The document provides information about thermochemistry and how to calculate the change in enthalpy (ΔH) of chemical reactions using standard enthalpy values and bond energies. It discusses how ΔH is measured experimentally using calorimetry and how Hess's law allows calculating ΔH from multiple reaction steps. Standard enthalpy values and bond dissociation energies allow predicting if reactions are endothermic or exothermic without direct experimentation.

2. General Chemistry II CHEM 152 Unit 3 Week 10

3. Week 10 Reading Assignment Chapter 6 – Sections 6.2 – 6.8 (thermochemistry)

4. HYDROCHLORIC ACID SODIUM HYDROXIDE Make a Prediction . Make a Prediction Will a reaction take place? . HCl + NaOH  NaCl + H 2 O ??????????? ?????????? YES!!!!!

5. What are the chances of the car just by itself Turning into the car on the right? Will this reaction occur by itself? 2 Fe 2 O 3  4 Fe + 3O 2 CHEMICAL REACTIONS have a natural DIRECTION How can we predict that direction???

7. During a chemical reaction chemical bonds are broken and new bonds are formed. The chemical nature of the substances present in the system changes. ENERGY IS ABSORBED OR RELEASED IN THE PROCESS Thermochemistry How can we determine whether the process requires or releases energy? Macro Micro

8. Surroundings System q > 0 heat ENDOTHERMIC Heat is transferred from surroundings to the system The surroundings get cooler The surroundings get warmer EXOTHERMIC Surroundings q < 0 heat System Heat is transferred from the system to the surroundings

11. Exothermic H 2 (g) + ½ O 2 (g)  H 2 O(l)  H = - 285.8 kJ/mol In this case, products have a lower potential energy than reactants. H 2 + ½ O 2 PE H 2 O(l) Reactants Products ENERGY RELEASED  H

12. Endothermic NH 4 NO 3 (s) + H 2 O(l)  NH 4 + (aq) + NO 3 - (aq) In this case, products have a higher potential energy than reactants. NH 4 NO 3 + H 2 O PE NH 4 + + NO 3 - Reactants Products ENERGY ABSORBED  H

13. Q (Calorimeter) = − Q (reaction) How could we MEASURE the energy transferred? Heat coming out of the reaction all goes into the water. CALORIMETER What happens to the temperature of the water?

14. 0.0150 mol of a substance was combusted in a bomb calorimeter. 500.0 g of water in a calorimeter had an increase of 4.90 ºC. What was the energy change? C p of water is 4.184 J/(g ºC) What was the  H of the combustion reaction in units of kJ/mol?

15. Energy is never lost so…. Estimating Δ H using lab experiments H Sn + Cl 2 SnCl 2 + Cl 2 SnCl 4 -325 kJ -186 kJ -511 kJ Sn + Cl 2  SnCl 2 Δ H = -325 kJ SnCl 2 + Cl 2  SnCl 4 Δ H= -186 kJ Sn + 2Cl 2  SnCl 4 Δ H = -511 kJ Hess’ Law – If a reaction can be written as a series of steps, the The Δ H of the overall reaction equals the sum of the the Δ H’s Of the steps.

16. Calculate  H  for: Ca(OH) 2(aq) + 2HCl (aq)  CaCl 2(aq) + 2H 2 O (l) using: CaO (s) + 2HCl (aq)  CaCl 2(aq) + H 2 O (l)  H  = ‒186 kJ CaO (s) + H 2 O (l)  Ca(OH) 2(s)  H  = ‒62.3 kJ Ca(OH) 2(s)  Ca(OH) 2(aq)  H  = ‒12.6 kJ

18. What if we can’t do experiments?

19. Standard Enthalpy Values Most  H values for a chemical reaction are labeled  H o measured under standard conditions P = 1 bar (close to 1 atmosphere) T = usually 25 o C (298.15 K) All species in standard states e.g., C = graphite and O 2 = gas  H o f = standard molar enthalpy of formation This is the enthalpy change when 1 mol of a compound is formed from elements under standard conditions.

