1. Energy Relationships in Chemical Reactions Chapter 6 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. 2 Energy is the capacity to do work. Radiant energy comes from the sun and is earth’s primary energy source Thermal energy is the energy associated with the random motion of atoms and molecules Chemical energy is the energy stored within the bonds of chemical substances Nuclear energy is the energy stored within the collection of neutrons and protons in the atom Potential energy is the energy available by virtue of an object’s position

3. 3 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

4. 4 Thermochemistry is the study of heat change in chemical reactions. The system is the specific part of the universe that is of interest in the study. open mass & energyExchange: closed energy isolated nothing

5. 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. 6 Thermodynamics is the scientific study of the interconversion of heat and other kinds of energy. State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. energy, pressure, volume, temperature  U = U final - U initial  P = P final - P initial  V = V final - V initial  T = T final - T initial

7. 7 First law of thermodynamics – energy can be converted from one form to another, but cannot be created or destroyed.  U system +  U surroundings = 0 or  U system = -  U 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

8. 8 Another form of the first law for  U system  U = q + w  U 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 w = -P  V when a gas expands against a constant external pressure

9. 9 A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. What is the work done in joules if the gas expands (a) against a vacuum and (b) against a constant pressure of 3.7 atm? w = -P  V (a)  V = 5.4 L – 1.6 L = 3.8 L P = 0 atm W = -0 atm x 3.8 L = 0 Latm = 0 joules (b)  V = 5.4 L – 1.6 L = 3.8 L P = 3.7 atm w = -3.7 atm x 3.8 L = -14.1 Latm w = -14.1 Latm x 101.3 J 1Latm = -1430 J

10. 10 Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.  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

11. 11 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.

12. 12 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.

13. 13 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 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

14. 14 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

15. Hess’s Law The total enthalpy of a reaction is independent of the reaction pathway. This means that if a reaction is carried out in a series of steps, the enthalpy change (  H) for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps. This idea is also known as Hess’s law. Here are some rules for using Hess’s law in solving problems. –Make sure to rearrange the given equations so that reactants and products are on the appropriate sides of the arrows. –If you reverse equations, you must also reverse the sign of  H. –If you multiply equations to obtain a correct coefficient, you must also multiply the  H by this coefficient. 15

16. 16 C (graphite) + 1/2O 2 (g) CO (g) CO (g) + 1/2O 2 (g) CO 2 (g) C (graphite) + O 2 (g) CO 2 (g) The total enthalpy of a reaction is independent of the reaction pathway. This means that if a reaction is carried out in a series of steps, the enthalpy change (  H) for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.

17. 17 Calculate the standard enthalpy of formation of CS 2 (l) given that: C (graphite) + O 2 (g) CO 2 (g)  H ° = -393.5 kJ/mol rxn S (rhombic) + O 2 (g) SO 2 (g)  H ° = -296.1 kJ/mol rxn CS 2 (l) + 3O 2 (g) CO 2 (g) + 2SO 2 (g)  H = -1072 kJ/mol 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 ° = -393.5 kJ/mol 2S (rhombic) + 2O 2 (g) 2SO 2 (g)  H ° = -296.1 kJ/mol x 2 rxn CO 2 (g) + 2SO 2 (g) CS 2 (l) + 3O 2 (g)  H ° = +1072 kJ/mol rxn + C (graphite) + 2S (rhombic) CS 2 (l)  H ° = -393.5 + (2x-296.1) + 1072 = 86.3 kJ/mol rxn

18. 18 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 ° ) as a reference point for all enthalpy expressions. f Standard enthalpy of formation (  H ° ) 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 ° (O 2 ) = 0 f  H ° (O 3 ) = 142 kJ/mol f  H ° (C, graphite) = 0 f  H ° (C, diamond) = 1.90 kJ/mol f

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20. 20 The standard enthalpy of reaction (  H ° ) is the enthalpy of a reaction carried out at 1 atm. rxn aA + bB cC + dD H°H° rxn d  H ° (D) f c  H ° (C) f = [+] - b  H ° (B) f a  H ° (A) f [+] H°H° rxn n  H ° (products) f =  m  H ° (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.)

