CHAPTER SIX: THERMOCHEMISTRY
1. Thermochemistry is a branch of chemistry concerned with energy changes accompanying chemical and physical processes.
2. Examples:
a. heat liberated in a chemical reaction.
b. work done in motion of a piston.
c. heat required to change physical state.
d. conditions required to attain equilibrium. (not in this chapter)
e. forces driving processes that occur spontaneously. (not in this chapter)
A. Energy and its Units. (Section 6.1)
Energy = capacity to do work or transfer heat.
1. Kinetic Energy - energy of motion (mechanical)
mv 2 E = k 2
m=mass v = speed €
2. Potential Energy - energy possessed by virtue of position or composition. (gravitational, electrical, chemical, nuclear, etc.)
Ep = mgh
g = acceleration of gravity = 9.81 m/s2 h= height
Chapter 6 Page 1 3. The SI Unit of Energy is the Joule (J).
1 Joule = 1 kg m2/s2
4. Problem: suppose a 50 kg object is traveling at 30 m/s. What is its kinetic energy?
2 50kg 30 m 2 mv ∗ ( s ) m 2 E = = = 22500kg 2 = 22500J ≈ 22kJ k 2 2 s
5. Problem: what is the potential energy of a 50 kg object suspended 30 m above ground level? € Ep = mgh = 50 kg * 9.81 m/s2 * 30 m = 14711 J ≈ 15 kJ
6. More common everyday unit of energy is the calorie (cal)
1 cal = energy required to raise the temperature of 1 gram of liquid water by 1 degree Celsius. 1 cal = 4.184 J
7. Nutritionists “Calorie” is really 1000 cal:
1 “Calorie” = 1000 cal = 1 kcal
8. Problem: About 1 joule of energy is required for the heart to beat once. Heart rate averages 1 beat/second. How much energy is required for entire day of heart function in calories?
1 Joule/beat x 1 beat/sec x 60 sec/min x 60 min/hr x 24 hr/1 day x 1 cal/4.184 = 2.065 x 104 cal.
How many nutritional Calories?
2.065 x 104 cal x 1 “Cal”/1000 cal = 20.65 “Cal”
9. Internal energy: consider the 50 kg object again. We calculated its kinetic energy by considering its motion as a whole. However, even when the object is at rest, since it consists of atoms and molecules that are in constant rapid ‘thermal’ motion, it still has energy. This is its Internal Energy U.
Etot = Ek + Ep + U
When we consider a lab sample, we consider its thermodynamic internal energy U.
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10. Law of Conservation of Energy:
Energy is neither created or destroyed but may be converted between various forms.
This is the essence of First Law of Thermodynamics.
B. Heat of Reaction. (Section 6.2)
1. We are interested in the heat liberated or absorbed in a chemical reaction or physical change.
2. Necessary terminology needed:
System = that sample of matter we are concerned with.
Surroundings = environment around the system.
Universe = system and surroundings.
Thermodynamic state = set of conditions that specify the properties of a system.
State variables = P, V, T, U, H, etc. (all capital lettered functions)
Change in a state variable X = ΔX.
example ΔU = Ufinal - Uinitial
Exothermic processes are those releasing energy to the surroundings in the form of heat.
Endothermic processes require absorption of energy from the surroundings to occur.
3. Heat = a form of energy called thermal energy, a manifestation of the random, chaotic molecular motion in a sample (extensive property).
4. Temperature = a measure of the intensity of heat, the “hotness” or “coldness” of a body, which relates to heat energy per amount of material (an intensive property).
5. Heat always flows spontaneously from a hotter body to a colder body, high T to low T.
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6. Can’t measure heat energy directly, but we can measure temperature change.
q = Heat flow across the system boundary
is (+) if heat flowed in is (-) if heat flowed out
Heat of reaction = q necessary to restore a system’s temperature after a reaction has taken place in the system.
7. Example: 20 mL of aqueous 1 M NaOH is poured into 20 mL of 1 M aqueous HCl. Both samples were initially at 25°C. The temperature rises several degrees, and as a result, heat flows out of the system into the surroundings until the solutions are eventually restored to 25 °C.
