1. Thermochemistry Chapter 6 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Acknowledgement Thanks to The McGraw-Hill Companies, Inc. For allowing usage in the classroom.

2. Thermochemistry is the study of heat change in chemical reactions. In physics, total energy = potential energy + kinetic energy

3. The system is the specific part of the universe that is of interest in the study. open mass & energyExchange: closed energy isolated nothing 6.2 Types of systems

4. Heat is the transfer of thermal energy between two bodies that are at different temperatures. Note: (1) Heat transfer direction: hotter  colder; (2) Hotter object loses heat (sign: negative); colder object absorbs heat (sign: positive). Energy Changes in Chemical Reactions Temperature is a measure of the thermal energy. 6.2 Temperature = Thermal Energy

5. Exothermic process is any process that gives off heat and thus if expressed in a chemical equation, we’ll see “heat or energy” as the product (or right) or see q < 0 or ∆H < 0 (that is negative value) – transfers thermal energy from the system to the surroundings. Examples: g  l  s; g  s; combustion; explosion; acid-base neutralization Endothermic process is any process in which heat has to be supplied to the system from the surroundings. Therefore if expressed in a chemical equation, we’ll see “heat or energy” shows up at the reactant side (or left) or see q > 0 or ∆H > 0 2H 2 (g) + O 2 (g) 2H 2 O (l) + energy H 2 O (g) H 2 O (l) + energy energy + 2HgO (s) 2Hg (l) + O 2 (g) 6.2 energy + 2H 2 O (l) → 2H 2 (g) + O 2 (g)

6. ExothermicEndothermic 6.2

7. Thermodynamics is the scientific study of the interconversion of heat and other kinds of energy. We study the changes in the state of a system, which is defined by the value of macroscopic properties,e.g. composition, energy, pressure, volume, temperature. State functions (always can be expressed by mathematical symbol of ∆) are properties that are determined by the state of the system, regardless of how that condition was achieved. E.g. energy, pressure, volume, temperature Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. 6.3  E = E final - E initial  P = P final - P initial  V = V final - V initial  T = T final - T initial

8. First law of thermodynamics – energy can be converted from one form to another, but can neither be created nor destroyed.  E system +  E surroundings = 0 or  E system = -  E surroundings C 3 H 8 + 5O 2 3CO 2 + 4H 2 O Exothermic chemical reaction! 6.3 Chemical energy lost by combustion = Energy gained by the surroundings system surroundings

9. Another form of the first law for  E system 6.3  E = q + w  E is the change in internal energy of a system q is the heat exchange between the system and the surroundings w is the work done on (or by) the system

10. What is the change in energy (J) of the gas when a gas expends and does P-V work on the surroundings equal to 325 J and at the same time absorbs 127 J of heat from the surroundings?  E = q + w = 127 J-325J = -198J Heat (q)/ work(W) are not state functions because its value depends on the path of the process and varies accordingly. What is the sign of w and q?  q= q final - q initial  E = q + w, E is a state function  w = w final - w initial X

11. Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. H =E+PV  H = n∑H (products) – m∑H (reactants)  H = heat given off or absorbed during a reaction at constant pressure; ∑ means addition. If there are two or more reactants and products, we must calculate the total (sum). M, n refer to stoichiometry---See section 6.6 for details. Enthalpy of reactions

12. Thermochemical Equations H 2 O (s) H 2 O (l)  H = 6.01 kJ/mol 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. 6.4

13. Thermochemical Equations CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l)  H = -890.4 kJ/mol 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. 6.4 Note: the per mol in the unit means that this the enthalpy change per mole of the reactant, CH 4 (g), as it is written. Preferably not to write /mol.

14. H 2 O (s) H 2 O (l)  H = 6.01 kJ/mol The stoichiometric coefficients always refer to the number of moles of a substance Thermochemical Equations- show enthalpy changes as well as the mole relationship. If you reverse a reaction, the sign of  H changes H 2 O (l) H 2 O (s)  H = - 6.01 kJ/mol If you multiply both sides of the equation by a factor n (can be an integer or a fraction), 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/mol 6.4

15. H 2 O (s) H 2 O (l)  H = 6.01 kJ/mol The physical states of all reactants and products must be specified in thermochemical equations. Thermochemical Equations 6.4 H 2 O (l) H 2 O (g)  H = 44.0 kJ/mol How much heat is evolved when 266 g of white phosphorus (P 4 ) burn in air? P 4 (s) + 5O 2 (g) P 4 O 10 (s)  H = -3013 kJ/mol 266 g P 4 1 mol P 4 123.9 g P 4 x -3013 kJ 1 mol P 4 x = -6470 kJ

16. Consider the reaction below: 2CH 3 OH(l) + 3O 2 (g)  4H 2 O(l) + 2CO 2 (g) ∆H = -1452.8 kJ/mol What is the value of ∆H for the reaction of 8H 2 O(l) + 4CO 2 (g)  4CH 3 OH (l) + 6O 2 (g)? If you reverse a reaction, the sign of ΔH changes. If you multiply both sides of the equation by a factor n, then ∆H must change by the same factor n. ∆H = 2*(+1452.8 kJ/mol) = +2905.6 KJ/mol

17. The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. Heat (q) absorbed or released: q = m x s x  t  t = t final - t initial 6.5 Calorimetry- the measurement of heat.

