1. a “Calorimeter”

2. Calorimetry Calculations When analyzing data obtained using a calorimeter, make these assumptions: Any thermal energy transferred from the calorimeter to the outside environment is negligible. Any thermal energy absorbed by the calorimeter itself is negligible. All dilute, aqueous solutions have the same density (1.00 g/mL) and specifi c heat capacity (4.18 J/(g∙°C)) as water thermal energy absorbed or released by a chemical system is given the symbol q. q water = mc  t 5.2

3.  H = H (products) – H (reactants)  H = heat given off or absorbed during a reaction at constant pressure H products < H reactants  H < 0 H products > H reactants  H > 0 5.2 Energy Changes during Exothermic & Endothermic Reaction in an Open System

4. Representing Molar Enthalpy Changes 1. The equations we use to represent energy changes are called thermochemical equations. 2. Endothermic enthalpy changes are reported as positive values; H 2 O (l) → H 2 + 1/2O 2(g) ∆H decomp. = + 285.8 KJ/mol H 2 O 3. Exothermic enthalpy changes are reported as negative values H 2(l) + 1/2O 2 (g) → H 2 O (l) ∆H comb = -285.8 KJ/mol H 2 Energy changes can be communicated in four different ways. Three of them are thermochemical equations and one uses diagram. These are: 1. By including an energy value as a term in the thermochemical equation e.g., CH 3 OH (l) + 3/2 O 2 → CO 2(g) + 2H 2 O (g) + 726 KJ(Exothermic Reaction)

5. Calculations Involving Thermal Energy Transfer Thermal Energy transfer by metal 5.2 How much heat is given off when an 869 g iron bar cools from 94 0 C to 5 0 C? Solution: Given: c of Fe = 0.444 J/g 0 C m = 869 g  t = t final – t initial = 5 0 C – 94 0 C = -89 0 C q = mc  t = 869 g x 0.444 J/g 0 C x –89 0 C = -34,000 J

6. 5.2 A student places 50.0 mL of liquid water at 21.00 °C into a coffee- cup calorimeter. She places a sample of gold at 100 °C into the calorimeter. The final temperature of the water is 21.33 °C. The specific heat capacity of water is 4.18 J/g∙°C and the density of water, d, is 1.00 g/mL. Calculate the quantity of thermal energy, q, absorbed by the water in the calorimeter SOLUTION: Given: V H2O =50.0 mL; c=4.18 J/(g∙°C); d H2O =1.00 g/mL; T initial =21.00 °C;T final =21.33 °C R:quantity of thermal energy absorbed, q d =m/v m= vd, m =50mL x 1.00g/mL = 50 g ∆t = 21.33 o C – 21.00 o C Calculations of metal in water

7. q = mc∆t =50 g x 4.18 J x 0.33 o C g. o C = 69 J Determining specific heat capacity of a substance Using the value of q above, calculate the specific heat capacity, c, of gold if its mass is 6.77 g. Final temperature of the gold is same as that of water in the calorimeter. Given:m Au =6.77 g; q = 69 J; t initial =100 o C; t final = 21.33 o C ∆t =21.33 o C – 100 o C = - 78.67 o C Required: c Au q = mc ∆t Rearrage and solve for c

8. Thermal Energy Transfer during a Neutralization reaction A 50.0 mL sample of a 1.0 mol/L aqueous solution of hydrochloric acid, HCl(aq), was mixed with 50.0 mL of a 1.0 mol/L aqueous solution of sodium hydroxide, NaOH(aq), at 25.0 °C in a calorimeter. After the solutions were mixed by stirring, the temperature was 31.9 °C. (a) Determine the quantity of thermal energy transferred by the reaction to the water, q, Assume that the specific heat capacity and density of both solutions is the same as that of liquid water (b) State whether the reaction was endothermic or exothermic. Given: Total volume = 100 mL; c water = 4.18J/g. o C d water = m =1g/mL v m = vd; m solution = 100 g ∆t = 31.9 o C – 25.0 o C = 6.9 o C; q = mc∆t =100 g x 4.18 J x 6.9 o C g. o C = 2.9 x10 3 J = 2.9 kJ

