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Chapter 6 Chemical Energy. Chapter 6 Table of Contents Copyright © Cengage Learning. All rights reserved 2 6.1 The Nature of Chemical Energy 6.2 Learning.

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Presentation on theme: "Chapter 6 Chemical Energy. Chapter 6 Table of Contents Copyright © Cengage Learning. All rights reserved 2 6.1 The Nature of Chemical Energy 6.2 Learning."— Presentation transcript:

1 Chapter 6 Chemical Energy

2 Chapter 6 Table of Contents Copyright © Cengage Learning. All rights reserved 2 6.1 The Nature of Chemical Energy 6.2 Learning to Solve Problems 6.3 Enthalpy 6.4 Hess’s Law 6.5 Standard Enthalpies of Formation 6.6Present Sources of Energy 6.7New Energy Sources

3 Section 6.1 The Nature of Chemical Energy Return to TOC Copyright © Cengage Learning. All rights reserved 3 The capacity to do work or to product heat.  kinetic E : produced by a moving object  potential E : object’s position (store E)  chemical E, radiant E …… All forms of E can be interconverted. The total E in the universe remain constant ╙→ The law of conservation of E Energy

4 Section 6.1 The Nature of Chemical Energy Return to TOC Copyright © Cengage Learning. All rights reserved 4

5 Section 6.1 The Nature of Chemical Energy Return to TOC Copyright © Cengage Learning. All rights reserved 5 Heat involves the transfer of energy between two objects due to a temperature difference. Work – force acting over a distance. Energy is a state function; work and heat are not:  State Function – property that does not depend in any way on the system’s past or future (only depends on present state). Energy

6 Section 6.1 The Nature of Chemical Energy Return to TOC Copyright © Cengage Learning. All rights reserved 6 System – part of the universe on which we wish to focus attention.  (reactants + products) Surroundings – include everything else in the universe.  rxn’s container, the room …… Chemical Energy system w (+) (-) (+) (-) q endothermic,  H>0 exothermic,  H<0  E = q + w (internal energy of a system)

7 Section 6.1 The Nature of Chemical Energy Return to TOC Copyright © Cengage Learning. All rights reserved 7 Concept Check Is the freezing of water an endothermic or exothermic process? Explain.

8 Section 6.1 The Nature of Chemical Energy Return to TOC Copyright © Cengage Learning. All rights reserved 8 Concept Check Classify each process as exothermic or endothermic. Explain. The system is underlined in each example. a)Your hand gets cold when you touch ice. b)The ice gets warmer when you touch it. c)Water boils in a kettle being heated on a stove. d)Water vapor condenses on a cold pipe. e)Ice cream melts. Exo Endo Exo Endo

9 Section 6.1 The Nature of Chemical Energy Return to TOC Copyright © Cengage Learning. All rights reserved 9 Concept Check For each of the following, define a system and its surroundings and give the direction of energy transfer. a)Methane is burning in a Bunsen burner in a laboratory. b)Water drops, sitting on your skin after swimming, evaporate.

10 Section 6.1 The Nature of Chemical Energy Return to TOC Copyright © Cengage Learning. All rights reserved 10 Concept Check Hydrogen gas and oxygen gas react violently to form water. Explain.  Which is lower in energy: a mixture of hydrogen and oxygen gases, or water?

11 Section 6.1 The Nature of Chemical Energy Return to TOC Thermodynamics : –The study of E & its interconversions (w, q……) Thermochemistry : –The study of heat changes in chemical rxns. A Common Type of Work in Chemical Processes Considerwork done by a gas : expansion work done to a gas : compression

12 Section 6.1 The Nature of Chemical Energy Return to TOC | w | = | F  d | = | P  A   h | = | P  V | expansion work  V > 0 & w < 0  w = - P  V Ex. 6.3 at p248  E = ? For a balloon 4.00  10 6 L 4.50  10 6 L Given: 1 L·atm = 101.3 J q = 1.3  10 8 J

13 Section 6.2 Atomic MassesLearning to Solve Problems Return to TOC Copyright © Cengage Learning. All rights reserved 13 Where are we going?  Read the problem and decide on the final goal. How do we get there?  Work backwards from the final goal to decide where to start. Reality check.  Does my answer make sense?  Is it reasonable? Conceptual Problem Solving

14 Section 6.3 The MoleEnthalpy Return to TOC (1) H = E + PV   H =  E +  (PV) ╙→ state function  H = H f - H i (2) at const. pressure :  E = q p + w = q p - P  V  q p =  E + P  V (3) from (1) & (2)  H = q p At const. P, the change in enthalpy (  H) of the sys. is equal to the energy flow as heat.

