Thermochemistry

1
Chapter 5

• Thermochemistry

2
Energy

• The capacity to do work (or produce heat)
• We cannot see or directly measure energy
• Energy is a substance-like quantity that is
  always conserved
• Energy is responsible for any change
• Without energy, there would be no chemistry

3
Types of Energy

• potential energy- (PE) due to position or
  composition
• ex. attractive or repulsive forces ( and ?)
• kinetic energy- (KE) due to motion of the object
  (thermal/translational)
• KE ½mv2 depends on mass and speed

4
SI Units of Energy

• calorie - amount of energy to raise 1 gram of
  water by 1 degree C
• Joule 1 cal 4.184 Joules
• In nutrition, 1 Calorie 1000 cal 1kcal

5
Transfer of Energy Work Heat

• Heat- (q) transfer of energy between two objects
  because of a temperature difference
• Heat always flows from Hot to Cold
• Work- (w) force acting over a distance (w F x
  d)
• Temperature- measure of the average kinetic
  energy of the particles

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Chemical Energy

• Exothermic Reactions
• Gives off energy as it progresses
• PE stored in chemical bonds is converted to
  thermal energy (random KE) through heat.
• Products are generally more stable (stronger
  bonds) than reactants.

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• Endothermic Reactions
• Energy is absorbed from the surroundings
• Energy flows into the system to increase the
  potential energy of the system
• Products are generally less stable (weaker bonds)
  than the reactants.

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Transfer of Energy
Combustion of Methane Gas is exothermic CH4
2O2 ---gt 2H2O CO2 energy
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Transfer of Energy
Reaction between nitrogen and oxygen is
endothermic N2 O2 energy ---gt
2 NO
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State Functions

• Dependent only on the current condition.
• Independent of the past or the future.
• Independent of the pathway taken to get to that
  state
• Ex A liter of water behind a dam has the same
  potential energy for work regardless of whether
  it flowed downhill to the dam, or was taken
  uphill to the dam in a bucket. The PE is a state
  function dependent only on the current position
  of the water, not on how the water got there.

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Thermodynamics

• Thermodynamics the study of energy and its
  transformations.
• First Law of Thermodynamics
• Energy can be neither created nor destroyed
  energy is conserved.
• Energy that is lost by the system must be gained
  by the surroundings, and vice versa.
• ?E q w

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Internal Energy

• Eint PE KE
• ?Eint q w
• can be changed by work, heat, or both

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Signs

• signs will always reflect the systems point of
  view unless otherwise stated
• q heat
• () for endothermic reactions
• (-) for exothermic reactions
• w work
• () work done on the system
• (-) system does work

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• DE change in energy
• q and w, then DE
• -q and w, then DE
• If the signs of q and w are opposite of one
  another, then the sign for DE will depend on the
  magnitude of q and w.

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Signs
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Work done by gases

• w -P DV
• Through Expansion
• DV () and w is (-)
• Through Compression
• DV (-) and w is ()

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Example 1

• Find the ?E for endothermic process where 15.6 kJ
  of heat flows and 1.4 kJ of work is done on
  system
• Since it is endothermic, what does that mean
  about the sign of q and w?
• q is () and w is ()

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Example 2

• Calculate the work of expansion of a gas from 46
  L to 64 L at a constant pressure of 15 atm.
• Since it is an expansion, ?V is and w is -

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Example 3

• A balloon was inflated from 4.00 x 106 L to 4.50
  x 106 L by the addition of 1.3 x 108 J of heat.
  Assuming the pressure is 1.0 atm, find the ?E in
  Joules.
• (1 Latm101.3 J)
• Since it is an expansion, ?V is and w is -

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Enthalpy (H)

• H Eint PV
• Since E, P, and V are all state functions so is
  H
• If system at constant P, then qp DH
• DH is the amount of energy exchanged between a
  system and its surroundings at constant P.

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• heat of reaction and change in enthalpy are used
  interchangeably for a reaction at constant P
• ?H Hproducts - Hreactants
• endo ?H exo - ?H

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Calorimetry

• science of measuring
• heat
• calorimeter- device used to experimentally find
  the heat associated with a chemical reaction
• substances respond differently when heated

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Heat Capacity

• (C) how much heat it takes to raise a substances
  T by one C or K
• the amount of energy depends on the amount of
  substance

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Heat Capacity

• specific heat capacity (c) heat capacity per
  gram
• in J/Cg or J/Kg
• molar heat capacity
• heat capacity per mole
• in J/Cmol or J/Kmol

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Constant-Pressure Calorimetry

• uses simplest calorimeter (like coffee-cup
  calorimeter) since it is open to air
• used to find changes in enthalpy (heats of
  reaction) for reactions occurring in a solution
  since qP ?H
• heat of reaction is an extensive property
  (dependent on amount of substance), so we usually
  write them per mole so they are easier to use

