1. Chapter 6 Thermochemistry

2. T HERMODYNAMICS 6 | 2 Thermodynamics The science of the relationship between heat and other forms of energy علم العلاقة بين االحرارة واشكال الطاقة الاخرى Thermochemistry An area of thermodynamics that concerns the study of the heat absorbed or evolved by a chemical reaction الكيمياء الحرارية هي مجال من مجالات الحرارة الديناميكية التي تهتم بامتصاص الحرارة او انتاجها خلال التفاعل الكيميائي

3. E NERGY 6 | 3 Energy The potential or capacity to move matter, also the capacity to do work. الطاقة هي القدرة على تحريك الاشياء او القيام بعمل ما. One form of energy can be converted to another form of energy: electromagnetic, mechanical, electrical, or chemical. يمكن تحول الطاقة من شكل لاخر مثل الطاقة المغناطيسية، الميكانيكية، طاقة الوضع، والطاقة الداخلية.

4. K INETIC E NERGY 6 | 4 Kinetic Energy, E K The energy associated with an object by virtue of its motion. الطاقة الحركية هي تلك المتعلقة بتأثير حركة الاجسام m = mass (kg) v = velocity (m/s)

5. E NERGY U NITS 6 | 5 The SI unit of energy is the joule, J, pronounced “jewel.” The calorie is a non-SI unit of energy commonly used by chemists. It was originally defined as the amount of energy required to raise the temperature of one gram of water by one degree Celsius. The exact definition is given by the equation: الكالوري هي كمية الطاقة اللازمة لرفع درجة حرارة واحد غرام من الماء درجة مئوية واحدة

6. E XAMPLE 1 6 | 6 A person weighing 75.0 kg (165 lbs) runs at 1.78 m/s (4.00 mph). What is the person’s kinetic energy? m = 75.0 kg V = 1.78 m/s E K = ½ mv 2

7. P OTENTIAL E NERGY 6 | 7 Potential Energy, E P The energy an object has by virtue of its position in a field of force, such as gravitaitonal, electric or magnetic field. طاقة الوضع او الكامنة هي طاقة التي يتاثر بها الجسم بتغيير وضعه في مجال من القوة مثل الجاذبية الارضية او المجال الكهربائي او المغناطيسي. Gravitational potential energy is given by the equation m = mass (kg) g = gravitational constant (9.80 m/s 2 ) h = height (m)

8. I NTERNAL E NERGY 6 | 8 Internal Energy, U The sum of the kinetic and potential energies of the particles making up a substance الطاقة الداخلية هي مجموع الطاقات الكامنة والحركية للجسيمات المكونة لمادة ما. Total Energy U = E K + E P

9. E NERGY C ONSERVATION 6 | 9 Law of Conservation of Energy Energy may be converted from one form to another, but the total quantity of energy remains constant. The total amount of energy in the universe is constant. قانون حفظ الطاقة، من الممكن ان تتحول الطاقة من شكل لاخر ولكن مجموع كميات الطاقة قبل وبعد التحويل تبقى ثابتة.

10. S YSTEM AND S URROUNDINGS 6 | 10 Thermodynamic System The substance under study in which a change occurs is called the thermodynamic system (or just system). هو النظام الذي تخضع فيه المادة للتغيير Thermodynamic Surroundings Everything else in the vicinity, including the solvent, is called the thermodynamic surroundings (or just the surroundings). هو كل شيء مجاور للمادة المذكورة اعلاه بما فيه المذيب

11. H EAT 6 | 11 Heat, q The energy that is transferred into or out of a system because of a difference in temperature between the thermodynamic system and its surroundings Heat in a system or its surroundings is manifested as particle motion. Heat flows spontaneously from a region of higher temperature to a region of lower temperature. q is defined as positive if heat is absorbed by the system (heat is added to the system) q is defined as negative if heat is evolved by a system (heat is subtracted from the system)

12. H EAT OF R EACTION 6 | 12 Heat of Reaction The value of heat ( q) produced or absorbed by a reaction system that is usually transferred to or from the system’s surroundings

13. E NDOTHERMIC AND E XOTHERMIC 6 | 13 Endothermic Process A chemical reaction or process in which heat is absorbed by the system from the surroundings ( q is positive). The reaction vessel (surroundings) will feel cool. التفاعل او العملية التي يمتص فيها النظام الحرارة من المحيط وفيها تكون القيمة ايجابية، بحيث يصبح محيط التفاعل باردا. Exothermic Process A chemical reaction or process in which heat is evolved by the system and transferred to the surroundings ( q is negative). The reaction vessel (surroundings) will feel warm. التفاعل او العملية التي يصدر عنها حرارة الى محيط التفاعل بحيث تكون القيمة سالبة والمحيط ساخنا.

