1. Thermochemistry Chapter 6

2. Thermochemistry is the study of heat change in chemical reactions.

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

4. Heat is the transfer of thermal energy between two bodies that are at different temperatures. Energy Changes in Chemical Reactions Temperature is a measure of the thermal energy. Temperature = Thermal Energy

5. Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. Endothermic process is any process in which heat has to be supplied to the system from the surroundings. 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) energy + H 2 O (s) H 2 O (l)

6. ExothermicEndothermic

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 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.  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 cannot be created or 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! Chemical energy lost by combustion = Energy gained by the surroundings system surroundings

9. Another form of the first law for  E system  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. The value depends on the path of the process and vary 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 = H (products) – H (reactants)  H = heat given off or absorbed during a reaction at constant pressure 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.

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. Note: the per mol in the unit means that this the enthalpy change per mole of the reaction as it is written.

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, 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

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 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 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 g x 0.444 J/g 0 C x ( –89 0 C) = -34,000 J

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 o.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 Standard enthalpy of formation and reaction

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  - 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(–285.8) ] – [ 2x49.04 ] = -5946 kJ/mol -5946 kJ 2 mol C 6 H 6 = - 2973 kJ/mol C 6 H 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