1. Energy Thermodynamics-study of energy and its interconversions Labs
#22 Calorimetry
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2. Energy Work Heat Capacity to do work (or produce heat)
Energy used to cause object with mass to move against a force
Force x distance (force acting over distance)
Heat
Involves transfer of energy between two objects
Energy used to cause temperature of object to increase
Chemicals may store potential energy in bonds that can be released as heat energy
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3. Potential energy- energy due to position or composition
Depends on position relative to pull of gravity
Stored energy released when process occurs that changes its position
When substances undergo chemical reactions, positions of atoms change, and energy is released or absorbed
Chemical energy stored within molecules as consequence of attractions between electrons and atomic nuclei (electrostatic potential energy, Eel)
Proportional to electrical charge on two interacting objects, Q1Q2, and inversely proportional to distance, d, separating them
Based on magnitude of electron charge (1.60 x C)
If both have same sign, Eel is positive
If opposite signs, Eel is negative Eel = KQ1Q2/d
K = constant of proportionality, 8.99 x 109 J-m/C2
Lower energy of system/more strongly opposite charges interact, more stable the of system
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4. Kinetic energy- energy due to the motion of an object
KE= ½ mv2
According to kinetic-molecular theory, all matter has thermal energy, because atoms and molecules are in constant motion
Vibrate/rotate/move from one point to another
More energy of molecular motion, higher temperature
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5. Temperature vs. Heat Temperature Heat
Random motion of particles in substance
Indicates direction heat energy will flow
Heat
Measure of energy content
What is transferred (due to temperature difference between system and surroundings)
Always flows from hotter to colder body
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6. State Functions (state property) property of system that depends only on its present state
Independent of pathway (energy, temperature, pressure, enthalpy, entropy)
Does not matter how you get there/how process carried out
Liter of water behind dam has same potential energy for work regardless of whether it flowed downhill to dam, or was taken uphill to dam in bucket (PE state function dependent only on current position of water, not how it got there)
If you climb mountain, you can take many paths to get to top, but once there, height is independent of path
If NaCl is produced by one equation or many equations, change in heat still the same
Work and heat are not state functions
Travel to top of mountain by many paths, so distance traveled is dependent on path taken
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8. Exothermic Reactions Give off energy (PE in chemical bonds)
Products generally more stable (stronger bonds) than reactants
Total chemical energy of products less than total chemical energy of reactants
Heat released during chemical reaction comes from decrease in chemical potential energy as reactants are converted to products
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9. Endothermic Reactions
Energy absorbed from surroundings
Energy (heat) flows into system, increasing its PE
Products generally less stable (weaker bonds) than reactants
Products have greater chemical energy than reactants
Difference in energy between reactants/products is equal to heat absorbed by system
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10. Thermodynamics (Euniverse is constant)
First law of thermodynamics (law of conservation of energy) energy can be converted from one form to another, but cannot be created or destroyed
(Euniverse is constant)
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11. Thermodynamic quantities
ΔE = Efinal -Einitial
Number/unit-gives magnitude of change
Sign-indicates direction of flow
+q (ΔE)
Efinal > Einitial
Endothermic reactions
Energy flows into system (gained energy)
System’s energy is increasing
-q (ΔE)
Efinal < Einitial
Exothermic reactions
Energy flows out of system (lost energy)
System’s energy is decreasing
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12. System Energy Internal energy, E, of system ΔE = q + w
Sum of KE/PE of all particles in system
Change energy of system by flow of work, heat or both
Heat gained/word done on system both positive quantities
Both increase internal energy of system, causing ΔE to be positive quantity
ΔE = q + w
ΔE = change in system’s internal energy
q = heat
w = work
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13. Sign conventions for q, w and ΔE
+ means system gains heat
- Means system loses heat
For w
+ means work done on system
- Means work done by system
For ΔE
+ means net gain of energy by system
- Means net loss of energy by system
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14. Work
Energy lost/gained by system by mechanical means, rather than by heat conduction
Systems do not "contain" work or heat, but energy
SI unit for energy (work, heat)-joule (J)
Non-SI unit for heat-calorie (cal)-amount of heat needed to raise temperature of 1 gram of water by 1°C from 14.5°C to 15.5°C
Heat or energy in calories can be converted to joules
1 cal = J and 1 kcal = kJ
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15. Since pressure = force / area, work = pressure  volume wsystem = PV
work = force  distance
Since pressure = force / area,
work = pressure  volume
wsystem = PV
Units of pressure
P = F/A = kg m s2/m2 = kg m-1 s-2 = 1 Pascal
Units of volume
P x V = kg m-1s-2 x m3 = kg m2 s-2 = 1 Joule
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16. Work done by gases: Work = P∆V for ideal gases (pressure constant)
From point of view of system, W = -P∆V
By gas (through expansion)
System expands, positive work done on surroundings, negative work done on system
ΔV is positive/w is negative
To gas (by compression)
System contracts, surrounding have done work on system, positive work done on system
ΔV is negative/w is positive
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17. Calculate the work (with proper sign) associated with the contraction of a gas from 75 L to 30 L (work is done “on the system”) at a constant external pressure of 6.0 atm in:
L atm
∆V = Vfinal –Vinitial = 30 L – 75 L = -45 L
W = -P∆V = -6.0 atm (-45 L) = +270 L atm
Joules (1 L atm = J)
+270 L atm J = +2.7 x 104 J
1 L atm
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18. Calculate the change in energy of the system if 38
Calculate the change in energy of the system if 38.9 J of work is done by the system with an associated heat loss of 16.2 J.
