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Chapters 5 and 19.  Energy = capacity to do work  Kinetic = energy of motion  Potential = energy of position relative to other objects  Work = energy.

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Presentation on theme: "Chapters 5 and 19.  Energy = capacity to do work  Kinetic = energy of motion  Potential = energy of position relative to other objects  Work = energy."— Presentation transcript:

1 Chapters 5 and 19

2  Energy = capacity to do work  Kinetic = energy of motion  Potential = energy of position relative to other objects  Work = energy used to cause an object with mass to move against a force  Force = push or pull exerted on an object  Heat = energy used to cause temp. to increase

3 F = the electric force between 2 charged objects k = Coulomb’s constant = 8.99 x 10 9 J-m/C 2 Q 1 and Q 2 are the electrical charges r is the distance of separation between the two objects

4  System = portion singled out for study  Open  Closed  Isolated  Surroundings = everything else

5  Work is energy used to cause an object to move against a force  w = F x d [Eq. 5.3]  Heat is energy transferred from a hotter object to a colder one

6  Energy is conserved.  Internal energy  Sum of all kinetic and potential energies of the system’s components  Change in energy is what we can know   E = E final – E initial  Magnitude of the change is important  Direction of change is important (+ or -)

7  Shows the direction and magnitude of the energy change for a chemical reaction

8   E is positive when energy is added to the system and negative when energy exits   E = q + w (q is heat and w is work)  Endothermic  System absorbs heat  Exothermic  System releases heat

9  Depends only on the present state of the system, and not the path it took to get there  Energy of a system Final

10  P-V work = pressure-volume work, which is involved in the expansion or compression of gases  w = -P  V  Enthalpy = heat flow in processes @ constant P where no other work is done except P-V work  Internal energy + P*V   H =  E + P  V  Change in enthalpy = heat gained/lost @ constant P

11  Enthalpy change that accompanies a reaction  Heat of reaction (  H rxn )   H =  H products –  H reactants

12 1. Enthalpy is an extensive property. 1. Directly proportional to the amount of reactant consumed in the reaction. 2. Enthalpy change for a rxn is equal in magnitude but opposite in sign to  H for the reverse rxn. 3. The enthalpy change for a rxn depends on the state of the reactants and products.

13 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 Thermochemical Equations: show enthalpy changes as well as the mass relationships 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

14 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. Thermochemical Equations H 2 O (l) H 2 O (g)  H = 44.0 kJ 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 266 g P 4 1 mol P 4 123.9 g P 4 x -3013 kJ 1 mol P 4 x = -6470 kJ

15  Hydrogen peroxide can decompose to water and oxygen by the following reaction:  2H 2 O 2 (l) → 2H 2 O (l) + O 2 (g)  H = -196 kJ Calculate the value of q when 5.00 g of hydrogen peroxide decomposes at constant pressure.

16  Measurement of heat flow  Calorimeter = device used to measure magnitude of temp. change that the heat flow produces  Heat capacity = C = amount of heat required to raise the temp. by 1K (or 1°C).  Extensive property; units are J/K or J/°C

17  Molar heat capacity (C m )= heat capacity of one mole of a substance  Specific heat (C s )= heat capacity of one gram of a substance  Intensive property  q = m * C s *  T

18 How much heat is given off when an 869 g iron bar cools from 94 0 C to 5 0 C? C s of Fe = 0.444 J/g 0 C  t = t final – t initial = 5 0 C – 94 0 C = -89 0 C q = m*Cs*  t = 869 g *0.444 J/g 0 C * –89 0 C= -34,000 J

19  Large beds of rocks are used in some solar- heated homes to store heat. Assume that the specific heat of the rocks is 0.82 J/g-K. Calculate the temperature change 50.0 kg of rocks would undergo if they emitted 450. kJ of heat.

20  Coffee cup calorimeter  Pressure of the system = atmospheric pressure  Assume that the calorimeter contains all heat generated in the reaction  Use q = m*C s *  T  H = q rxn No heat enters or leaves!

21  When 50.0 mL of 0.100 M silver nitrate and 50.0 mL of 0.100 M hydrochloric acid are mixed in a constant- pressure calorimeter, the temperature of the mixture increases from 22.30°C to 23.11°C. The temperature increase is caused by the following reaction: AgNO 3 (aq) + HCl (aq) → AgCl (s) + HNO 3 (aq)  Calculate  H for this reaction in kJ/mol AgNO 3, assuming that the combined solution has a mass of 100.0 g and a specific heat of 4.18 J/g-°C.

22  Bomb calorimetry  Bomb calorimeter = used to study combustion reactions  q rxn = -C cal *  T

23 Constant-Volume Calorimetry No heat enters or leaves! q sys = q water + q bomb + q rxn q sys = 0 q rxn = - (q water + q bomb ) q water = m x C s x  t q bomb = C bomb x  t Reaction at Constant V  H ~ q rxn  H = q rxn Measured in a constant-volume bomb calorimeter

24  A 0.5865-g sample of lactic acid (HC 3 H 5 O 3 ) is burned in a calorimeter whose heat capacity is 4.812 kJ/°C. The temperature increases from 23.10°C to 24.95°C. Calculate the heat of combustion of lactic acid (a) per gram and (b) per mole.


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