Thermochemistry

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Thermochemistry

• Chapter 5 BLB 11th

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Expectations

• Heat enthalpy same or different?
• Heat calculations
• Temp. change
• Phase change (11.4)
• Reactions
• Enthalpy calculations
• Read the chapter, study, and apply!

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5.1 The Nature of Energy

• Chemistry ? ? ? Energy
• Potential stored energy chemical
• Kinetic released energy energy of motion
  thermal
• Electrostatic potential interaction between
  charged particles

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Examples of Kinetic Energy
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Energy, cont.

• Units of energy
• Joule (J) SI unit of energy
• calorie (cal)
• amount of energy required to raise the
  temperature of exactly 1 gram of pure water by
  1C (from 14.5C to 15.5C)
• 1 cal 4.184 J (exactly)
• Calorie (dietary calorie),Cal
• 1 Cal 1000 cal 1 kcal

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Energy, cont.

• System and Surroundings
• System component(s) of interest open, closed,
  or isolated
• Surroundings everything outside of the system

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Energy, cont.

• Transferring Energy
• Work (w) energy used to move an object against
  a force w F x d
• Heat (q) energy transferred from a hotter
  object to a cooler one
• Energy capacity to do work or to produce heat
  ?E q w

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Combustion
Heat? Work?
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5.2 The First Law of Thermodynamics

• Energy can be neither created nor destroyed.
• Energy is conserved.
• Internal energy, E sum of all the kinetic and
  potential energy

What kinds of energy are in here? What
changes could occur?
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5.2 The First Law of Thermodynamics

• More interested in the change in energy
• ?E Efinal Einitial
• Need to give number, units, and sign for all
  thermodynamic quantities.
• ?E gt 0 system has gained energy endergonic
• ?E lt 0 system has lost energy exergonic
• Note Opposite change occurs with respect to
  the surroundings.

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Energy, heat work

• ?E q w
• Sign of ?E depends upon sign and magnitude of q
  and w.

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Exercise 5.3
Endothermic of exothermic? Sign of work and
heat? Net gain in system energy?
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Sample Exercise 5.2

• System loses 1150 J of heat to the surroundings.
• The piston move downwards doing 480 J of work on
  the system.
• ?E ?

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Calculate ?E (in J) exothermic or endothermic?

• Balloon heating by adding 900 J of heat and
  expands doing 422 J of work on atmosphere.
• 50 g of H2O cooled from 30C to 15C losing 3140
  J of heat.
• Reaction releases 8.65 kJ of heat, no work done.

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Heat or Thermal Energy (q)

• Exothermic system ? surroundings
• Heat energy released to surroundings
• q lt 0
• e.g. combustion reaction, crystallization
• Surroundings get warmer
• Endothermic system ? surroundings
• Heat energy flows into the system
• q gt 0
• e.g. melting, boiling, dissolution of NH4NO3
• Surroundings get colder

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Heat, cont.

• Evidenced by a change in temperature
• Spontaneously transferred from the hotter to the
  cooler object
• Atoms or molecules with more energy move faster
• Temperature-dependent
• Extensive property (depends on amount)
• Total energy of system is the sum of the
  individual energies of all the atoms and
  molecules of the system.

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Work (w)

• Force acting over a distance
• w F x d -P?V
• Compression work ? surroundings
• Work is done on the system.
• ?V lt 0
• w gt 0
• Expansion work ? surroundings
• Work is done on the surroundings.
• ?V gt 0
• w lt 0

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Heat Work, cont.

• Work and heat are pathways by which energy can be
  transferred.
• State function depends only on the systems
  present state independent of the pathway
  internal energy, P, V, ?E, ?H, ?S are state
  functions
• Energy is a state function, as is enthalpy.

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5.3 Enthalpy

• Enthalpy heat flow at constant pressure from
  Gr. enthalpien to warm
• Enthalpy change (?H) energy transferred as heat
  at constant pressure ?H Hproducts Hreactants
• H E PV
• For a change _at_ constant pressure
• ?H ?E P?V
• ?H ?E - w qP
• ?H lt 0 exothermic reactants ? products heat
• ?H gt 0 endothermic reactants heat ? products

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q gt 0
q lt 0
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5.4 Enthalpies of Reaction, ?Hrxn