Get the sign correct:
Q = - because heat is lost
W = - because work is done by the system
Solve:
ΔE = q + w = J + (-38.9 J) = J
The system has lost 55.1 J of energy.
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19. A piston is compressed from a volume of 8. 3 L to 2
A piston is compressed from a volume of 8.3 L to 2.8 L against a constant pressure of 1.9 atm. In the process, there is a heat gain by the system of 350 J. Calculate the change in energy of the system.
w = -P∆V
w = -1.9 atm (-5.5 L) = L atm
10.45 L atm J = J (q)
L atm
ΔE = q + w = = 1409 J = 1400 J
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20. Homework: Read 6.1, pp. 241-248 Q pp. 280-281, #9, 10, 18, 22, 24, 28
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21. Enthalpy (from Greek “to warm”)
Formerly called “heat content” so it is still symbolized as H
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22. Enthalpy H = E + PV E = internal energy of system
P = pressure of system
V = volume of system
Since internal energy, pressure and volume all state functions, enthalpy also state function (change in H independent of pathway)
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23. Enthalpy = ∆H = ∆E + ∆PV E = qP + w Since w = -P∆V
E = qP  PV (qP = heat at constant P)
qP = H only at constant pressure
So H = E + PV
Therefore, at constant pressure, qP = E + PV
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24. Enthalpy of reaction (rxn) or enthalpy change of system
Heat transferred to (absorbed) or from (evolved) system at constant pressure)
Difference between enthalpies of products and reactants
ΔH = enthalpy of system after reaction – enthalpy of system before reaction
Difference between sum of enthalpies of products and sum of enthalpies of reactants
ΔH = ∑Hproducts – ∑Hreactants
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25. Changes in enthalpy can be positive or negative
Endothermic reaction
H(products) > H(reactants): H is positive, ΔH>0
Heat added (absorbed by system) is converted into potential energy of system
Exothermic reaction
H(reactants) > H(products): H is negative, ΔH<0
Loss of chemical potential energy accounts for the amount of heat released (given off by system)
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26. 3/25/2017

27. Enthalpy Diagrams
Each line represents set of reactants or products for balanced chemical reaction
When going from one line to another, atoms must balance
Enthalpy associated with condensation of water
CO2(g) on both sides cancel to yield
Relative distance of each line reflect relative enthalpy difference (ΔH) between reactants/products
If enthalpy change in going from reactants to products is negative, line for products must be below reactants
Length of distance must be proportional
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28. Upon adding solid sodium hydroxide pellets to water, the following reaction takes place: NaOH(s)  NaOH(aq). For this reaction at constant pressure, ∆H = -43 kJ/mol. Answer the following questions regarding the addition of 14 g of NaOH to water:
Does the beaker get warmer or colder?
If ∆H < 0, heat is given off by system-beaker gets warmer.
Is the reaction exo- or endothermic?
If heat is given off by system, reaction is exothermic.
What is the enthalpy change for the dissolution?