Heat of reaction = q
Is q positive or negative?
C. Statement of First Law.
1. U ≡ total internal energy possessed by a system.
2. q = heat absorbed by system during a process.
3. w = work done on system during a process.
4. ΔU = q + w
5. Scenario:
w = work involved expanding against its surroundings = -PΔV
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6. If system is heated but not allowed to respond in any way, then:
ΔU = q
7. So all heat absorbed in a process in which system is not allowed to expand against its surroundings (constant volume) goes into increasing its internal energy.
All energy input remains in the system.
D. Enthalpy Changes. (Section 6.3)
1. Most processes we care about occur at constant Pressure, not constant Volume.
2. Example: Heating an open beaker of water.
3. Heat absorbed by system heated at constant pressure.
↑↑ expands slightly against surroundings
Tiny amount of energy input q does not end up increasing U
q > ΔU
4. Enthalpy H = “heat content” of a system.
enthalpy change = ΔH = q occurring under constant P
5. ΔH slightly > ΔU.
6. Later, when the system is asked to give back its energy as heat, it can give back more than ΔU; it can give back ΔH.
q
Chapter 6 Page 5 7. Enthalpy and Energy are nearly synonymous, but ΔH is more directly measurable for heat transfers under constant P conditions.
H = U + PV
ΔH = ΔU + Δ (PV)
But if reaction happening at constant P last term becomes:
ΔH = ΔU + PΔV
PΔV = the work done = the energy expended by the system to expand against its surroundings in order to keep the pressure constant.
This term is typically small compared to ΔH or ΔU.
E. Thermochemical Equations. (Section 6.4)
1. The ΔH of reaction always refers to enthalpy change of a reaction as it is written, interpreting coefficients as moles:
2H2(g) + O2(g) → 2H2O(l) ΔH = -571.6 kJ
Says: “571.6 kJ are liberated when 2 moles H2 gas react with 1 mole O2 gas to form 2 moles liquid H2O.”
2. What is ΔH of this reaction as written?
1 H2(g) + 2 O2(g) → H2O(l) ΔH = -285.8 kJ (half of -571.6)
3. What is ΔH of this reaction?
1 H2O(l) → H2(g) + 2 O2(g) ΔH = +285.8 kJ
4. Physical states (s), (l), (g), (aq) of reactants and products are important and must be specified.
Problem: How much energy (in kJ) is required to produce 10.0 grams of molecular oxygen in the electrolytic decomposition of water above?
1 mol O2 285.8kJ 10.0g O2 ∗ ∗ =179kJ 32.0 g O2 0.50 mol O2
Chapter 6 Page 6
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F. Heat Transfer and Measurement of Heat. (Section 6.6)
1. Heat Capacity of a sample = amount of heat required to raise the temperature of the whole sample by 1 ° Celsius (or Kelvin).
Units = energy/°C
q = CΔT where C = heat capacity
heat capacity thus relates heat to temperature.
An intensive or extensive property of a substance? Which? extensive
OK, let’s factor out the extensive part, the size-dependent part:
C = n Cm n = moles
Cm = molar heat capacity (intensive)
So now q = nCmΔT
Chapter 6 Page 7 Another way to factor out system size:
C = m Cs
m= mass in grams
Cs = specific heat capacity (heat capacity per gram of material)
Such that q = mCsΔT
2. The Specific Heat of a substance = the amount of heat required to raise the temperature of 1 gram of that substance by 1 degree Celsius. Units = energy/gram-°C.
An intensive or extensive property of a substance? Which? intensive
specific heat = Cs = (amount of heat required) (mass in g)(temp rise in °C)
Thus for liquid water, the specific heat = 1 cal/g-°C or = 4.184 J/g-°C
G. Measurement of Heat of Reaction – Calorimetry.
1. Calorimetry measures heat liberated or absorbed in reactions, called the “heat of reaction,” ΔH.
2. Usually performed at constant P, so ΔH measured, not ΔU.
3. Directly measures ΔT of reaction system.
4. Must relate this ΔT to energy absorbed or released by chemical reaction.
5. In general: q = CΔT
where we define
q = rxn heat flowing from reacting system into calorimeter and its contents
C = heat capacity of calorimeter and its contents ΔT = T change of calorimeter and its contents due to rxn
Chapter 6 Page 8 6. C = heat capacity of calorimeter + contents, is found by adding known amount of heat
energy (qcalibrate) to setup and observing ΔTcalibrate.
qcalibrate = C * ΔTcalibrate
Thus we can obtain C of the calorimeter.
