CHEM 1105 S10 March 11 & 14, 2014 Today s topics: Thermochemistry (Chapter 6) Basic definitions Calorimetry Enthalpy Thermochemical equations Calculating heats of reaction Hess s Law Energy and Heat Some definitions: Energy (E) capacity to do work or transfer heat Heat (q) energy transferred through temperature differences Energy units: Joule = J = kg m 2 /s 2 calorie = the amount of energy required to raise 1 gram of water by 1 o C 1
calorie (cal) = the amount of energy required to raise 1 gram of water by 1 o C 1 kcal = 10 3 cal kcal = Calorie = food Calorie 1 cal = 4.184 J 1 kcal = 1 Calorie = 4.184 kj 1 kj = 10 3 J Calorimetry Calorimetry = measurements of heat changes Heat changes are observed through changes in temperature q = C. ΔT C = heat capacity (of an object or system) ΔT = T f T i (final temp initial temp) q = m. C s. ΔT C s = specific heat capacity (for a pure substance) (sometimes called s or specific heat ) 2
A 28.0 g sample of Ni at 22.5 C is allowed to cool down on a piece of dry ice. It cools to a temperature of -9.7 C. How much heat is lost in the process? C s for Ni = 0.444 J/g C. Enthalpy Enthalpy = H = energy from chemical reactions Change in Enthalpy (ΔH) Heat is evolved or absorbed in all chemical reactions q rxn = reaction heat q p = heat of reaction at constant pressure q p = ΔH = H(products)- H(reactants) Enthalpy is a state function (independent of path) 3
Exothermic and endothermic reactions Heat is evolved or absorbed in all chemical reactions ΔH = H(products)- H(reactants) Exothermic reaction: heat evolved heat flows from reaction mixture to surroundings products have lower energy than reactants energy loss appears as heat ΔH is negative Endothermic reaction: heat absorbed heat flows from surroundings to reaction mixture products have higher energy than reactants energy gain takes heat from surroundings ΔH is positive 144 J were needed to raise the temperature of 32.3 g of an oil from 20.0 C to 25.0 C. Calculate the specific heat of the oil. (Answer: 0.89 J/g C) 4
500. g of gold at 75.0 C were added to 100. ml of water at 20.0 C in a calorimeter. Calculate the final temperature. Assume that there is no heat exchange with the surroundings (i.e., all the heat lost by the gold is gained by the water) and that the calorimeter has zero heat capacity. C s for water = 4.18 J/g C; C s for gold = 0.132 J/g C (Answer: 27.5 C) Thermochemical equations Thermochemical equation: a chemical equation that indicates gain or loss of heat. 4 Fe(s) + 3 O 2 (g) 2 Fe 2 O 3 ΔH = -1644 kj H 2 O NH 4 NO 3 (s) NH 4 + (aq) + NO 3 - (aq) ΔH = +25.7 kj Note: when ΔH is given with a chemical reaction, this means energy PER MOLE of reaction. 5
Combustion of glucose (blood sugar): C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O(l) ΔH = -2826 kj ΔH = -2826 kj/mol, or burning one mole of glucose gives off 2826 kj Example: John burns 635 Calories when running for one hour. If he uses only glucose for energy, how many grams of glucose is he burning? 635 Calories = 635 kcal 635 kcal (4.184 kj/1 kcal )(1 mol/ 2826 kj )(180.16 g/mol) = 169 g glucose Measuring ΔH using calorimetry In calculating for a reaction by measuring q, the heat given off in the reaction corresponding to the number of moles in the reaction must be calculated 6
Bomb calorimeter Setup Some heat from the reaction warms water; therefore: q water = m C s DT Some heat from the reaction warms the calorimeter bomb ; therefore: q cal = C cal DT Total heat evolved q total = q water + q cal Calculate ΔH for the reaction C 6 H 4 O 2 (s) + 6O 2 (g) 6 CO 2 (g) + 2H 2 O(l) from the following data: 2.30 g of C 6 H 4 O 2 (s) were burned in a bomb calorimeter (C cal = 3.27 kj/ C) containing 1000. g of water at 19.22 C. After the reaction, the temperature of the calorimeter and contents rose to 27.07 C. (Use C s for water = 4.18 J/g C) 7
Coffee-Cup Calorimeter Setup Calculations are similar to those with bomb calorimeter You must use the information about the solutions to calculate the stoichiometry Heat from reaction transferred to or from solution (With aqueous solutions, use C s and density of water) Calculate ΔH for the reaction 2NaOH(aq) + H 2 SO 4 (aq) Na 2 SO 4 (aq) + 2H 2 O(l) from the following data: 40.0 ml of 1.000 M NaOH were placed in a calorimeter (C cal = 72.0 J/ C) at 22.0 C and 20.0 ml of 1.500 M H 2 SO 4 at 22.0 C were added. The temperature of the mixture rose to 29.0 C. 8
Standard Enthalpy of Formation ( H fo ) The change in enthlapy associated with the formation of a substance from its elements in their standard states. Measured under standard conditions P = 1 atm, T = 25 o C (298K) NOTE: NOT the same as STP for gases for solutions: 1 molar Table 6.5 in book, or Appendix IIB, units of kj/mol Depends on state (state will be given for substance) The DH f o values for all elements in their standard states is 0 (zero). DH f o = 0 for O 2 (g), Fe(s), C(s, graphite), Br 2 (l), etc. 9
Write the thermochemical equation corresponding to the DH f o for: CH 4 (g) DH f o = -74.6 kj/mol (from table 6.5) Na 2 CO 3 (s), DH f o = -1130.7 kj/mol Using ΔH f o values to calculate ΔH for a chemical reaction H rxn = H o f (products) - Ho f (reactants) Example: Calculate H for the following reaction: 2NH 3 (g) + 3Cl 2 (g) N 2 (g) + 6HCl(g) 10
Using the information from a Table 6.5 or a similar table, calculate the heat of combustion of methanol: CH 3 OH(g) + 3/2 O 2 (g) CO 2 (g) + 2 H 2 O(g) 2ClF 3 (g) + 2NH 3 (g) N 2 (g) + 6HF(g) + Cl 2 (g); ΔH = -1195.6 kj Use data from Table 6.5 to calculate ΔH f for ClF 3 (g) 11
Hess s Law Hess s Law states that the total enthalpy change for a reaction is the same whether the reaction occurs in one or several steps. Remember, since enthalpy is a state function, ΔH is independent of the path. Example: given the following thermochemical equations: Sn(s) + Cl 2 (g) SnCl 2 (s) ΔH = -349.8 kj SnCl 2 (s) + Cl 2 (g) SnCl 4 (l) ΔH = -195.4 kj calculate ΔH for the reaction: Sn(s) + 2Cl 2 (g) SnCl 4 (l) 12
Based on the thermochemical equations for the three reactions shown below: BCl 3 (g) + 3H 2 O(l) H 3 BO 3 (s) + 3HCl(g) ΔH = -112.5 kj B 2 H 6 (g) + 6H 2 O(l) 2H 3 BO 3 (s) + 6H 2 (g) ΔH = -493.4 kj ½ H 2 (g) + ½ Cl 2 (g) HCl(g) ΔH = -92.3 kj calculate ΔH for the following reaction: B 2 H 6 (g) + 6Cl 2 (g) 2BCl 3 (g) + 6HCl(g) 13