15 Section 6.3 The MoleEnthalpy Return to TOC (4) for chemical rxn :  H = H f - H i = H products - H reactants  H > 0endothermic rxn  H < 0exothermic rxn

16 Section 6.3 The MoleEnthalpy Return to TOC (1) (per mole) avg., random, translational E (2) molar heat capacity : C J/mol ‧ K (3) Heating an ideal gas when  V = 0 ① w = - P  V = 0 ② C V = 3/2 R (heat required for 1 mole gas↑  T = 1K at const. V) ③ q = n × C V ×  T (J) Thermodynamics of ideal gases

17 Section 6.3 The MoleEnthalpy Return to TOC (4) Heating an ideal gas at  P = 0 E required 1 mole gas ↑ by 1K at const. P E= E (translation) + E (PV work) = = = = C p 9.3 Thermodynamics of ideal gases - II

18 Section 6.3 The MoleEnthalpy Return to TOC (6) Energy & Enthalpy (H) H = E + PV  H =  E +  (PV) =  E + n R  T = n C V  T + n R  T = n (C V + R)  T = n C P  T Enthalpy : the heat flow into or out of a system at const. pressure  q P = C V  T Thermodynamics of ideal gases

19 Section 6.3 The MoleEnthalpy Return to TOC Question 84 at p 279  &    P = 0 w = - P  V q = q P =  H = nC P  T  E = q + w = nC V  T where  T=  PV/nR V i = 10.0 L P i = 2.0 atm V f = 30.0 L P f = 1.0 atm V m = 30.0 L P m = 2.0 atm V m ’ = 10.0 L P m ’ = 1.0 atm    

20 Section 6.3 The MoleEnthalpy Return to TOC Question 84 at p 279  &    V = 0 w = 0 q = q V =  E = nC V  T  H = nC P  T where  T=  PV/nR V i = 10.0 L P i = 2.0 atm V f = 30.0 L P f = 1.0 atm V m = 30.0 L P m = 2.0 atm V m ’ = 10.0 L P m ’ = 1.0 atm    

21 Section 6.3 The MoleEnthalpy Return to TOC summary w  w q  q  E =  E  H =  H  Not  State functions V i = 10.0 L P i = 2.0 atm V f = 30.0 L P f = 1.0 atm V m = 30.0 L P m = 2.0 atm V m ’ = 10.0 L P m ’ = 1.0 atm    

22 Section 6.4 Hess’s Law Return to TOC (1)H is a state function.  H = H f - H i (2)A + B → C + D  H C + D → A + B -  H (3)2A + 2B → 2C + 2D2  H

23 Section 6.4 Hess’s Law Return to TOC Copyright © Cengage Learning. All rights reserved 23 This reaction also can be carried out in two distinct steps, with enthalpy changes designated by ΔH 2 and ΔH 3. N 2 (g) + O 2 (g) → 2NO(g) ΔH 2 = 180 kJ 2NO(g) + O 2 (g) → 2NO 2 (g) ΔH 3 = – 112 kJ N 2 (g) + 2O 2 (g) → 2NO 2 (g) ΔH 2 + ΔH 3 = 68 kJ ΔH 1 = ΔH 2 + ΔH 3 = 68 kJ N 2 (g) + 2O 2 (g) → 2NO 2 (g) ΔH 1 = 68 kJ

24 Section 6.4 Hess’s Law Return to TOC Copyright © Cengage Learning. All rights reserved 24 The Principle of Hess’s Law

25 Section 6.4 Hess’s Law Return to TOC Calorimetry: ΔH measurement at p466

26 Section 6.4 Hess’s Law Return to TOC Copyright © Cengage Learning. All rights reserved 26 Consider the following data: Calculate ΔH for the reaction Example

27 Section 6.4 Hess’s Law Return to TOC ex. 6.6 2B (S) + 3H 2(g) → B 2 H 6(g)  H = ? Given:  H (a) 2B ( S) + 3/2 O 2(g) → B 2 O 3 (S) - 1273 kJ (b) B 2 H 6(g) + 3O 2(g) → B 2 O 3(S) + 3H 2 O (g) - 2035 KJ (c) H 2(g) + 1/2 O 2(g) → H 2 O (l) - 286 KJ (d) H 2 O (l) → H 2 O (g) 44 KJ Solve: (a) - (b) + 3(C) + 3(d)

28 Section 6.5 Standard Enthalpies of Formation Return to TOC (1)  H depend on Temp. & Pressure standard  25 ℃ 1atm (2)  H o rxn : standard H change in reaction (3)  H o f :  H o of a rxn for the formation of one mole of a compound directly from its elements. (4) Table 6.1 & Appendix 4 △ H° f = 0 for all the elements (5)  H o rxn =  H o f (products) -  H o f (reactants)

29 Section 6.5 Standard Enthalpies of Formation Return to TOC Copyright © Cengage Learning. All rights reserved 29 A Schematic Diagram of the Energy Changes for the Reaction CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(l) ΔH° reaction = –(–75 kJ) + 0 + (–394 kJ) + (–572 kJ) = –891 kJ

30 Section 6.5 Standard Enthalpies of Formation Return to TOC Copyright © Cengage Learning. All rights reserved 30 Exercise Calculate  H° for the following reaction: 2Na(s) + 2H 2 O(l) → 2NaOH(aq) + H 2 (g) Given the following information:  H f ° (kJ/mol) Na(s)0 H 2 O(l) –286 NaOH(aq) –470 H 2 (g)0  H° = –368 kJ

31 Section 6.6 Present Sources of Energy Return to TOC Copyright © Cengage Learning. All rights reserved 31 Energy Sources Used in the United States

32 Section 6.6 Present Sources of Energy Return to TOC Copyright © Cengage Learning. All rights reserved 32 The Earth’s Atmosphere

33 Section 6.7 New EnergySources Return to TOC Copyright © Cengage Learning. All rights reserved 33 Coal Conversion Hydrogen as a Fuel Other Energy Alternatives  Oil shale  Ethanol  Methanol  Seed oil


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