7. Finally, reaction is performed with exactly the same setup, and ΔTrxn is measured:
qrxn = C * ΔTrxn Solve for qrxn:
Answer: ΔHrxn = -qrx = heat of rxn for given amount of reactant materials
Report: ΔHrxn negative if rxn is exothermic ΔHrxn positive if rxn is endothermic
8. Problem: When 3.40 kJ of heat is added to a calorimeter containing 50.0 g water, T rises from 24.0 to 36.5°C. In a second experiment in the same calorimeter, when 25.0
mL of 1.0 M H2SO4 is added to 25.0 mL of 2.0 M NaOH (still ~50.0 g water), T rises from 24.0 to 31.2°C. What is ΔHrxn of the reaction:
H2SO4 + 2 NaOH ---> 2 H2O + Na2SO4(aq) ΔH = ? kJ
First get calorimeter heat capacity:
qcalibrate = C ΔTcalibrate 3.40 kJ = C * 12.5° C = 3.40 kJ/12.5° = 0.272 kJ/°C
Now look at the reaction itself:
qrxn = C * ΔTrxn = 0.272 kJ/°C * 7.2°C
qrxn = +1.96 kJ
So ΔHrxn = -1.96 kJ for the amount of materials reacted here.
But what about when 1 mol H2SO4 reacts with 2 moles NaOH, as the problem asks?
The reaction that actually ran involved only 1.0 M H2SO4 * 0.025 L = 0.025 moles acid.
Final answer = -1.96 kJ / 0.025 mol = -78 kJ/mol
Chapter 6 Page 9 H. Hess’ Law of Heat Summation. (Section 15-8)
1. ΔH of a reaction is the same whether it occurs in one step or by any series of steps.
2. Elevation illustration.
3. Example: Use Hess’ Law to find ΔHrxn° of:
1 (Rxn A) CO(g) + 2 O2(g) → CO2(g) Given:
(Rxn B) C(s) + O2(g) → CO2(g) ΔH = -393.5 kJ
1 (Rxn C) C(s) + 2 O2(g) → CO(g) ΔH = -110.5 kJ
Is there some way to combine (Rxn B) and (Rxn C) to produce (Rxn A)?
(Rxn B) C(s) + O2(g) → CO2(g) -393.5 kJ 1 -(RxnC) -[C(s) + 2 O2(g) → CO(g)] -(-110.5 kJ)
1 (Rxn A) CO(g) + 2 O2(g) → CO2(g) -393.5-(-110.5)
= -283 kJ
4. Pictorially:
Chapter 6 Page 10 I. Standard States and Standard ΔH’s. (Section 6.8)
1. Useful to tabulate ΔH of known rxns because they can be algebraically combined (by Hess’ Law) to calculate ΔH of unmeasured rxns.
2. Define a thermochemical standard state:
a. pressure = 1 atm
b. all pure substances (elements and compounds) in their pure unmixed states.
c. solutes are at 1 molar concentration.
d. T is not part of S.S. def., must be specified separately.
3. Symbolism ΔH° (superscript denotes standard state conditions)
4. Most important types of tabulated quantities:
a. ΔHf° = “heats of formation”
b. ΔHc° = “heats of combustion”
J. Standard Molar Heat of Formation, ΔHf°.
1. Refers to enthalpy change in forming 1 mole of a substance from its elements in their reference states.
2. Reference state of an element:
• pure • most stable allotropic form (solid, liquid, gas) • for Carbon, Carbon-graphite, not diamond
• for Bromine, Br2(l) • for Hydrogen, Oxygen, Nitrogen, Chlorine, and Fluorine it is
diatomic gases: H2, O2, etc. • for Sulfur, S8(s) • for Phosphorus, P4(s) • metals are all solids - M(s)
3. ΔHf° of an element in its reference state is defined to be zero.
