Thermochemistry

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• Chapter 12
• Thermochemistry

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Thermochemisty

• Thermochemistry is the study of the changes in
  in chemical reactions.
• Almost all chemical reactions either or
  energy, because bond requires energy and
  bond releases energy.
• 11-2

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Thermochemisty

• Reactions release energy (heat).
• Reactions absorb energy (heat).
• Because heat and energy are related, the base
  unit of energy, the (J), also serves as the
  unit of heat.
• 11-3

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Enthalpy

• Enthalpy is the absorbed or released during a
  chemical reaction. It is represented by H, and
  in enthalpy is ?H.
• ?H Hproducts Hreactants

11-4
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Enthalpy

• Standard Enthalpy Change ( ) is the change in
  enthalpy when reactants in their standard
  change to products in their standard .
• 11-5

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Exothermic Reactions

• C3H8(g) 5O2(g) ? 3CO2(g) 4H2O (-2043KJ)

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Exothermic Reactions

• C3H8(g) 5O2(g) ? 3CO2(g) 4H2O (-2043KJ)
• In this reaction, energy is . ?H is negative
  because products have a value for H than the
  reactants. ?H is always for exothermic
  reactions.
• 11-7

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Endothermic Reactions

• C(s) H2O(g) 113KJ ? CO(g) H2(g)

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Endothermic Reactions

• C(s) H2O(g) 113KJ ? CO(g) H2(g)
• In this reaction, energy is . ?H is because
  the products have a higher value for H than the
  reactants. ?H is always for endothermic
  reactions.
• 11-9

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Enthalpy

• Sign of ?H Process Heat
• Positive endothermic absorbed
• Negative exothermic released
• The enthalpy change for a reaction is
  proportionately or depending on the
  of reactants and products.
• 11-10

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Enthalpy

• Ex. How much heat will be released when 5.00g of
  H2O2 decomposes?
• 2H2O2(l) ? 2H2O(l) O2(g) ?H -190 KJ
• First, convert grams to moles
• 5.00g H2O2 x 1.0mol H2O2 mol H2O2
• 34.0g
• Now, multiply the number of moles by the ?H for
  the reaction, which is per 2 moles of H2O2
• mol H2O2 x -190 KJ KJ
• 2 mol H2O2
• 11-11

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Enthalpy

• 1) Now you try one How much heat will be
  released when 6.44g of Sulfur reacts with an
  excess of O2?
• 2S 3O2 ? 2SO3 ?H -791.4 kJ
• First, convert grams to moles
• 6.44g S x 1.0mol S mol S
• 32.07g
• Now, multiply the number of moles by the ?H for
  the reaction, which is per 2 moles of H2O2
• mol S x -791.4 KJ KJ
• mol S
• 11-12

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Hesss Law

• Hesss Law states that if a of
  reactions are added together, the enthalpy change
  for the reaction will be the of the enthalpy
  changes for the individual steps.
• ?Hnet ?H 1 ?H 2

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Hesss Law

• Ex. N2(g) O2(g) ? 2NO(g) ?H 181kJ
• 2NO(g) O2(g) ? 2NO2(g) ?H -113kJ
• N2(g) 2O2(g) ? 2NO(g) 2NO2(g)
• ?Hnet (181kJ) (-113kJ)
• kJ
• 11-14

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Hesss Law

• Rules for applying Hesss Law
• 1) If you an equation by a , you must
  multiply the ?H by the same
• CH4(g) 2O2(g) ? CO2(g) 2H2O(g)
• ?H -802 kJ
• 2CH4(g) 4O2(g) ? 2CO2(g) 4H2O(g)
• ?H -1604 kJ
• 2) If an equation is , the sign of ?H is
  also .
• 11-15

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Hesss Law

• Ex. From the following enthalpy changes
• Rxn 1) H2S(g) 3/2 O2(g) ? H2O(l) SO2(g)
• ?H -563kJ
• Rxn 2) CS2(l) 3O2(g) ? CO2(g) SO2(g)
• ?H -1075kJ
• Calculate the value of ?H for
• Final Rxn) CS2(l) 2H2O(l) ? CO2(g) 2H2S(g)
• In order for Rxn 1 and Rxn 2 to add up to the
  final rxn, you must Rxn 1 and multiply it by
  (so for the ?H of Rxn 1, you must the sign and
  multiply it by )

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Hesss Law

• 2H2O(l) 2SO2(g) ? 2H2S(g) 3O2(g) ?H 1126kJ
• CS2(l) 3O2(g) ? CO2(g) SO2(g) ?H
  -1075k
• CS2(l) 2H2O(l) ? CO2(g) 2H2S(g) ?H
  kJ
• 2) Your turn From the following enthalpy
  changes
• C(s) ½ O2(g) ? CO(g) ?H -110.5kJ
• CO(g) ½ O2 ? CO2(g) ?H -283.0kJ
• Calculate the value of ?H for the reaction
• C(s) O2(g) ? CO2(g) 11-17

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Hesss Law

• C(s) ½ O2(g) ? CO(g) ?H -110.5kJ
• CO(g) ½ O2 ? CO2(g) ?H -283.0kJ
• C(s) O2(g) ? CO2(g) ?H -393.5
• 11-18

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Entropy

• Entropy ( ) a quantitative measure of the , or
  randomness, in the substances involved in a rxn.
  (pg. 756)
• (The prefers increasing disorder!)
• Solid ? Liquid ? Gas increasing S
• 12-19

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Gibbs Free Energy

• The of a reaction depends on
• ?H and ?S. Gibbs Free Energy ( ) is a way to
  measure whether a reaction will occur
  spontaneously or not.
• ?G Spontaneous
• ?G - Not spontaneous
• ?G 0 Equilibrium

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Calorimetry

• Now for some definitions
• Calorimetry the study of heat and heat .
• Heat Capacity the amount of heat needed to
  the temperature of an object oC.
• Specific Heat the amount of heat needed to
  raise the temp. of of a substance
  oC.
• Calorie the amount of heat that increases the
  temp. of g of 1oC.
• ( joules 1cal)

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Calorimetry

• Now for some definitions
• Calorimeter a well container filled with a
  known mass of in which a rxn is carried out.
  The heat in the reaction is the heat by the
  water.
• Heat q
• qrxn -qsur
• qsur m (mass of water) x (specific
  heat) x Tf-Ti

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Calorimetry

• qsur m (mass of water) x c (specific heat) x
  Tf-Ti
• Ex. When a 12.8g sample of KCl dissolves in 75.0
  g of water in a calorimeter, the temp. drops from
  31.0oC to 21.6 oC. Calculate ?H.
• qsur (75.0g)(4.184J/goc)(21.6 31.0)
• -2950J
• qrxn -qsur
• 2950J
• Now calculate moles of KCl
• 12.8g KCl x 1mol KCl mol KCl
• 74.55g
• ?H 2950J/ mol KCl kJ 11-23

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Calorimetry

• qsur m x c x Tf-Ti
• 3) When a 25.7g sample of NaI dissolves in 80.0
  g of water in a calorimeter, the temp. increases
  from 20.5oC to 24.4 oC. Calculate ?H.
• qsur ( g)( J/goc)(24.4 20.5)
• J
• qrxn -qsur
• J
• Now calculate moles of NaI
• 25.7g NaI x 1mol NaI mol NaI
• g
• ?H / mol NaI kJ
• 11-24

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• The
• End!