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Chemistry 1011

Chemistry 1011. TOPIC Thermochemistry TEXT REFERENCE Masterton and Hurley Chapter 8. 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.

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Chemistry 1011

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  1. Chemistry 1011 TOPIC Thermochemistry TEXT REFERENCE Masterton and Hurley Chapter 8 Chemistry 1011 Slot 5

  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. Chemistry 1011 Slot 5

  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. Chemistry 1011 Slot 5

  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. Chemistry 1011 Slot 5

  5. 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 Chemistry 1011 Slot 5

  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 Chemistry 1011 Slot 5

  7. 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 Chemistry 1011 Slot 5

  8. Thermochemical Equations Rule #3 H is independent of the path of the reaction – it depends only on the initial and final states Chemistry 1011 Slot 5

  9. Hess’ law DH = DH1 + DH2 Chemistry 1011 Slot 5

  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 Chemistry 1011 Slot 5

  11. 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 Chemistry 1011 Slot 5

  12. Enthalpy Diagram C(s) + O2(g) DH1-110.5kJ CO(g) + 1/2O2(g) DH3 -393.5kJ DH2 +283.0kJ CO2(g) Chemistry 1011 Slot 5

  13. 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) Chemistry 1011 Slot 5

  14. 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) Chemistry 1011 Slot 5

  15. Find DH for the reaction:CH4(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 Chemistry 1011 Slot 5

  16. 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. Chemistry 1011 Slot 5

  17. 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 Chemistry 1011 Slot 5

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