14. E NDOTHERMIC AND E XOTHERMIC In an endothermic reaction: The reaction vessel cools. تكون اوعية التفاعل باردة Heat is absorbed. الحرارة تمتص Energy is added to the system. تضاف الطاقة الى النظام q is positive. قيمة موجبة In an exothermic reaction: The reaction vessel warms. اوعية التفاعل دافئة Heat is evolved. تصدر الحرارة منه Energy is subtracted from the system. تطرح الطاقة خارج النظام q is negative. قيمة سالبة 6 | 14

15. E NTHALPY 6 | 15 Enthalpy, H An extensive property of a substance that can be used to obtain the heat absorbed or evolved in a chemical reaction. Enthalpy is a state function (defined below) that is defined as the heat change for a process measured under constant pressure. خاصية واسعة النطاق للمادة التي يمكن استخدامها للحصول على مقدار الحرارة التي امتصت او انتجت في التفاعل الكيميائي، والمحتوى الحراري عندها هو الدالة الوظيفية ( معرفة في الاسفل ) والتي تعرف بتغير الحرارة في عملية كيميائية تقاس تحت ضغط ثابت. Extensive Property A property that depends on the amount of substance. Mass and volume are extensive properties. الخاصية واسعة النطاق هي الخاصية التي تعتمد على كمية المادة. الكتلة والحجم هي خواص واسعة.

16. S TATE F UNCTIONS 6 | 16 A state function is a property of a system that depends only on its present state, which is determined by variables such as temperature and pressure, and is independent of any previous history of the system. State functions are represented by capitalized single letter symbols الدالة الوظيفية هي خاصية من خواص النظام تعتمد فقط على وضعه الحالي الموجود فيه، ويمكن معرفته من خلال العديد من المتغيرات كالحرارة والضغط، وهي مستقلة ومختلفة تماما عن الحالة السابقة للنظام قبل التفاعل او العمليات الكيميائية عليه. وتمثل الدالة الوظيفية بحرف كبير مفرد.

17. E NTHALPY OF R EACTION 6 | 17 Enthalpy of Reaction The change in enthalpy for a reaction at a given temperature and pressure  H = H (products) – H (reactants) Note:  means “change in.” Enthalpy change is equal to the heat of reaction at constant pressure:  H = q P

18. E NTHALPY OF R EACTION The diagram illustrates the enthalpy change for the reaction 2Na( s ) + 2H 2 O( l )  2NaOH( aq ) + H 2 ( g ) 6 | 18 The reactants are at the top. The products are at the bottom. The products have less enthalpy than the reactants, so enthalpy is evolved as heat (exothermic). The signs of both q and  H are negative.

19. I NTERNAL E NERGY C HANGE 6 | 19 The change in internal energy ( U f – U i =  U), for a system can be defined as follows  U = q + w Where q = heat change for a process in a system And w = work done by the system on its surroundings or by the surroundings on the system To use this relationship you must assign proper signs to the q and w values لاستخدام هذه المعادلة يجب تعيين اشارة مناسبة للحرارة ومقدار الجهد من قبل النظام على المحيط

20. E NTHALPY AND I NTERNAL E NERGY A common form of work observed in a chemical system is P  V, sometimes called pressure-volume work, where a system expands or contracts under constant pressure in the surroundings. اذن ما هو الشكل المعروف من الجهد للنظام خلال التفاعل؟ هو الضغط مضروب في فرق الحجم، نسميه في بعض الاحيان جهد الضغط والحجم والذي فيه يتمدد او يتقلص النظام تحت ضغط محيطي ثابت. Using this as a state function, along with enthalpy change for a heat change, the internal energy change for a process can be calculated, using only state functions, as:  U =  H – P  V Often the amount of work exchanged with the surroundings is so small, it can be ignored. 6 | 20

21. PV W ORK 6 | 21 Copyright © Houghton Mifflin Company. All rights reserved. For the reaction 2Na( s ) + 2H 2 O( l )  2NaOH( aq ) + H 2 ( g ) –PV–PV The H 2 gas had to do work to raise the piston. For the reaction as written at 1 atm, -P  V = -2.5 kJ. In addition, 368.6 kJ of heat are evolved.