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Constant-Pressure Calorimetry

• when 2 reactants are mixed and T increases, the
  chemical reaction must be releasing heat so is
  exothermic
• the released energy from the reaction increases
  the motion of molecules, which in turn increases
  the T

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Constant-Pressure Calorimetry

• If we assume that the calorimeter did not leak
  energy or absorb any itself (that all the energy
  was used to increase the T), we can find the
  energy released by the reaction
• E released by rxn E absorbed by soln
• ?H qP m c ?T

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Constant-Volume Calorimetry

• uses a bomb calorimeter
• weighed reactants are placed inside the rigid,
  steel container and ignited
• water surrounds the reactant container so the T
  of it and other parts are measured before and
  after reaction

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Constant-Volume Calorimetry

• E released by rxn ?T x Ccalorimeter

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Example 1

• When 1 mol of CH4 is burned at constant P, 890 kJ
  of heat is released. Find ?H for burning of 5.8 g
  of CH4 at constant P.
• 890 kJ is released per mole of CH4
• CH4 2O2 ---gt CO2 2H2O ?H ?890 kJ

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Example 2

• When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0C
  is mixed with 1.00 L of 1.00 M Na2SO4 solution at
  25.0C in a coffee-cup calorimeter, solid BaSO4
  forms and the T increases to 28.1C. The specific
  heat of the solution is 4.18 J/gC and the
  density is 1.0 g/mL. Find the enthalpy change per
  mole of BaSO4 formed.

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Example 2

• Write the net ionic equation for the reaction
• Ba2 (aq) SO42- (aq) ? BaSO4(s)
• Is the energy released or absorbed? What does
  that mean about ?H and q?
• exothermic ??H and qP
• How can we calculate ?H or heat?
• heat q m c ?T
• How can we find the m?
• use density and volume

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Example 2

• Find the mass
• Find the change in T
• Calculate the heat created

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Example 2

• since it is a one-to-one ratio and the moles of
  reactants are the same, there is no limiting
  reactant
• 1.0 mol of solid BaSO4 is made so
• ?H -2.6x104 J/mol -26 kJ/mol

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Example 3

• Compare the energy released in the combustion of
  H2 and CH4 carried out in a bomb calorimeter with
  a heat capacity of 11.3 kJ/C. The combustion of
  1.50 g of methane produced a T change of 7.3C
  while the combustion of 1.15 g of hydrogen
  produced a T change of 14.3C. Find the energy of
  combustion per gram for each.

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Example 3

• methane CH4
• hydrogen H2

The energy released by H2 is about 2.5 times the
energy released by CH4
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Hesss Law

• If a reaction is carried out in a series of
  steps, ?H for the reaction will equal the sum of
  the enthalpy changes for the individual steps.

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Examples

• One Step
• N2(g) 2O2 (g) ? 2NO2 (g) DH1 68kJ
• Two Step
• N2(g) O2 (g) ? 2NO (g) DH2 180kJ
• 2NO(g) O2 (g) ? 2NO2 (g) DH3 -112kJ
• N2(g) 2O2 (g) ? 2NO2 (g) DH2 DH3 68kJ

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Hesss Law Rules

• If the reaction is reversed, then the sign on DH
  is reversed
• The magnitude of DH 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 DH is multiplied by the same integer

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Using Hesss Law

1. Work backward from the final reaction
2. Reverse reactions as needed, being sure to also
   reverse DH
3. Remember that identical substances found on both
   sides of the summed equation cancel each other

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Example 2

• Using the enthalpies of combustion for graphite
  (-394 kJ/mol) and diamond (-396 kJ/mol), find the
  ?H for the conversion of graphite to diamond.
• Cgraphite (s) ? Cdiamond (s) ?H?

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Example 2 Cgraphite (s) ? Cdiamond (s) ?H?

• (1) Cgraphite(s) O2(g) ? CO2(g) ?H-394kJ/mol
• (2) Cdiamond(s) O2(g) ? CO2(g) ?H-396kJ/mol
• to get the desired equation, we must reverse 2nd
  equation
• (1) Cgraphite(s) O2(g) ? CO2(g)
  ?H-394kJ/mol
• (2) CO2(g) ? Cdiamond(s) O2(g) ?H 396kJ/mol
• Cgraphite (s) ? Cdiamond (s) ?H-394 396
• ?H2 kJ/mol