• ?H is an extensive property value depends upon
  the BALANCED equation.
• H2O(g) ? H2(g) ½ O2(g)
• ?H 242 kJ per mole of H2O
• 2 H2O(g) ? 2 H2(g) O2(g)
• ?H 484 kJ per 2 moles of H2O

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5.4 Enthalpies of Reaction

• For reverse reactions
• ?H values are equal in magnitude, but opposite
  in sign.
• For water ?Hvap 44.0 kJ/mol
• ?Hcond -44.0 kJ/mol

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For the combustion of methane
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5.4 Enthalpies of Reaction

• ?H is dependent upon physical state.
• ?Hf values
• C6H6(g) 82.9 kJ/mol
• C6H6(l) 49.0 kJ/mol
• H2O(l) ? H2O(g) ?H 44 kJ

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CH3OH(g) ? CO(g) 2 H2(g) ?H 90.7 kJ

• Exothermic or endothermic?
• Heat transferred for 1.60 kg CH3OH?
• If 64.7 kJ of heat were used, how many grams of
  H2 would be produced?

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CH3OH(g) ? CO(g) 2 H2(g) ?H 90.7 kJ

• ?H of reverse reaction?
• Heat (in kJ) released when 32.0 g of CO(g)
  reacts completely?

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5.5 Calorimetry

• Calorimetry science of measuring heat flow
• Calorimeter a device used to measure heat flow

Coffee-cup calorimeter ?
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Heat Capacity and Specific Heat

• Heat capacity (C) - amount of heat required for a
  1C temperature change
• J/C J/K
• extensive property

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Heat Capacity and Specific Heat

• Specific heat capacity (Cs) heat capacity for 1
  g
• J/gC or J/gK
• Molar heat capacity heat capacity for 1 mole
  J/molC or J/molK

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Heat Capacity and Specific Heat

• Specific heat values (more on p. 181)
• Fe 0.45 J/gK
• glass 0.84 J/gK
• water 4.18 J/gK (highest of all liquids and
  solids except ammonia)

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Calculating heat (q)

• To calculate the quantity of heat transferred
• q s x m x ?T

q heat (J) s specific heat (J/gK) m mass
(g) ?T change in temp. (K or C)
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Calculate the heat (in J) required to raise the
temperature of 62.0 g toluene from 16.3C to
38.8C. The specific heat of toluene is 1.13
J/gK.
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Calculate the specific heat of lead if 78.2 J of
heat were required to raise the temperature of a
45.6-g block of lead by 13.3C.
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Coffee-cup Calorimetry

• Constant-pressure, ?H qP and ?E qP w
• Assume no heat is lost to surroundings.
• Usually exothermic (qrxn lt 0)
• Applications
• Heat transfer between objects
• Reactions in aqueous solutions
• Use specific heat of water (4.18 J/gK).
• Use mass (or moles) of solution.

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A 15.0-g piece of nickel at 100.0C is dropped
into a coffee-cup calorimeter containing 55.0 g
H2O at 23.0C. What is the final temperature of
the water and nickel after reaching thermal
equilibrium? The specific heat capacity of nickel
is 0.444 J/gK and of water is 4.18 J/gK.
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Coffee-cup Solution Calorimetry

• heat lost by reaction heat gained by solution
• -qrxn qsoln
• qrxn -(Cs,soln x msoln x ?T)
• Enthalpy of reaction (?Hrxn) per mole
• ?Hrxn qrxn/mol of specified reactant

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In a coffee-cup calorimeter, 2.50 g of MgO was
combined with 125 mL of 1.0 M HCl. The
temperature increased by 9.6C. Calculate the
enthalpy of reaction per mole of MgO for the
following reaction.Mg2(aq) H2O(l) ? MgO(s)
2 H(aq)
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Bomb Calorimetry (p. 183)

• Constant-volume
• No work is done (?V 0), so ?E qV
• Used for combustion reactions
• The bomb components absorb the heat lost by the
  reaction.
• Heat capacity of the bomb (Ccal) needed to
  calculate the heat of combustion (reaction)
• qrxn -(Ccal x ?T)

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Bomb Calorimeter
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A 1.320-g sample of a new organic substance is
combusted in a bomb calorimeter (Ccal 8.74
kJ/K). The temperature of the bomb increased
from 22.14C to 26.82C. What is the heat of
combustion per gram of the substance?
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Phase Changes (Fig. 11.17, p. 449)
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Phase Changes (p. 450)
Endothermic ?
Cs 1.84 J/gK
?Hvap 40.67 kJ/mol
? Exothermic
Cs 4.18 J/gK
Cs 2.09 J/gK
?Hfus 6.01 kJ/mol
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Heat transfer phase changes

• To calculate the quantity of heat transferred
  during a change of state
• q ?Hprocess x m
• or
• q ?Hprocess x mol
• No change in temperature, so no ?T.
• For a complete process, add together the heat
  transferred for each segment.
• See Sample Exercise 11.4, p. 451.