21. 21 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) H°H° rxn n  H ° (products) f =  m  H ° (reactants) f  - H°H° rxn 6  H ° (H 2 O) f 12  H ° (CO 2 ) f = [+] - 2  H ° (C 6 H 6 ) f [] H°H° rxn = [ 12(–393.5) + 6(–285.8) ] – [ 2(49.04) ] = ? ? 2 mol = ?

22. Calorimetry

23. 23 Heat Transfer, q Heat (q) = the transfer of energy which causes the temperature of an object to change units: joules (j), calories (cal) –A calorie is the amount of energy needed to raise the temperature of 1.00 g water by 1°C. 1cal = 4.184 joules Heat spontaneously moves regions of high temperature to regions of lower temperature. A metal spoon at 25°C is placed in boiling water. What happens?

24. 24 The Sign Convention of q Endothermic systems require the surroundings to add energy to the system. q is positive (+) Exothermic reactions release energy to the surroundings. q is negative (-) Energy changes are measured from the point of view of the system

25. 25 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. C = m x s Heat (q) absorbed or released: q = m x s x  t q = C x  t  t = t final - t initial The heat capacity** (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. **Heat Capacity (C) is sometimes called the calorimeter constant

26. 26 How much heat is given off when an 869 g iron bar cools from 94 o C to 5 o C? s of Fe = 0.444 J/g o C  t = t final – t initial = 5 o C – 94 o C = -89 o C q = ms  t = 869 g x 0.444 J/g o C x –89 o C= -34,000 J

27. 27 Calorimetry Calorimeter: a closed container used to measure temperature changes in physical and chemical processes. From the temperature changes we can calculate the heat of the reaction, q –q v ; heat measured under constant volume conditions –q p : heat measured under constant pressure conditions q = m x s x  t

28. 28 Two of the most common types of calorimeters are the coffee cup calorimeter and the bomb calorimeter.

29. 29 Coffee Cup Calorimeter The open system allows the pressure to remain constant. Thus we measure q p = ΔH, the enthalpy change. ΔH =change in heat at constant pressure.

30. 30 The 1 st Law of Thermodynamics Keeps Track of Heat Transfer If we monitor the heat transfers (q) of all materials involved, we can predict that their sum will be zero. By monitoring the surroundings, we can predict what is happening to our system. Heat transfers until thermal equilibrium, thus the final temperature is the same for all materials.

31. 31 Example of Using 1 st Law for Heat Transfer A 43.29 g sample of solid is transferred from boiling water (T=99.8°C) to 152 g water at 22.5°C in a Styrofoam coffee cup calorimeter. The T water increased to 24.3°C. Calculate the specific heat of the solid assuming no heat was lost or gained by the calorimeter cup. q sample + q water + q cup = 0 q cup is neglected in problem = 0 Hence q sample = - q water

32. 32 q sample = - q water

33. 33 Chemistry in Action:

34. 34 Chemistry in Action: Fuel Values of Foods and Other Substances 1 cal = 4.184 J 1 Cal = 1000 cal = 4184 J Substance  H combustion (kJ/g) Apple-2 Beef-8 Beer-1.5 Gasoline-34

35. A metal alloy was heated for 2 minutes in a hot water bath at 101.20  C before being transferred into an aluminum calorimeter cup containing water at 23.42 °C. A rise in the temperature of the water in the calorimeter cup was observed. In fact the maximum temperature reached was 28.25 °C. Using the data below, calculate the specific heat of the metal object in cal/g°C. Note that the specific heats of water and aluminum are 1.00 cal/g  C and 0.22 cal/g  C respectively. Wt. Alloy 243.42 g Wt. water and calorimeter cup 213.14 g Wt. Calorimeter cup 45.37 g Initial temp. of water 23.42  C Initial temp. of calorimeter cup 23.42 °C Initial temp. of metal alloy 101.20 °C Final temp. of water 28.25  C 35

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