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Example 3

• Find ?H for the synthesis of B2H6, diborane
• 2B(s) 3H2(g) ? B2H6(g) ?H ?
• Given
• 2B(s) 3/2O2(g) ? B2O3(s) ?H1-1273kJ
• B2H6(g) 3O2(g) ? B2O3(s) 3H2O(g)
  ?H2-2035kJ
• H2(g) ½O2(g) ? H2O(l) ?H3-286kJ
• H2O(l) ? H2O(g) ?H444 kJ

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Example 3

• 2B(s) 3H2(g) ? B2H6(g) ?H ?
• Start by paying attention to what needs to be on
  reactants and products side
• (1) 2B(s) 3/2O2(g) ? B2O3(s) ?H1-1273kJ
• -(2) B2O3(s) 3H2O(g) ? B2H6(g) 3O2(g)
  -?H22035kJ
• (3) H2(g) ½O2(g) ? H2O(l) ?H3-286kJ
• (4) H2O(l) ? H2O(g) ?H444 kJ

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Example 3

• Underline what you want to keep- that will help
  you figure out how to cancel everything else
• (1) 2B(s) 3/2O2(g) ? B2O3(s) ?H1-1273kJ
• -(2) B2O3(s) 3H2O(g) ? B2H6(g) 3O2(g)
  -?H22035kJ
• (3) H2(g) ½O2(g) ? H2O(l) ?H3-286kJ
• (4) H2O(l) ? H2O(g) ?H444 kJ

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Example 3

• Need 3 H2 (g) so 3 x (3)
• Need 3 H2O to cancel so 3 x (4)
• (1) 2B(s) 3/2O2(g) ? B2O3(s) ?H1-1273kJ
• -(2) B2O3(s) 3H2O(g) ? B2H6(g) 3O2(g)
  -?H2-(-2035kJ)
• 3x(3) 3H2(g) 3/2O2(g) ? 3H2O(l) 3?H33(-286kJ)
• 3x(4) 3H2O(l) ? 3H2O(g) 3?H43(44 kJ)
• 2B(s) 3H2(g) ? B2H6(g)
• ?H -1273 -(-2035) 3(-286) 3(44) 36kJ
• 
  

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Standard Enthalpy of Formation ?Hf

• change in enthalpy that accompanies the formation
  of one mole of a compound from its elements in
  standard states
• means that the process happened under
  standard conditions so we can compare more easily

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Standard States

• For a COMPOUND
• for gas P 1 atm
• For (s) or (l) pure liquid or solid
• in solution concentration is 1 M
• For an ELEMENT
• form that exists in at 1 atm and 25C
• O O2(g) K K(s) Br Br2(l)

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Writing Formation Equations

• always write equation where 1 mole of compound is
  formed (even if you must use non-integer
  coefficients)
• NO2(g)
• ½N2(g) O2(g) ? NO2(g)
• ?Hf 34 kJ/mol
• CH3OH(l)
• C(s) 2H2(g) ½ O2(g) ?CH3OH(l)
• ?Hf -239 kJ/mol

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Using Standard Enthalpies of Formation
?Hrxn ?n ?Hf (products) ? ?n ?Hf (reactants)

• where
• n number of moles of products/reactants
• ? means sum of
• ?Hf is the standard enthalpy of formation for
  reactants or products
• ?Hf for any element in standard state is zero so
  elements are not included in the summation

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Example 1

• Calculate the standard enthalpy change for the
  reaction that occurs when ammonia is burned in
  air to make nitrogen dioxide and water
• 4NH3(g) 7O2(g) ? 4NO2(g) 6H2O(l)
• ?Hfo values (in appendix in back of book (pg.
  1041)
• Ammonia -46.19 kJ/mol
• Oxygen 0 (because it is in its standard state)
• Nitrogen Dioxide 33.84 kJ/mol
• Water (l) - 285.8 kJ/mol

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Example 1

• (4 x 33.84 kJ/mol) (6 x -285.8) (4 x
  -46.19) (7 x 0)
• -1394.68 kJ/mol
• -1395 kJ/mol

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Example 2

• Calculate the standard enthalpy change for the
  following reaction (DH Iron (III) Oxide -1676
  kJ and DH Aluminum Oxide -826 kJ)
• 2Al(s) Fe2O3(s) ? Al2O3(s) 2Fe(s)

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Example 3

• Compare the standard enthalpy of combustion per
  gram of methanol with per gram of gasoline (it is
  C8H18).
• Write equations
• 2CH3OH(l) 3O2(g) ? 2CO2(g) 4H2O(l)
• 2C8H18(l) 25O2(g)?16CO2(g) 18H2O(l)

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Example 3

• Calculate the enthalpy of combustion per mole
• 2CH3OH(l) 3O2(g) ? 2CO2(g) 4H2O(l)

2C8H18(l) 25O2(g)?16CO2(g) 18H2O(l)
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Example 3

• Convert to per gram using molar mass
• so octane is about 2x more effective