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11.40 Calculate the total heat transferred to
raise the temperature of 50.0 g of the
fluorocarbon, C2Cl3F3, from a liquid at 10.00C
to a gas at 85.00C.Data b.p. 47.6C, ?Hvap
27.49 kJ/mol, s(liquid) 0.91 J/gK, s(gas)
0.67 J/gK
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5.6 Hesss Law

• If a reaction is carried out in a series of
  steps, ?Hrxn will equal the sum of the ?H values
  of the individual steps.
• Hesss Law works because ?H is a state function,
  i.e. it only depends upon the initial reactant
  and the final product states.

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Hesss Law Example
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Hesss Law Example
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Calculation of ?H
C3H8 (g) 5 O2 (g) ?? 3 CO2 (g) 4 H2O (l)

• Imagine this as occurring
• in three steps

C3H8 (g) ?? 3 C (graphite) 4 H2 (g)
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Calculation of ?H
C3H8 (g) 5 O2 (g) ?? 3 CO2 (g) 4 H2O (l)

• Imagine this as occurring
• in three steps

C3H8 (g) ?? 3 C (graphite) 4 H2 (g) 3 C
(graphite) 3 O2 (g) ?? 3 CO2 (g)
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Calculation of ?H
C3H8 (g) 5 O2 (g) ?? 3 CO2 (g) 4 H2O (l)

• Imagine this as occurring
• in three steps

C3H8 (g) ?? 3 C (graphite) 4 H2 (g) 3 C
(graphite) 3 O2 (g) ?? 3 CO2 (g) 4 H2 (g) 2
O2 (g) ?? 4 H2O (l)
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Calculation of ?H
C3H8 (g) 5 O2 (g) ?? 3 CO2 (g) 4 H2O (l)

• The sum of these equations is

C3H8 (g) ?? 3 C (graphite) 4 H2 (g) 3 C
(graphite) 3 O2 (g) ?? 3 CO2 (g) 4 H2 (g) 2
O2 (g) ?? 4 H2O (l)
C3H8 (g) 5 O2 (g) ?? 3 CO2 (g) 4 H2O (l)
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Based on the following reactions
?H, kJN2(g) O2(g) ? 2 NO(g)
180.72 NO(g) O2(g) ? 2 NO2(g)
-113.12 N2O(g) ? 2 N2(g) O2(g)
-163.2Calculate the ?Hrxn for
the following reactionN2O(g) NO2(g) ? 3 NO(g)
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Based on the following reactions
?H, kJC2H2(g) 5/2 O2(g) ? 2 CO2(g) H2O(l)
-1300.C(s) O2(g) ? CO2(g)
-394 H2(g) ½ O2(g) ? H2O(l)
-286Calculate the ?Hrxn for
the following reaction2 C(s) H2(g) ? C2H2(g)
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5.7 Standard Enthalpies of Formation

• Standard state of a substance pure form at
  atmospheric pressure (1 atm) and temperature of
  interest (usually 25C).
• Standard enthalpy - ?H
• Standard enthalpy of formation, ?Hf
• - the change in enthalpy for the formation of 1
  mole of a substance from its elements in their
  standard states, kJ/mol of product.
• ?Hf of an element in its most stable form 0
  kJ/mol

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Standard Enthalpies of Formation(see Appendix C,
p. 1112)
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Calculating Enthalpies of Reaction

• ?Hrxn Sn ?Hf(products) - S n ?Hf(reactants)

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5.71 (c) Calculate ?Hrxn for the following
reactionN2O4(g) 4 H2(g) ? N2(g) 4 H2O(g)
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Calculate ?Hrxn for the following reaction2
KOH(s) CO2(g) ? K2CO3(s) H2O(g)
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5.8 Foods and Fuels

• Glucose is our bodys fuel source.
• Carbs and fats are metabolized into glucose.
• Excess fat is stored.
• Whats the big deal? Take a look at the fuel
  value.

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Nonbiological Fuel
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