18. How much heat is given off when an 869 g iron bar cools from 94 0 C to 5 0 C? s of Fe = 0.444 J/g 0 C  t = t final – t initial = 5 0 C – 94 0 C = -89 0 C q = ms  t = 869 gx(0.444 J/g 0 C)x(–89 0 C) = -34,000 J 6.5

19. A sheet of gold weighing 10.0 g and at a temperature of 18.0 o C is placed flat on a sheet of iron weighing 20.0 g and at a temperature of 55.6 o C. What is the final temperature ( o C) of the combined metals? Assume that no heat is lost to the surroundings. The specific heat of iron and gold are 0.444 and 0.129 J/g. o C, respectively. No heat is lost to the surroundings: heat absorbed by the colder object gold = - heat released by the hotter object iron. q gold+ q iron =0, q gold =- q iron q = m x s x Δt, 10.0g*0.129J/g. o C(T-18.0)= 20.0g*0.444 J/g. o C*(55.6-T), T=50.7 o C

20. Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest? Establish an arbitrary scale with the standard enthalpy of formation (  H 0 ) as a reference point for all enthalpy expressions. f Standard enthalpy of formation (  H 0 ) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. “ 0 ”-standard state at 1atm; “f”-formation f The standard enthalpy of formation of any element in its most stable form is zero.  H 0 (O 2 ) = 0 f  H 0 (O 3 ) = 142 kJ/mol f  H 0 (C, graphite) = 0 f  H 0 (C, diamond) = 1.90 kJ/mol f 6.6 Standard enthalpy of formation and reaction

21. 6.6

22. The standard enthalpy of reaction (  H 0 ) is the enthalpy of a reaction carried out at 1 atm. rxn aA + bB cC + dD H0H0 rxn d  H 0 (D) f c  H 0 (C) f = [+] - b  H 0 (B) f a  H 0 (A) f [+] H0H0 rxn n  H 0 (products) f =  m  H 0 (reactants) f  - 6.6 Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

24. Benzene (C 6 H 6 ) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol. 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) H0H0 rxn n  H 0 (products) f =  m  H 0 (reactants) f  - H0H0 rxn 6  H 0 (H 2 O) f 12  H 0 (CO 2 ) f = [+] - 2  H 0 (C 6 H 6 ) f [] H0H0 rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ -5946 kJ 2 mol C 6 H 6 = - 2973 kJ/mol C 6 H 6 6.6

25. Calculate the standard enthalpy change (kJ/mol) for the reaction 2Al(s) + Fe 2 O 3 (s)  2Fe(s) + Al 2 O 3 (s) Given that 2Al(s) + 3/2O 2 (g)  Al 2 O 3 (s) ∆H = -1669.8 kJ/mol 2Fe(s) + 3/2O 2 (g)  Fe 2 O 3 (s)∆H = -822.2 kJ/mol Indirect method:Hess’s Law 2Al(s) + 3/2O 2 (g)  Al 2 O 3 (s) ∆H = -1669.8 kJ/mol +Fe 2 O 3 (s)  2Fe(s) + 3/2O 2 (g)∆H = 822.2 kJ/mol 2Al(s) + Fe 2 O 3 (s)  2Fe(s) + Al 2 O 3 (s) ∆H = -1669.8 +822.2=-847.6 kJ/mol

26. Calculate the standard enthalpy of formation of CS 2 (l) given that: C (graphite) + O 2 (g) CO 2 (g)  H 0 = -393.5 kJ rxn S (rhombic) + O 2 (g) SO 2 (g)  H 0 = -296.1 kJ rxn CS 2 (l) + 3O 2 (g) CO 2 (g) + 2SO 2 (g)  H 0 = -1072 kJ rxn 1. Write the enthalpy of formation reaction for CS 2 C (graphite) + 2S (rhombic) CS 2 (l) 2. Add the given rxns so that the result is the desired rxn. rxn C (graphite) + O 2 (g) CO 2 (g)  H 0 = -393.5 kJ 2S (rhombic) + 2O 2 (g) 2SO 2 (g)  H 0 = -296.1x2 kJ rxn CO 2 (g) + 2SO 2 (g) CS 2 (l) + 3O 2 (g)  H 0 = +1072 kJ rxn + C (graphite) + 2S (rhombic) CS 2 (l)  H 0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ rxn 6.6