Chemistry 1011 Slot 51 Chemistry 1011 TOPIC Thermochemistry TEXT REFERENCE Masterton and Hurley Chapter 8

Chemistry 1011 Slot 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  H for the reaction. –Use Hess's Law to determine the heat of reaction given appropriate equations and thermochemical data.

Chemistry 1011 Slot 53 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 54 Thermochemical Equations Balanced chemical equations that show the enthalpy relation between products and reactants H 2 (g) + Cl 2 (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) + O 2 (g);  H= +182 kJ Endothermic reaction: 182 kJ of heat must be absorbed to decompose 2 moles HgO.

Chemistry 1011 Slot 55 Thermochemical Equations The sign of  H 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  H applies when products and reactants are at the same temperature, usually 25 o C

Chemistry 1011 Slot 56 Thermochemical Equations Rule #1  H is directly proportional to the amount of reactants and products. When one mole of ice melts,  H = kJ. When one gram of ice melts,  H = kJ.g -1 H 2(g) + Cl 2(g)  2HCl (g) ;  H = -185kJ 1 / 2 H 2(g) + 1 / 2 Cl 2(g)  HCl (g) ;  H = -92.5kJ

Chemistry 1011 Slot 57 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) + O 2 (g);  H= +182 kJ 2Hg(l) + O 2 (g)  2HgO(s) ;  H= -182 kJ

Chemistry 1011 Slot 58 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 59 Hess’ law  H =  H1 +  H2

Chemistry 1011 Slot 510 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 1.C(s) + 1 / 2 O 2 (g)  CO(g);  H 1 = ? If we know  H for 2.CO(g) + 1 / 2 O 2 (g)  CO 2 (g);  H 2 = kJ 3.C(s) + O 2 (g)  CO 2 (g);  H 3 = kJ

Chemistry 1011 Slot 511 Calculating the Unknown  H Write equation #3 C(s) + O 2 (g)  CO 2 (g);  H 3 = kJ Reverse equation #2 CO 2 (g)  CO(g) + 1 / 2 O 2 (g); -  H 2 = kJ Add the two equations C(s) + 1 / 2 O 2 (g)  CO(g);  H 1 = kJ kJ = kJ

Chemistry 1011 Slot 512 Enthalpy Diagram C(s) + O 2 (g) CO 2 (g) CO(g) + 1 / 2 O 2 (g)  H kJ  H kJ  H kJ

Chemistry 1011 Slot 513 Some Other Examples Given 1. 1 / 2 N 2(g) + 3 / 2 H 2(g)  NH 3(g) ;  H = -46.1kJ 2.C(s) + 2H 2(g)  CH 4(g) ;  H = -74.7kJ 3.C(s) + 1 / 2 H 2(g) + 1 / 2 N 2(g)  HCN(g);  H = kJ Find  H for the reaction: 4.CH 4(g) + NH 3(g)  HCN(g) + 3H 2(g)

Chemistry 1011 Slot 514 Some Other Examples Given 1 / 2 N 2(g) + 3 / 2 H 2(g)  NH 3(g) ;  H = -46.1kJ C(s) + 2H 2(g)  CH 4(g) ;  H = -74.7kJ C(s) + 1 / 2 H 2(g) + 1 / 2 N 2(g)  HCN(g);  H = kJ Find  H for the reaction: CH 4(g) + NH 3(g)  HCN(g) + 3H 2(g)

Chemistry 1011 Slot 515 Find  H for the reaction: CH 4(g) + NH 3(g)  HCN(g) + 3H 2(g Add Reverse Eq #2 to Reverse Eq #1 CH 4(g)  C(s) + 2H 2(g) ;  H = +74.7kJ NH 3(g)  1 / 2 N 2(g) + 3 / 2 H 2(g) ;  H = +46.1kJ CH 4(g) + NH 3(g)  C(s) + 2H 2(g) + 1 / 2 N 2(g) + 3 / 2 H 2(g) ;  H = kJ Add Eq #3 C(s) + 1 / 2 H 2(g) + 1 / 2 N 2(g)  HCN(g);  H = kJ CH 4(g) + NH 3(g)  HCN(g) + 3H 2(g;  H = kJ

Chemistry 1011 Slot 516 Some Other Examples A 1.00g sample of table sugar (sucrose), C 12 H 22 O 11, is burned in a bomb calorimeter containing 1.50 x 10 3 g water. –The temperature of the calorimeter and water rises from o C to o C. –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 517 Solution Heat absorbed by the water = m.c.  t = 1.50 x 10 3 g x J.g -1.K -1 x 2.32K = 14.6 kJ Heat absorbed by the calorimeter = C.  t = 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