Chemistry 1011

1
Chemistry 1011

• TOPIC
• Thermochemistry
• TEXT REFERENCE
• Masterton and Hurley Chapter 8

2
8.4 Thermochemical Equations

• YOU ARE EXPECTED TO BE ABLE TO
• Define molar enthalpy of reaction, molar heat of
  fusion and molar heat of vaporization.
• Carry out calculations relating heat absorbed or
  released in a chemical reaction, the quantity of
  a reactant or product involved, and DH for the
  reaction.
• Use Hess's Law to determine the heat of reaction
  given appropriate equations and thermochemical
  data.

3
Molar Heat of Fusion and of Vaporization

• The latent heat of fusion (vaporization) of a
  substance is the heat absorbed or released when 1
  gram of the substance changes phase.
• The molar heat of fusion (vaporization) of a
  substance is the heat absorbed or released when
  1mole of the substance changes phase.

4
Thermochemical Equations

• Balanced chemical equations that show the
  enthalpy relation between products and reactants
• H2(g) Cl2(g) ? 2HCl(g) ?H -185 kJ
• Exothermic reaction 185 kJ of heat is evolved
  when 2 moles of HCl are formed.
• 2HgO(s) ? 2Hg(l) O2(g) ?H 182 kJ
• Endothermic reaction 182 kJ of heat must be
  absorbed to decompose 2 moles HgO.

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Thermochemical Equations

• The sign of DH indicates whether the reaction is
  endothermic () or exothermic ()
• Coefficients in the equations represent the
  numbers of moles of reactants and products
• Phases must be specified (s), (l), (g), (aq)
• The value quoted for DH applies when products and
  reactants are at the same temperature, usually
  25oC

6
Thermochemical Equations Rule 1

• ?H is directly proportional to the amount of
  reactants and products.
• When one mole of ice melts, ?H 6.00 kJ.
• When one gram of ice melts, ?H 0.333 kJ.g-1
• H2(g) Cl2(g) ? 2HCl(g) DH -185kJ
• 1/2H2(g) 1/2Cl2(g) ? HCl(g) DH -92.5kJ

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Thermochemical Equations Rule 2

• ?H for a reaction is equal in magnitude but
  opposite in sign to ?H for the reverse reaction.
• 2HgO(s) ? 2Hg(l) O2(g) ?H 182 kJ
• 2Hg(l) O2(g) ? 2HgO(s) ?H -182 kJ

8
Thermochemical Equations Rule 3
?H is independent of the path of the reaction
it depends only on the initial and final states
9
Hess law

• DH DH1 DH2

10
Using Hess Law

• Hess Law is used to calculate ?H for reactions
  that are difficult to carry out directly.
• In order to apply Hess Law we need to know ?H
  for other related reactions.
• For example, we can calculate ?H for
• C(s) 1/2O2(g) ? CO(g) ?H1 ?
• If we know ?H for
• CO(g) 1/2O2 (g) ? CO2(g) ?H2 -283.0 kJ
• C(s) O2(g) ? CO2(g) ?H3 -393.5 kJ

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Calculating the Unknown DH

• Write equation 3
• C(s) O2(g) ? CO2(g) ?H3 -393.5 kJ
• Reverse equation 2
• CO2(g) ? CO(g) 1/2O2 (g) -?H2 283.0 kJ
• Add the two equations
• C(s) 1/2O2(g) ? CO(g)
• ?H1 -393.5kJ 283.0kJ -110.5kJ

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Enthalpy Diagram
C(s) O2(g)
DH1-110.5kJ
CO(g) 1/2O2(g)
DH3 -393.5kJ
DH2 283.0kJ
CO2(g)
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Some Other Examples

• Given
• 1/2N2(g) 3/2H2(g) ? NH3(g) DH -46.1kJ
• C(s) 2H2(g) ? CH4(g) DH -74.7kJ
• C(s) 1/2H2(g) 1/2 N2(g) ? HCN(g) DH
  135.2kJ
• Find DH for the reaction
• CH4(g) NH3(g) ? HCN(g) 3H2(g)

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Some Other Examples

• Given
• 1/2N2(g) 3/2H2(g) ? NH3(g) DH -46.1kJ
• C(s) 2H2(g) ? CH4(g) DH -74.7kJ
• C(s) 1/2H2(g) 1/2 N2(g) ? HCN(g) DH
  135.2kJ
• Find DH for the reaction
• CH4(g) NH3(g) ? HCN(g) 3H2(g)

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Find DH for the reactionCH4(g) NH3(g) ?
HCN(g) 3H2(g

• Add Reverse Eq 2 to Reverse Eq 1
• CH4(g) ? C(s) 2H2(g) DH 74.7kJ
• NH3(g) ? 1/2N2(g) 3/2H2(g) DH 46.1kJ
• CH4(g) NH3(g) ? C(s) 2H2(g) 1/2N2(g)
  3/2H2(g) DH 120.8kJ
• Add Eq 3
• C(s) 1/2H2(g) 1/2 N2(g) ? HCN(g) DH
  135.2kJ
• CH4(g) NH3(g) ? HCN(g) 3H2(g DH 256.0kJ

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Some Other Examples

• A 1.00g sample of table sugar (sucrose),
  C12H22O11, is burned in a bomb calorimeter
  containing 1.50 x 103g water.
• The temperature of the calorimeter and water
  rises from 25.00oC to 27.32oC.
• The heat capacity of the metal components of the
  bomb is 837J.K-1.
• The specific heat of water is 4.184J.g-1.K-1.
• Calculate
• (a) the heat evolved by the 1.00g of sucrose, and
• (b) the heat evolved per mole of sucrose.

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Solution

• Heat absorbed by the water m.c.Dt
• 1.50 x 103 g x 4.184 J.g-1.K-1 x 2.32K
• 14.6 kJ
• Heat absorbed by the calorimeter C.Dt
• 837 J.K-1 x 2.32K 2.03 kJ
• Total heat absorbed by the calorimeter and
  contents
• - total heat released by 1.00g sucrose
• - 16.6kJ
• Heat evolved per mole of sucrose
• - 16.6 kJ x 342 g.mol-1 - 5680 kJ