1. CHM 108 SUROVIEC SPRING 2014 Chapter 6 Energy Transfer

2. I. Principles of Heat Flow Energy is the capacity to do work Work is the result of a force acting through a distance

3. A. Different Types of Energy Kinetic energy – energy associated with motion of an object Thermal energy – energy associated with temperature of an object Potential energy – energy associated with position of an object Chemical energy – energy associated with position of an electron around an atom

4. B. Law of Conservation of Energy Energy cannot be created or destroyed Energy can be transferred from one kind to another

5. II. First Law of Thermodynamics Total energy of the universe is constant and energy is neither created or destroyed A. Internal Energy (IE)  Sum of kinetic and potential energies of all the particles that compose the system  Is a state function

6. A. Internal Energy Internal energy change =  E

7. B. Exchanging Energy A system can exchange energy with the surroundings through heat and work

8. B. Exchanging Energy

9. III. Quantifying Heat and Work A. Heat Heat = exchange of thermal energy between a system and its suroundings

10. B. Heat Capacity When a system absorbs heat (q) its temperature changes by  T. Heat capacity = quantity of heat required to raise T by 1 o C. (J/ o C) Specific heat capacity = quantity of heat required to raise T of 1 gram of material by 1 o C. (J/g o C)

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12. Example Suppose that you have the idea to start making stained glass. Given that glass turns to a pliable liquid around 1500 o C and you are starting at 25 o C, how much heat does a 2.50 g piece of glass absorb?

13. C. Pressure – Volume work How does work change with the volume changes?

14. Example Inflating a balloon requires P-V work 1. If I inflate a balloon from 0.200L to 0.985L at an external pressure of 1.00 atm, how much work in L  atm is done? 2. Given that 101.3 J = 1 L  atm, what is this value of work in Joules?

15. IV. Measuring ΔE The way to measure this is with constant volume calorimetry

16. A. Calorimetry Calorimetry measures change in thermal energy between a system and its surroundings. At constant volume the change in temperature is related to heat absorbed by entire calorimeter

17. Example When 2.09g of sucrose (C 12 H 22 O 11 ) is combusted the temperature rose from 23.89 o C to 28.45 o C. What is the  rxn for this in kJ and then in kJ/mole?

18. V. Enthalpy In most cases we cannot hold the volume constant We are interested with how much energy is given off at constant pressure

19. V. Enthalpy The signs of  H and  E are similar in meaning

20. Example If we burn 1 mole of fuel and constant pressure it produces 3452 kJ of heat and does 11 kJ or work. What are the values of  H and  E?

21. A. Stoichiometry involving  H Enthalpy for chemical reactions =  H rxn For the reaction below, if you start with 13.2 kg of C 3 H 8 what is q in kJ? C 3 H 8 (l) + 5O 2 (g)  3CO 2 (g) + 4H 2 O (l)  H rxn = -2044 kJ

22. VI. Measuring  H rxn Coffee cup calorimetry is a way to measure  H rxn. Knowing the mass of solution, the heat evolved will cause a  T. Combing that with C s of solution you can determine q.

23. Example We want to know the  H rxn of the following reaction: Ca(s) + 2HCl (aq)  CaCl 2 (aq) + H 2 (g) Given 0.204 g of Ca(s) in 100.0 mL of HCl (aq), the temperature rose from 24.8 ° C to 33.9 °C. The density of the solution is 1.00 g/mL and the C s,soln is 4.18 J/g  ° C

24. 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 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 IV. Thermochemical Equations H 2 O (s) H 2 O (l)  H = 6.01 kJ The physical states of all reactants and products must be specified in thermochemical equations. H 2 O (l) H 2 O (g)  H = 44.0 kJ

25. 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 The final reaction for formation of CS 2 is: C (graphite) + 2S (rhombic) CS 2 (l)

26. 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. f The standard enthalpy of formation of any element in its most stable form is zero.  H f 0 (O 2 ) = 0  H f 0 (O 3 ) = 142 kJ/mol  H 0 (C, graphite) = 0 f  H f 0 (C, diamond) = 1.90 kJ/mol V. Enthalpy of formation

27. VII. Standard Heats of Formation Example Write an equation for the formation of C 6 H 12 O 6 from elements. The  H f 0 for C 6 H 12 O 6 is -1273.3 kJ/mole

28. B. Calculating  H f 0 for a reaction To calculate  H rxn 0 subtract the heats of formations of the reactant multiplied by their stoichiometric coefficient from heats of formation of the products multiplied by their stoichiometric coefficents.

29. The standard enthalpy of reaction (  H 0 rxn ) is the enthalpy of a reaction carried out at 1 atm. 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. A. Hess’s Law

30. 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) Example Compound  H f 0 (kJ/mole) C 6 H 6 (l)+49.04 O 2 (g)0 CO 2 (g)-393.5 H 2 O (l)-187.6