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Ch 16 Reaction Energy.  Standard: –7.d. Students know how to solve problems involving heat flow and temperature changes, using known values of specific.

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Presentation on theme: "Ch 16 Reaction Energy.  Standard: –7.d. Students know how to solve problems involving heat flow and temperature changes, using known values of specific."— Presentation transcript:

1 Ch 16 Reaction Energy

2  Standard: –7.d. Students know how to solve problems involving heat flow and temperature changes, using known values of specific heat.  Objective: –We will define heat, give its units, and perform specific-heat calculations.

3 Thermochemistry   Thermochemistry: the study of the transfer of energy as heat that occurs during chemical reactions and changes in state.   Heat: q, is energy transferred from one object to another because of a temperature difference between them.   Heat always flows from a warmer object to a cooler object and will continue to flow until they are in equilibrium.

4  Endothermic Process: one that absorbs heat from the surroundings (+q).  Exothermic Process: one that releases heat to it’s surroundings (-q).  Calorimeter: the insulated device used to measure the absorption or release of heat in chemical or physical processes.

5 Specific Heat  Heat flow is measured in joules (J).  One joule of heat raises the temperature of 1g of pure water 0.2390°C.  Specific Heat: the amount of heat needed to increase the temp of 1 g of the substance 1°C or 1 Kelvin.  Water’s specific heat is 4.18 J/(g●K)

6 Calculating Specific Heat (C)  Divide the heat input, q (Joules) by the temperature change, ΔT (°C) times the mass of the substance, m (g). C = q = heat m x ΔT mass x change in temp

7 Example  The temperature of a 95.4 g piece of copper increases from 25.0°C to 48.0°C when the copper absorbs 849 J of heat. What is the specific heat, C, of copper? q = 849 J m = 95.4g ΔT = (48.0°C -25.0°C)=23.0°C C = q = 849 J m x ΔT 95.4 g x 23.0°C C = 0.387 J/(g x °C)

8 Example 2  How much heat, q, is required to raise the temperature of 400.0 g of silver 45°C? The specific heat of silver is 0.24 J/(g x °C). ΔT = 45°Cm = 400.0 g C= 0.24 J/(g x °C) q = C x m x ΔT q = 0.24 J/(g x °C) x 400.0 g x 45°C q = 4320 J

9 Enthalpy of Reaction   Enthalpy Change is the amount of energy absorbed by a system as heat during a process at constant pressure. – –The enthalpy change is always the difference between the enthalpies of the products and the reactants and is called Enthalpy of Reaction. ∆H = H products - H reactants

10   ∆H is negative for an exothermic reaction because the system loses heat.   ∆H is positive for an endothermic reaction because the system gains heat.

11 Example 3   Enthalpy of Reaction for the formation of water vapor. – –What we already know: 2H 2 (g) + O 2 (g)  2H 2 O(g) – –This equation does not tell us that energy is released as heat during the reaction. – –Thermochemical Equation: 2H 2 (g) + O 2 (g)  2H 2 O(g) + 483.6kJ – –Writing with ∆H: 2H 2 (g) + O 2 (g)  2H 2 O(g) ∆H = -483.6kJ

12   Enthalpy of Reaction for the decomposition of water vapor. – –What we already know: 2H 2 O(g)  2H 2 (g) + O 2 (g) – –This equation does not tell us that energy as heat is absorbed during the reaction. – –Thermochemical Equation: 2H 2 O(g) + 483.6kJ  2H 2 (g) + O 2 (g) – –Writing with ∆H: 2H 2 O(g)  2H 2 (g) + O 2 (g) ∆H = +483.6kJ

13   Compounds that release a large amount of energy as heat when they are formed are very stable.   Compounds that release a very small amount of energy as heat or absorb a large amount of energy as heat when they are formed are sometimes unstable and may decompose or react violently.

14 Chapter 17 Reaction Kinetics

15 Ch 17.1 Reaction Kinetics

16  Standard: –8.a. Students know the rate of reaction is the decrease in concentration of reactants or the increase in concentration of products with time.  Objective: –We will interpret chemical reactions and define activated complex. We will draw energy diagrams.

17 Chemical Reactions  Activation Energy: the minimum energy that colliding particles must have in order to react.  Activated Complex: an unstable arrangement of atoms that forms momentarily at the peak of the activation-energy barrier.  This is also called the Transition State.

18 Energy Diagrams   ∆E forward = energy of products – energy of reactants   ∆E reverse = energy of reactants – energy of products   E a = energy of activated complex – energy of reactants   E a ’ = energy of activated complex – energy of products

19   ∆E forward is positive for endothermic and negative for exothermic   ∆E reverse is negative for endothermic and positive for exothermic

20 Examples!!

21 Ch 17.2

22  Standard: –8.b. Students know how reaction rates depend on such factors as concentration, temperature and pressure. –8.c. Students know the role a catalyst plays in increasing the reaction rate.  Objective: –We will discuss the factors that influence reaction rate and define a catalyst.

23 Rate Influencing Factors  The rate of a chemical reaction depends upon temperature, concentration, particle size, and the use of a catalyst.  The nature of the reactants and their bonds is also a factor, but not one that can be easily changed so we will not talk about it.

24 TEMPERATURE  Raising the temperature speeds up the reaction and lowering the temperature slows down the reaction.  The higher the concentration, the more likely collisions will take place, which increases the reaction rate. CONCENTRATION

25 PARTICLE SIZE  The smaller the particle size, the more surface area, which increases the reaction rate.  Adding a catalyst will increase the rate of reaction, in some cases, better than increasing the temperature.  Inhibitor: a substance that interferes with the action of a catalyst. –These will slow down or even stop a reaction. CATALYST


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