22. T HERMOCHEMICAL E QUATIONS 6 | 22 Thermochemical Equation The thermochemical equation is the chemical equation for a reaction (including phase labels) in which the equation is given a molar interpretation, and the enthalpy of reaction for these molar amounts is written directly after the equation هو معادلة التفاعل الحراري والتي فيها توصف المولات المشاركة في التفاعل والمحتوى الحراري لتلك الكمية من المولات بحيث تكتب مباشرة بعد التفاعل. For the reaction of sodium metal with water, the thermochemical equation is 2Na( s ) + 2H 2 O( l )  2NaOH( aq ) + H 2 ( g );  H = – 368.6 kJ

23. E XAMPLE 2 6 | 23 Sulfur, S 8, burns in air to produce sulfur dioxide. The reaction evolves 9.31 kJ of heat per gram of sulfur at constant pressure. Write the thermochemical equation for this reaction.

24. E XAMPLE 2 (C ONT ) 6 | 24 S 8 (s) + 8O 2 (g)  8SO 2 (g) We first write the balanced chemical equation: Next, we convert the heat per gram to heat per mole. Now we can write the thermochemical equation: S 8 (s) + 8O 2 (g)  8SO 2 (g);  H = – 2.39 × 10 3 kJ Note: The negative sign indicates that heat is evolved; the reaction is exothermic.

25. T HERMOCHEMICAL E QUATIONS 6 | 25 Manipulating a Thermochemical Equation When the equation is multiplied by a factor, the value of  H must be multiplied by the same factor. When a chemical equation is reversed, the sign of  H is reversed. اذا ضربت المعادلة بعامل يضرب المحتوى الحراري بنفس العامل، واذا اعيد التفاعل بشكل عكسي تعكس اشارة المحتوى الحراري ايضا.

26. E XAMPLE 3 6 | 26

27. E XAMPLE 3 (C ONT ) 6 | 27 c.CO( g ) + 3H 2 ( g )  CH 4 ( g ) + H 2 O( g ) This reaction is the reverse of the given reaction. It is exothermic.  H = -206 kJ CH 4 (g) + H 2 O(g)  CO(g) + 3H 2 (g);  H = 206 kJ

28. E XAMPLE 3 (C ONT ) 6 | 28 d.2CO( g ) + 6H 2 ( g )  2CH 4 ( g ) + 2H 2 O( g ) This reaction is reverse and double the given reaction. It is exothermic.  H = -412 kJ Equations c and d are exothermic. Equation d is the most exothermic reaction. CH 4 (g) + H 2 O(g)  CO(g) + 3H 2 (g);  H = 206 kJ

29. E XAMPLE 4 6 | 29 When sulfur burns in air, the following reaction occurs: S 8 ( s ) + 8O 2 ( g )  8SO 2 ( g );  H o = – 2.39 x 10 3 kJ Write the thermochemical equation for the dissociation of one mole of sulfur dioxide into its elements.

30. E XAMPLE 4 (C ONT ) 6 | 30 S 8 ( s ) + 8O 2 ( g )  8SO 2 ( g );  H = – 2.39 × 10 3 kJ We want SO 2 as a reactant, so we reverse the given reaction, changing the sign of  H : 8SO 2 ( g )  S 8 ( g ) + 8O 2 ( g ) ;  H = +2.39 × 10 3 kJ We want only one mole SO 2, so now we divide every coefficient and  H by 8: SO 2 (g)  1 / 8 S 8 (g) + O 2 (g) ;  H = +299 kJ

31. S TOICHIOMETRY AND E NTHALPY 6 | 31 Applying Stoichiometry to Heats of Reaction

32. E XAMPLE 5 6 | 32 You burn 15.0 g sulfur in air. How much heat evolves from combusting this amount of sulfur? The thermochemical equation is: S 8 (s) + 8O 2 (g)  8SO 2 (g);  H = -2.39 x 10 3 kJ