Chapter 6 Page 11 4. See Table 6.2. ΔHf° of CH4(g) = -74.81 kJ refers to this reaction as written:
C(s) + 2H2(g) → CH4(g) liberates 74.81 kJ
5. Easily remembered formula: (a consequence of Hess’ Law)
° ° ° ΔHrxn = ∑nΔH f − ∑nΔH f products reac tan ts
6. Example use of formula:
CH (g) + 2O (g) CO (g) + 2H O(g) H ° = ? € 4 2 → 2 2 Δ rxn
ΔHrxn° = [ΔHf°(CO2(g)) + 2 ΔHf°(H2O(g))] -
[ΔHf°(CH4(g)) + 2 ΔHf°(O2(g))]
= [-393.5 + 2(-241.8)] - [-74.8 + 2(0)]
= -802.3 kJ
7. Can be used to find ΔH° of any rxn for which ΔHf° data is available for all species present.
K. Bond Energies. (Optional Section)
1. Energy liberated or absorbed in chemical reactions is due to energy of breaking and forming bonds.
2. Bond Energy (B.E.) = energy needed to break one mole of particular type of bond.
3. The stronger the bond, the larger the B.E.
4. Simple case:
Reaction H2(g) → 2H(g); ΔHrxn° = +436 kJ Therefore B.E. of H-H = +436 kJ
5. More complicated case:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
ΔHrxn° = -800 kJ (experimental)
ΔHrxn° = 4 BE(C-H) + 2 BE(O=O) – 2 BE – 4BE(O-H)
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6. In general:
ΔHrxn° = ∑ B.E. (reactants) - ∑ B.E. (products)
7. Difficulty: C-H bond in one hydrocarbon slightly different than C-H bond in another. True of all other bond types as well.
8. So B.E. is an average and is useful for estimating ΔHrxn°.
L. Applications of Thermochemistry:
1. Nutrition: Given that the human body typically requires 2400 Calories (nutritionist) per day to operate, how many grams of glucose must be metabolized daily to meet this demand, given:
C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(l) ΔH=-2803 kJ
4.184kJ 1mol glucose 180g glucose 2400kcal ∗ ∗ ∗ = 645g glucose 1kcal 2803kJ 1mol glucose
2. By the same token, how many fat grams are burned off in a day of fasting, if 2400 € Calories are expended by exercise, given that fat can be represented adequately as glyceryl trimyristate:
C45H86O6(s) + 127/2 O2(g) 45 CO2 + 43 H2O(l) ΔH=-27820kJ
4.184kJ 1mol "fat" 723.1g "fat" 2400kcal ∗ ∗ ∗ = 260.9g "fat" 1kcal 27820kJ 1mol "fat"
3. Given that 1 watt = is 1 J/s, what is the wattage of consumption of energy by the € human body given 2400 Calories are expended per day? (Most of this is used by your brain.)
2400Cal 1kcal 4.184kJ 103 J 1day 1hr 1min ∗ ∗ ∗ ∗ ∗ ∗ =116 J day 1Cal 1kcal kJ 24hr 60 min 60 sec s
Chapter 6 Page 13 € 4. If your automobile ran on pure ethanol, how many liters per day would you need to operate it, given that you travel 1000 miles/month in a car getting 20 miles per gallon. Ethanol density = 0.80 g/mL.
1000mi 1month 1gal 4qt 1L ∗ ∗ ∗ ∗ = 6.29 L month 30days 20mi 1gal 1.06qt day
5. How much heat is released into the surroundings by combustion of that much ethanol?
€ 3 10 mL 0.80g 3 6.29L ∗ ∗ = 5.0 ×10 g ethanol L mL
ΔH C2H5OH + 3 O2 → 2 CO2 + 3 H2O(l)
€ ΔH = 2 ΔHf (CO2) + 3 ΔHf (H2O(l)) – ΔHf (C2H5OH)
€ = 2 * (-393.5 kJ) + 3 * (-285.8 kJ) – (-277.7 kJ) = -1367 kJ
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