20.  H o f , Standard Enthalpy of Formation H 2 (g) + 1/2 O 2 (g)  H 2 O(l)  H o f = -285.8 kJ/mol H 2 (g) + O 2 (g)  H 2 O 2 (l)  H o f = -187.8 kJ/mol H 2 O 2 (l)  H 2 O(l) + ½ O 2 (g) ? By convention,  H o f = 0 for elements in their standard states H 2 , O 2 H 2 O(l) H 2 O 2 (l) PE -285.8 kJ -187.8 kJ

21. Molar Enthalpies of Formation Substance Name  H o f (kJ/mol) CH 4 (g) Methane -74.8 CO(g) Carbon monoxide -110.5 CO 2 (g) Carbon dioxide -393.5 H 2 O(g) Water vapor -241.8 H 2 O(l) Liquid water -285.8 CH 3 OH(l) Methanol -238.7 C 2 H 5 OH(l) Ethanol -277.7 BaCO 3 (s) Barium carbonate -1216.3 CaO(s) Calcium oxide -635.1 NH 3 (g) Ammonia -46.1

24. Using Standard Enthalpy Values H 2 O(g) + C(graphite)  H 2 (g) + CO(g)  H o rxn =   H o f (products) -   H o f (reactants)  H o rxn =  H o f (H 2 ) +  H o f (CO) -  H o f (H 2 O) –  H o f (C)  H o rxn = 0 +(–110.5 kJ) – (-241.8 kJ) – 0  H o rxn = + 131.3 kJ (Endothermic)

25. H 2 O(g) + C(graphite)  H 2 (g) + CO(g)  H o rxn =   H o f (products) -   H o f (reactants)  H o rxn = + 131.3 kJ (Endothermic) If we find ∆H for the heats of formation If we find ∆H from a calorimeter measuring the reaction ∆ H = + 131.3 kJ ∆ H is the same no matter what route we take.

26. Using Standard Enthalpy Values Nitroglycerin is a powerful explosive because it decomposes exothermically and four different gases are formed: 2 C 3 H 5 (NO 3 ) 3 (l)  3 N 2 (g) +½ O 2 (g) +6 CO 2 (g) +5 H 2 O(g) For liquid nitroglycerin  H o f =-364.0 kJ/mol. Calculate the energy transfer when 1 mole of nitroglycerin explodes Substance Name  H o f (kJ/mol) CO 2 (g) Carbon dioxide -393.5 H 2 O(g) Water vapor -241.8 H 2 O(l) Liquid water -285.8

27. Using Bond Energy Data We can estimate  H by comparing the amount of energy needed to separate atoms in the reactants ( BREAKING BONDS ) and the energy released in forming new compounds ( MAKING BONDS ) Remember ∆H is a function of state BOND ENERGY (kJ/mol) H—H 436 C—C 346 C=C 602 C  C 835 N  N 945 The energy required to break a bond is a measure of “bond strength” The same energy is released in making the bond.

34. Using Bond Energies Which of these two molecules will generate more energy during combustion? Oxygenated bonds require more energy to break than other types of bonds (C-H) Sucrose Stearic Acid

35. Does burning several logs release more Or less energy than burning one log? There is a relationship between the enthalpy of a reaction And the moles of material. Given by the balanced equation

36. Energy and Moles What if I use up 6.00 moles of Oxygen in the following reaction? How much energy is released? H 2 (g) + 1/2O 2 (g)  H 2 O(l) Δ H= - 286kJ Conversion factor 6.00mol O 2 ( -286kJ ) = -3430 kJ ( ½ mol O 2 ) Proportions 6.00 mol O 2 = X solving x = -3430kJ ½ mol O 2 -286kJ Δ H= - 3430kJ OR +3430kJ are RELEASED

38. Summary Activity 12C (s,graphite) + 11H 2(g) + O 2(g)  C 12 H 22 O 11(s) No one has been able to get this reaction to run… so how could we get a value for the heat of formation of sucrose? We can combust sucrose: C 12 H 22 O 11(s) + 12O 2(g)  12CO 2(g) + 11H 2 O (l) The standard heat of combustion  H c ° of sucrose is – 5640.9 kJ/mol. Can you now calculate the heat of formation of sucrose using Hess’s Law?  H f º(CO 2 (g)) = -393.5 kJ/mol  H f º(H 2 O(l)) = -285.8 kJ/mol 11 2

39. Summary Activity Nitrogen monoxide has recently been found to be involved in a wide range of biological processes. The gas reacts with oxygen to give brown NO 2 gas. 2NO (g) + O 2(g)  2NO 2 (g) Is the reaction endothermic or exothermic? If 1.25 g of NO is converted completely to NO 2 , what quantity of heat is absorbed or released?  Hº = -114.1 kJ M (NO) = 30.01 g/mol