9. Molar Enthalphy Change Energy changes involving 1 mole of a substances…(∆H) Types of molar Enthalpies Solution (∆H sol ) NaBr (s) →Na (aq) + Br (aq) Combustion (∆H comb ) CH 4(g) + 2O 2(aq) → CO 2(aq) + H 2 O (l) Vapourization (∆H vap ) CH 3 OH (l) → CH 3 OH (g) Freezing (∆H fr ) H 2 O (l) → H 2 O(s) Neutralization (∆H neut ) 2NaOH (aq) + H 2 SO 4(aq) →Na 2 SO 4(aq) + H 2 O (l) Formation (∆H f ) C (s) +2H 2(g) + 1/2O 2(g) → CH 3 OH (l) ∆H = n∆Hr Where n is the amount and ∆Hr is the molar enthalpy change of the reaction 5.2

10. Molar Enthalpy Calculations Calculate ∆H for Vaporization Reactions Ethanol, CH 3 CH 2 OH (l), is used to disinfect the skin prior to an injection. If a 1.00 g sample of ethanol is spread across the skin and evaporated, what is the expected enthalpy change? The molar enthalpy of vaporization of ethanol is 38.6 kJ/mol. Given: M ethanol =1.00 g; ∆H vap =38.6 kJ/mol M ethanol = 24 +6 +16 46g/mol R = ∆H? n = 1 g x 1 mol/46 g A= ∆H = n∆H vap = 0.022 ∆H = 0.022 mol x 38kJ/mol = 0.836 KJ 5.2

11. H 2 O (s) H 2 O (l)  H = 6.01 kJ Is  H negative or positive? System absorbs heat Endothermic  H > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 0 0 C and 1 atm. 5.2 Potential Energy Diagram Thermochemical Equations

12. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l)  H = -890.4 kJ Is  H negative or positive? System gives off heat Exothermic  H < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 25 0 C and 1 atm. 5.2 Thermochemical Equations

13. H 2 O (s) H 2 O (l)  H = 6.01 kJ The stoichiometric coefficients always refer to the number of moles of a substance Calculations in Thermochemical Equations If you reverse a reaction, the sign of  H changes H 2 O (l) H 2 O (s)  H = - 6.01 kJ If you multiply both sides of the equation by a factor n, then  H must change by the same factor n. 2H 2 O (s) 2H 2 O (l)  H = 2 x 6.01 = 12.0 kJ 5.2

14. 5.3:Bond energies Bonds in compounds are broken when energies are supplied. The energy required to break a chemical bond is called bond dissociation energy. Bond dissociation energies have positive values Bond dissociation energies are reported as average bond energies because it depends on types and # of bonds in the molecule. E.g. bond energy for methane: 413KJ/mol Calculated as follows:CH 4 →CH 3 + H---- (435KJ/mol) CH 3 → CH 2 + H----(453KJ/mol) CH 2 → CH + H ----(425KJ/mol) CH → C + H ----- (339KJ/mol) Total = 1652 4 This value is convenient for calculating Av. = 413 5.3

15. Enthalpy and Bond Energies During a chemical reaction, the bonds in the reactants must first break, energy must be added —an endothermic process. Bond breaking have positive signs Making new bonds in the products releases energy—an exothermic Process, energy terms associated with bond making have negative signs. ΔH=sum of energies required + to Break old bonds (positive values) Sum of energies released In the formation of new bonds (negative values) This leads to the equation Δ H= Σn x D (bonds broken) – Σn x D(bonds formed) Energy required energy released 5.3

16. Σ (sigma) means “the sum of,” n is the amount (in moles) of a particular bond type, and D is the bond energy per mole of bonds. D is obtained from reference tables Steps in using bond energies to predict ∆H for a reaction 1. Determine how many of each type of bond must be broken in the reactants. 2. Determine the number of bonds of each type that form in the products. 3. Use bond energy data from Table 1 (page 307) to calculate the total energy required to break the reactant bonds, followed by the total energy released by the formation of product bonds. 4. The energy change, ∆H, of the reaction is the difference between these two sums 5.3