33. E XAMPLE 5 (C ONT ) 6 | 33 S 8 ( s ) + 8O 2 ( g )  8SO 2 ( g );  H = -2.39 x 10 3 kJ Molar mass of S 8 = 256.52 g q = – 1.40 × 10 2 kJ

34. E XAMPLE 6 6 | 34 The daily energy requirement for a 20-year-old man weighing 67 kg is 1.3 x 10 4 kJ. For a 20-year-old woman weighing 58 kg, the daily requirement is 8.8 x 10 3 kJ. If all this energy were to be provided by the combustion of glucose, C 6 H 12 O 6, how many grams of glucose would have to be consumed by the man and the woman per day? C 6 H 12 O 6 (s) + 6O 2 (g)  6CO 2 (g) + 6H 2 O(l);  H o = -2.82 x 10 3 kJ

35. E XAMPLE 5 (C ONT ) 6 | 35 C 6 H 12 O 6 ( s ) + 6O 2 ( g )  6CO 2 ( g ) + 6H 2 O( l );  H = -2.82 x 10 3 kJ For a 20-year-old man weighing 67 kg: = 830 g glucose required (2 significant figures)

36. E XAMPLE 5 (C ONT ) 6 | 36 For a 20-year-old woman weighing 58 kg: = 560 g glucose required (2 significant figures)

37. H EATS OF R EACTION 6 | 37 Measuring Heats of Reaction We will first look at the heat needed to raise the temperature of a substance because this is the basis of our measurements of heats of reaction.

38. H EAT C APACITY AND S PECIFIC H EAT 6 | 38 Heat Capacity, C, of a Sample of Substance The quantity of heat needed to raise the temperature of the sample of substance by one degree Celsius (or one kelvin) Molar Heat Capacity The heat capacity for one mole of substance Specific Heat Capacity, c (or specific heat ) The quantity of heat needed to raise the temperature of one gram of substance by one degree Celsius (or one kelvin) at constant pressure

39. H EAT T RANSFER C ALCULATIONS 6 | 39 The heat required can be found by using the following equations. Using heat capacity: q = C  T Using specific heat capacity: q = cm  T Where m is the mass of sample under study and  T is the temperature change observed

40. C ALORIMETRY 6 | 40 A calorimeter is a device used to measure the heat absorbed or evolved during a physical or chemical change. Two examples are shown below.

41. E XAMPLE 6 6 | 41 A piece of zinc weighing 35.8 g was heated from 20.00°C to 28.00°C. How much heat was required? The specific heat of zinc is 0.388 J/(g°C). m = 35.8 g c = 0.388 J/(g°C)  T = 28.00°C – 20.00°C = 8.00°C q = cm  T q = 111 J (3 significant figures)

42. E XAMPLE 7 6 | 42 Nitromethane, CH 3 NO 2, an organic solvent burns in oxygen according to the following reaction: CH 3 NO 2 (g) + 3 / 4 O 2 (g)  CO 2 (g) + 3 / 2 H 2 O(l) + 1 / 2 N 2 (g) You place 1.724 g of nitromethane in a calorimeter with oxygen and ignite it. The temperature of the calorimeter increases from 22.23°C to 28.81°C. The heat capacity of the calorimeter was determined to be 3.044 kJ/°C. Write the thermochemical equation for the reaction.

43. E XAMPLE 7 (C ONT ) 6 | 43 We first find the heat evolved for the 1.724 g of nitromethane, CH 3 NO 2. Now, covert that to the heat evolved per mole by using the molar mass of nitromethane, 61.04 g.  H = – 709 kJ/mol

44. E XAMPLE 7 (C ONT ) 6 | 44 We can now write the thermochemical equation: CH 3 NO 2 (l) + ¾O 2 (g)  CO 2 (g) + 3 / 2 H 2 O(l) + ½N 2 (g);  H = – 709 kJ/mol

45. H ESS ’ S L AW 6 | 45 Hess’s Law of Heat Summation For a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation equals the sum of the enthalpy changes for the individual steps.

