<<

Unit 1 - Lecture 5 Chapter 6 : Flow and The Molecular Nature of and Change Fifth Edition

Martin S. Silberberg

Copyright ! The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Thermochemistry: Energy Flow and Chemical Change Figure 6.1 A chemical system and its surroundings.

6.1 Forms of Energy and Their Interconversion

the surroundings 6.2 : of Reaction and Chemical Change

the system

Energy diagrams for the transfer of (E) is the study of and its transformations. Figure 6.2 between a system and its surroundings.

Thermochemistry is a branch of thermodynamics that deals with the heat involved with chemical and physical changes.

Fundamental premise

When energy is transferred from one object to another, it appears as and/or as heat.

For our work we must define a system to study; everything else then becomes the surroundings.

The system is composed of particles with their own internal (E or U). "E = E - E = E - E Therefore the system has an internal energy. When a change occurs, the final initial products reactants internal energy changes. Figure 6.4 A system losing energy as work only. Figure 6.3 A system transferring energy as heat only.

Zn(s) + 2H+(aq) + 2Cl-(aq)

Energy, E

Work (w) done on "E<0 surroundings (w < 0)

2+ - H2(g) + Zn (aq) + 2Cl (aq)

Table 6.1 The Sign Conventions* for q, w and "E

q + w = "E "Euniverse = "Esystem + "Esurroundings = 0

+ + +

+ - depends on sizes of q and (J) 1 J = 1 kg*m2/s2 w - + depends on sizes of q and w (cal) 1 cal = 4.184 J - - - 1 Btu = 1055 J * For q: + means system gains heat; - means system loses heat. * For w: + means word done on system; - means work done by system.

Sample Problem 6.1 Determining the Change in Internal Energy of a System PROBLEM: When gasoline burns in a car engine, the heat released Figure 6.5 causes the products CO2 and H2O to expand, which pushes the pistons outward. Excess heat is removed by Some interesting the car’s cooling system. If the expanding do 451 J quantities of energy. of work on the pistons and the system loses 325 J to the surroundings as heat, calculate the change in energy ("E) PLAN: Define system and surroundings,in J, kJ, and assign kcal. signs to q and w and calculate "E. The answer should be converted from J to kJ and then to kcal. SOLUTION: q = - 325 J w = - 451 J

"E = q + w =-325 J + (-451 J) =-776 J

kJ kcal -776 J = -0.776 kJ -0.776 kJ = -0.185 kcal 103J 4.184 kJ Figure 6.6 Two different paths for the energy change of a system. Figure 6.7

Pressure- work.

Figure 6.8 The Meaning of Enthalpy Enthalpy diagrams for exothermic and endothermic processes.

CH4(g) + 2O2(g) CO2(g) + H2O(s) w = - P"V 2H O(g) H O(l) "H ! "E in 2 2 CH4 + 2O2 H2O(l) H Hfinal H = E + PV 1. Reactions that do not involve initial H gases. H where H is enthalpy 2. Reactions in which the number of "H < 0 heat out "H > 0 heat in "H = "E + P"V moles of does not change. Enthalpy, Enthalpy, 3. Reactions in which the number of CO + H O(s) moles of gas does change but q is 2 H 2 H q = "E + P"V = "H 2H O final initial p >>> P"V. 2

A B

Sample Problem 6.2 Drawing Enthalpy Diagrams and Some Important Types of Enthalpy Change Determining the Sign of "H PROBLEM: In each of the following cases, determine the sign of "H, state whether the reaction is exothermic or endothermic, heat of ("Hcomb) and draw and enthalpy diagram. (a) H2(g) + 1/2O2(g) H2O(l) + 285.8 kJ C4H10(l) + 13/2O2(g) 4CO2(g) + 5H O(g) (b) 40.7 kJ + H2O(l) 2 PLAN: Determine whetherH2O( heatg) is a reactant or a product. As a heat of formation ("Hf) reactant, the products are at a higher energy and the K( ) + 1/2Br ( ) KBr( ) reaction is endothermic. The opposite is true for an s 2 l s SOLUTION: (b) The reaction is (a) The reaction is heat of fusion ("Hfus) endothermic. exothermic.(reactants) H2(g) + 1/2O2(g) H2O(g) (products NaCl(s) ) NaCl(l) "H = -285.8 "H = +40.7 kJ heat of vaporization ("Hvap) kJ (reactants) H2O(l) (products H2O(l) ) C H (l) C H (g) EXOTHERMI ENDOTHERMI 6 6 6 6 C C