∆H = -43 kJ 1 mol NaOH g NaOH = -15kJ
mol g NaOH
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29. Calorimetry
Measurement of heat flow (released or absorbed) in reaction by calorimeter
Experimental technique used to determine heat exchange (q) associated with reaction
All calorimetry is based on measuring temperature change (t) of medium such as water
Techniques/equipment used depend on nature of process being studied
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30. At constant pressure, q = ∆H At constant volume, q = ∆E
In both cases, heat gain or loss is determined
Amount of heat exchanged in reaction depends upon
Net temperature change during reaction
Amount of substances (more you have, more heat can be exchanged)
Heat capacity (C) of substance
Three ways to express heat capacity
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31. Heat Capacity (C)
Amount of heat (q) required to raise temperature of given mass of substance by 1oC
Units: J/oC or J/K
Ratio of heat absorbed to increase in T
C = q/Δt (amt. of heat flowing in/out of substance)
Δt is temperature change given by tfinal – tinitial
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32. Specific Heat Capacity (Specific heat)
Heat capacity of one gram of substance
Unit = J/oC g or J/K g
Energy required to raise temp of 1 gram of substance by 1°C
Can calculate amount of heat absorbed by given amount of water/by object of given mass
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33. Molar Heat Capacity Heat capacity of one mole of substance
Unit = J/oC-mol or J/K mol
Energy required to raise temp of 1 mole of substance by 1°C
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34. Constant Pressure Calorimeter (solutions)
Determines changes in enthalpy (heats of reactions) for reactions occurring in solutions
“coffee-cup” calorimeter
Constant atmospheric pressure
No physical boundary between system/surroundings
Reactants/products are system
Water in which they dissolve/calorimeter are surroundings
Assume calorimeter prevents gain/loss of heat from solution (heat does not escape)
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35. Calculating Heat of Rxn, ΔH
When reactants (each at same T) are mixed, T of mixed solution is observed to increase or decrease
Calculating Heat of Rxn, ΔH
Energy released by reaction = energy absorbed by solution
Energy released = s x m x ΔT
qsoln = (sp. Heat solution) x (g of solution) x ΔT = -qrxn
Heat of rxn is extensive property-dependent on amount of substance
For dilute aqueous solutions, specific heat of solution approximately that of water (4.18 J/g-K)
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36. 10.0 g NaOH 1 mol NaOH -43 kJ 1000 J = -10,750 J 40.0 g NaOH mol 1kJ
If 10.0 g of solid NaOH is added to 1.00 L of water (specific heat capacity = 4.18 J/oC) at 25.0oC in a constant pressure calorimeter, what will be the final temperature of the solution? (assume the density of the final solution is 1.05 g/mL). We will use info from previous problem where enthalpy change per mole of NaOH is -43 kJ/mol.
We need to know three things:
Mass of solution: L x 1050 g/L = 1050 g
Heat capacity of solution: J/oC
Enthalpy of reactant:
10.0 g NaOH 1 mol NaOH kJ J = -10,750 J
40.0 g NaOH mol kJ
We want to know the change in temperature: Energy released = s x m x ΔT
-10,750 J = (4.18 J/oC)(1050 g)(ΔT) = -2.45
= 27.45oC
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37. Constant Volume Calorimetry (Bomb Calorimetry)
Coffee cup calorimeter is not suitable for some reactions
Greater precision may be required, or a device that allows higher temperatures
Bomb calorimeters are used to measure heat evolved in combustion reactions of foods and fuels
Heat capacity of calorimeter must be known, generally in kJ/°C
Measurements of water/calorimeter related to heat evolved by reaction q and to H
Energy released by reaction = energy absorbed by solution
=specific heat capacity x mass of solution x increase in temperature
Energy released = s x m x ΔT
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38. Stainless steel "bomb“ loaded with small amount of combustible substance and O2 at 30 atm of pressure, and is immersed in known amount of water
Heat evolved during combustion absorbed by water/calorimeter assembly
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39. 3/25/2017

40. Calculate heat capacity of bomb using data for glucose:
The heat of combustion of glucose is 2800 kJ/mol. A sample of glucose weighing 5.00 g was burned with excess oxygen in a bomb calorimeter. The temperature of the bomb rose 2.4oC. What is the heat capacity of the calorimeter? A 4.40 g sample of propane was then burned with excess oxygen in the same bomb calorimeter. The temperature of the bomb increased 6.85oC. Calculate ΔEcombustion of propane.
Calculate heat capacity of bomb using data for glucose:
Convert to moles: g 1 mol = 2.78 x 10-2 moles glucose
180.0 g
Heat capacity = 2800 kJ/mol x 10-2 moles = kJ/oC
(kJ/ oC) oC
Use this heat capacity to determine heat of combustion of propane: (kJ/mol)
32.4 kJ oC = kJ/mol
oC mol
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41. Homework:
Read 6.2, pp
Q pp , #12, 32, 34, 36, 38, 42, 46
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42. Hess’s Law
In going from particular set of reactants to particular set of products, change in enthalpy (ΔH) is same whether reaction takes place in one step or in series of steps
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43. Using Hess’s Law Work backward from final reaction
Reverse reactions as needed, being sure to also reverse ΔH
Identical substances on both sides of summed equation cancel each other
Can multiply entire equation by factor (3, 2, ½, or 1/3)-this includes ΔH
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44. Example: Step One: N2(g) + 2O2(g)  2NO2(g) ΔH1 = 68 kJ Step Two:
2NO(g) + O2 (g)  2NO2 g) ΔH3 = -112 kJ
N2(g) + 2O2(g)  2NO2(g) ΔH2 + ΔH3 = 68 kJ
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45. 3/25/2017

46. Example:
S(s) + O2(g)  SO2(g)           H°rxn = –296.1 kJ H2(g) + ½ O2(g)  H2O(l)            H°rxn = –285.8 kJ H2S(g) + 3/2O2(g)  SO2(g) + H2O(l)     H°rxn = –561.7 kJ
These equations can be arranged so that their sum is the formation reaction of hydrogen sulfide.