46. E XAMPLE 8 6 | 46 Suppose we want  H for the reaction 2C( graphite ) + O 2 ( g )  2CO( g ) It is difficult to measure directly. However, two other reactions are known: 2CO 2 ( g )  2CO( g ) + O 2 ( g );  H = – 566.0 kJ C(graphite) + O 2 (g)  CO 2 (g);  H = -393.5 kJ In order for these to add to give the reaction we want, we must multiply the first reaction by 2. Note that we also multiply  H by 2.

47. E XAMPLE 8 (C ONT ) 6 | 47 2C( graphite ) + O 2 ( g )  2CO( g ) 2C( graphite ) + 2O 2 ( g )  2CO 2 ( g );  H = -787.0 kJ 2CO 2 ( g )  2CO( g ) + O 2 ( g );  H = – 566.0 kJ Now we can add the reactions and the  H values. 2 C(graphite) + O 2 (g)  2 CO(g);  H = –1353.0 kJ

48. E XAMPLE 9 6 | 48  H sub =  H fus +  H vap

49. E XAMPLE 10 6 | 49 What is the enthalpy of reaction,  H, for the reaction of calcium metal with water? Ca(s) + 2H 2 O(l)  Ca 2+ (aq) + 2OH - (aq) + H 2 (g) This reaction occurs very slowly, so it is impractical to measure  H directly. However, the following facts are known: H + (aq) + OH - (aq)  H 2 O(l);  H = –55.9 kJ Ca(s) + 2H + (aq)  Ca 2+ (aq) + H 2 (g);  H = –543.0 k J

50. E XAMPLE 10 (C ONT ) 6 | 50 Ca(s) + 2H 2 O(l)  Ca 2+ (aq) + 2OH - (aq) + H 2 (g) First, identify each reactant and product: H + (aq) + OH - (aq)  H 2 O(l);  H = – 55.9 kJ Ca(s) + 2H + (aq)  Ca 2+ (aq) + H 2 (g);  H = – 543.0 kJ Each substance must be on the proper side. Ca(s), Ca 2+ (aq), and H 2 (g) are fine. H 2 O(l) should be a reactant. OH - (aq) should be a product. Reversing the first reaction and changing the sign of its  H accomplishes this.

51. E XAMPLE 10 (C ONT ) 6 | 51 Ca(s) + 2H 2 O(l)  Ca 2+ (aq) + 2OH - (aq) + H 2 (g) H 2 O(l)  H + (aq) + OH - (aq);  H = + 55.9 kJ Ca(s) + 2H + (aq)  Ca 2+ (aq) + H 2 (g);  H = – 543.0 kJ The coefficients must match those in the target equation. The coefficient on H 2 O and OH - should be 2. We multiply the first reaction and its  H by 2 to accomplish this.

52. E XAMPLE 10 (C ONT ) 6 | 52 Ca(s) + 2H 2 O(l)  Ca 2+ (aq) + 2OH - (aq) + H 2 (g) 2H 2 O(l)  2H + (aq) + 2OH - (aq);  H = + 111.8 kJ Ca(s) + 2H + (aq)  Ca 2+ (aq) + H 2 (g);  H = – 543.0 kJ We can now add the equations and their  H’s. Note that 2H + (aq) appears as both a reactant and a product.

53. E XAMPLE 10 (C ONT ) 6 | 53 Ca(s) + 2H 2 O(l)  Ca 2+ (aq) + 2OH - (aq) + H 2 (g) 2H 2 O(l)  2H + (aq) + 2OH - (aq);  H = + 111.8 kJ Ca(s) + 2H + (aq)  Ca 2+ (aq) + H 2 (g);  H = – 543.0 kJ Ca(s) + 2H 2 O(l)  Ca 2+ (aq) + 2OH - (aq) + H 2 (g);  H = – 431.2 kJ

54. E NTHALPIES OF F ORMATION 6 | 54 Standard Enthalpies of Formation The term standard state refers to the standard thermodynamic conditions chosen for substances when listing or comparing thermodynamic data: 1 atm pressure and the specified temperature (usually 25°C). These standard conditions are indicated with a degree sign (°). When reactants in their standard states yield products in their standard states, the enthalpy of reaction is called the standard enthalpy of reaction,  H °. (  H ° is read “delta H zero.”)

