
Change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states.
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56. Table 6.4 is not from Zumdahl’s book

58. A Schematic Diagram of the Energy Changes for the Reaction CH4(g) + 2O2(g) CO2(g) + 2H2O(l)
∆H◦f of the compounds involved are: CH4: –75 kJ , CO2: -394 kJ, H2O: –286 kJ ∆H◦rxn = ∆H◦f products - ∆H◦f reactants = {[ x (-286)] –(-75)} = –891 kJ Answer: –891 kJ
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59. Petroleum, Natural Gas, and Coal Wood Hydro Nuclear
Fossil Fuels
Petroleum, Natural Gas, and Coal
Wood
Hydro
Nuclear
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60. Energy Sources Used in the United States
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61. The Earth’s Atmosphere
Transparent to visible light from the sun.
Visible light strikes the Earth, and part of it is changed to infrared radiation.
Infrared radiation from Earth’s surface is strongly absorbed by CO2, H2O, and other molecules present in smaller amounts in atmosphere.
Atmosphere traps some of the energy and keeps the Earth warmer than it would otherwise be.
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62. The Earth’s Atmosphere
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63. Other Energy Alternatives Oil shale Ethanol Methanol Seed oil
Coal Conversion
Hydrogen as a Fuel
Other Energy Alternatives
Oil shale
Ethanol
Methanol
Seed oil
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