The first one is unchanged: S(s) + O2(g)  SO2(g) The second one is unchanged: H2(g) + ½ O2(g)  H2O(l) The third one is reversed: SO2(g) + H2O(l)  H2S(g) + 3/2 O2(g) Their sum is overall equation: S(s) + H2(g) H2S(g)
We must reverse the sign of the enthalpy change of the third reaction before summing the enthalpy changes.
S(s) + O2(g)  SO2(g)            H°rxn = –296.1 kJ
H2(g) + O2(g)  H2O(l)           H°rxn = –285.8 kJ
SO2(g) + H2O(l)  H2S(g) + 3/2O2(g) H°rxn = kJ S(s) + H2(g)  H2S(g)           H°f = –20.2 kJ
Summation of the three reaction steps yields the net reaction, and summation of the H° values yields the enthalpy change of the net reaction.
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48. Homework: Read 6.3, pp. 256-260 Q pp. 280-283, #14, 52, 54, 56, 58
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49. Enthalpy of formation (heat of formation)
Tables for
Enthalpies of vaporization (ΔH for converting LG)
Enthalpies of fusion (ΔH for melting solids)
Enthalpies of combustion (ΔH for combusting substance in oxygen)
Enthalpy (heat) of formation (ΔHf) associated with process of forming products from reactants
Conditions influencing enthalpy changes
Temperature
Pressure
State of reactants/products (s/l/g/aq)
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50. Standard enthalpy of formation
Change in enthalpy that accompanies formation of one mole of compound from its elements with all substances in their standard states
Form most stable at 298K/1 atm
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51. Standard Enthalpy of Formation (ΔHf°)
“o” symbol indicates standard conditions.
Standard state for a substance is precisely defined reference state
Elements/diatomic gases = 0
For a compound
Gaseous substance at pressure of exactly 1 atmosphere
Condensed state (pure liquid/solid)
Substance present in solution of exactly 1M concentration
For an element
Form in which the element exists under conditions of 1 atmosphere and 25oC [O2(g), Na(s), Hg(l), etc.].
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53. 3/25/2017

54. SO2(g) + ½ O2(g)  SO3(g) H° = –99 kJ
The enthalpy change in kilojoules applies to the reaction of the amount of substance shown in the balanced equation.
SO2(g) + ½ O2(g)  SO3(g)           H° = –99 kJ
When 2 mol SO2 react with 1 mol O2, 198 kJ of heat will be released
If the above equation was divided by 2 and written then the enthalpy change (–198 kJ) must be divide by 2 also
Remember that the heat evolved or absorbed by a chemical reaction is an extensive property: that is, it depends on the amount of reactants
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55. The following reaction absorbs heat
If we write the reverse of a chemical reaction, the magnitude of H is the same, but its sign changes.
The following reaction absorbs heat
CaCO3(s)  CaO(s) + CO2(g)           H° = 178 kJ
But the reverse reaction must release the same amount of heat
CaO(s) + CO2(g)  CaCO3(s)           H° = –178 kJ
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56. In the combustion of methane, water is a product
We must always specify the physical states of all reactants and products, because they help determine the magnitude of the enthalpy change.
In the combustion of methane, water is a product
The enthalpy change is different when H2O is formed as a gas, than when H2O is a liquid
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)      H° = –802 kJ
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)       H° = –890 kJ
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57. 3C(graphite) + 3O2(g) -> 3CO2(g) DH�f = �1180.5 kJ
3C(graphite) + 4H2(g) -> C3H8(g) DH�f = kJ (reverse equation/sign)
3C(graphite) + 3O2(g) -> 3CO2(g) DH�f = � kJ
C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)
ΔH = ∑Hproducts – ∑Hreactants
[3(-393.5) + 4(-285.8)] – [( ) + 0] = kJ
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58. Calculate ΔH° for the following reaction:
2C3H6(g) + 9O2(g)  6CO2 (g) + 6H2O
ΔHreactionO = ΣΔHfO (Products) - ΣΔHfO (Reactants)
Appendix 4 in the textbook lists them
[6(-286 kJ/mol) + 6( kJ/mol)] – [2(20.9 kJ/mol) + 9(0 kJ/mol)] –remember, elements are 0
ΔH° = kJ
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59. Homework:
Read 6.4 –6.6, pp
Q pg. 284, #60, 62, 64, 66, and try 72 (extra credit)
Submit quizzes by to me:
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