55. E NTHALPIES OF F ORMATION 6 | 55 Elements can exist in more than one physical state, and some elements exist in more than one distinct form in the same physical state. For example, carbon can exist as graphite or as diamond; oxygen can exist as O 2 or as O 3 (ozone). These different forms of an element in the same physical state are called allotropes. The reference form is the most stable form of the element (both physical state and allotrope).

56. E NTHALPIES OF F ORMATION 6 | 56 The standard enthalpy of formation,  H f °, is the enthalpy change for the formation of one mole of the substance from its elements in their reference forms and in their standard states.  H f ° for an element in its reference and standard state is zero. For example, the standard enthalpy of formation for liquid water is the enthalpy change for the reaction H 2 ( g ) + 1 / 2 O 2 ( g )  H 2 O( l )  H f ° = –285.8 kJ Other  H f ° values are given in Table 6.2 and Appendix C.

57. E XAMPLE 11 6 | 57 C( graphite ) + 2Cl 2 ( g )  CCl 4 ( l );  H f ° = –135.4 kJ We first identify each reactant and product from the target equation. Standard enthalpies of formation can be used to calculate the standard enthalpy for a reaction,  H°. CH 4 (g) + 4Cl 2 (g)  CCl 4 (l) + 4HCl(g);  H° = ? Table 6.2 shows the  H f ° values: 1 / 2 H 2 (g) + 1 / 2 Cl 2 (g)  HCl(g);  H f ° = -92.3 kJ CH 4 (g)  C(graphite) + 2H 2 (g);  H f ° = +74.9 kJ

58. E XAMPLE 11 (C ONT ) 6 | 58 C( graphite ) + 2Cl 2 ( g )  CCl 4 ( l );  H f ° = –135.4 kJ Each needs to be on the correct side of the arrow and is. Next, we’ll check coefficients. Target equation: CH 4 (g) + 4Cl 2 (g)  CCl 4 (l) + 4HCl(g);  H° = ? Table 6.2 shows the  H f ° values for contributing formation equations: 1 / 2 H 2 (g) + 1 / 2 Cl 2 (g)  HCl(g);  H f ° = -92.3 kJ CH 4 (g)  C(graphite) + 2H 2 (g);  H f ° = +74.9 kJ

59. E XAMPLE 11 (C ONT ) 6 | 59 C( graphite ) + 2Cl 2 ( g )  CCl 4 ( l );  H f ° = –135.4 kJ Cl 2 and HCl need a coefficient of 4. Multiplying the second equation and its  H by 4 does this. Target equation: CH 4 (g) + 4Cl 2 (g)  CCl 4 (l) + 4HCl(g);  H° = ? Contributing  H f ° values from Table 6.2: 1 / 2 H 2 (g) + 1 / 2 Cl 2 (g)  HCl(g);  H f ° = -92.3 kJ CH 4 (g)  C(graphite) + 2H 2 (g);  H f ° = +74.9 kJ

60. E XAMPLE 11 (C ONT ) 6 | 60 C( graphite ) + 2Cl 2 ( g )  CCl 4 ( l );  H f ° = –135.4 kJ Now, we can add the equations. Target Equation: CH 4 (g) + 4Cl 2 (g)  CCl 4 (l) + 4HCl(g);  H° = ? Contributing equations and values from Table 6.2: 2H 2 (g) + 2Cl 2 (g)  4HCl(g);  H f ° = -369.2 kJ CH 4 (g)  C(graphite) + 2H 2 (g);  H f ° = +74.9 kJ

61. E XAMPLE 11 (C ONT ) 6 | 61 C( graphite ) + 2Cl 2 ( g )  CCl 4 ( l );  H f ° = –135.4 kJ Target equation. CH 4 (g) + 4Cl 2 (g)  CCl 4 (l) + 4HCl(g);  H° = ? Contributing equations and values from Table 6.2: 2H 2 (g) + 2Cl 2 (g)  4HCl(g);  H f ° = -369.2 kJ CH 4 (g) + 4Cl 2 (g)  CCl 4 (g) + 4HCl(g);  H° = –429.7 kJ CH 4 (g)  C(graphite) + 2H 2 (g);  H f ° = +74.9 kJ

62. E XAMPLE 12 6 | 62 What is the heat of vaporization of methanol, CH 3 OH, at 25°C and 1 atm? Use standard enthalpies of formation (Appendix C). We have demonstrated the use of contributing equations and enthalpy values in calculating the standard enthalpy change for a reaction. We will now consider a shortcut method, assuming that combining equations is always a valid approach. This method works, not only for simple problems like this one, but for all problems untilizing standard enthalpies of formation

63. E XAMPLE 12 (C ONT ) 6 | 63 We want  H° for the reaction: CH 3 OH(l)  CH 3 OH(g)  H vap = +38.0 kJ

64. E XAMPLE 13 6 | 64 Methyl alcohol, CH 3 OH, is toxic because liver enzymes oxidize it to formaldehyde, HCHO, which can coagulate protein. Calculate  H o for the following reaction: 2CH 3 OH(aq) + O 2 (g)  2HCHO(aq) + 2H 2 O(l) Standard enthalpies of formation, : CH 3 OH(aq):-245.9 kJ/mol HCHO(aq):-150.2 kJ/mol H 2 O(l):-285.8 kJ/mol

65. E XAMPLE 13 (C ONT ) 6 | 65 We want  H° for the reaction: 2CH 3 OH(aq) + O 2 (aq)  2HCHO(aq) + 2H 2 O(l)

66. E NERGY AND M ETABOLISM 6 | 66 Foods fuels three needs of the body: They supply substances for the growth and repair of tissue. They supply substances for the synthesis of compounds used in the regulation processes. They supply energy. Foods do this by a combustion process.

67. E NERGY AND M ETABOLISM 6 | 67 For glucose, a carbohydrate: C 6 H 12 O 6 ( s ) + 6O 2 ( g )  6CO 2 ( g ) + 6H 2 O( l );  H ° = –2803 kJ For glycerol trimyristate, a fat: C 45 H 86 O 6 ( s ) + 127 / 2 O 2 ( g )  45CO 2 ( g ) + 43H 2 O( l );  H °= –27,820 kJ The average value for carbohydrates is 4.0 kcal/g and for fats is 9.0 kcal/g.

68. E NERGY P RODUCTION 6 | 68 Fossil fuels originated millions of years ago when aquatic plants and animals were buried and compressed by layers of sediment at the bottoms of swamps and seas. Over time this organic matter was converted by bacterial decay and pressure to petroleum (oil), gas, and coal.

69. E NERGY P RODUCTION Coal, which accounts for 22.9% of total U.S. energy consumption, varies in terms of the amount of carbon it contains and so varies in terms of the amount of energy it produces in combustion. Anthracite (hard coal) was laid down as long as 250 million years ago and can contain as much as 80% carbon. Bituminous coal, a younger variety, contains between 45% and 65% carbon. 6 | 69

70. E NERGY P RODUCTION Natural gas, which accounts for 22.7% of total U.S. energy consumption, is convenient because it is fluid and can be easily transported. Purified natural gas is primarily methane, CH 4, plus small amounts of ethane, C 2 H 6 ; propane, C 3 H 8 ; and butane, C 4 H 10. Petroleum is a mixture of compounds. Gasoline, which is obtained from petroleum, is a mixture of hydrocarbons (compounds of carbon and hydrogen). 6 | 70

71. E NERGY P RODUCTION The main issue with natural gas and petroleum is their relatively short supply. It has been estimated that petroleum deposits will be 80% depleted by 2030. Natural gas deposits may be depleted even sooner. Coal deposits, however, are expected to last for several more centuries. This has led to the development of commercial methods for converting coal to the more easily handled liquid and gaseous fuels. 6 | 71

72. E NERGY P RODUCTION Coal gasification is one way. Steam is passed over hot coal: C( s ) + H 2 O( g )  CO( g ) + H 2 ( g ) The mixture containing carbon monoxide can be converted by various methods into useful products. The mixture of carbon monoxide and hydrogen can be converted by various methods into useful products such as methane, CH 4. CO( g ) + 3H 2 ( g )  CH 4 ( g ) + H 2 O( g ) 6 | 72

73. E NERGY P RODUCTION Rockets are self-contained missiles propelled by the ejection of gases from an orifice. Usually these are hot gases expelled from the rocket from the reaction of a fuel with an oxidizer. One factor in determining the appropriate fuel/oxidizer combination is its mass of the mixture. 6 | 73