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How are they related?. Energy Encountered Daily What is Energy?  Defined as the ability to do work or create heat.  Many types of energy  Thermal.

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Presentation on theme: "How are they related?. Energy Encountered Daily What is Energy?  Defined as the ability to do work or create heat.  Many types of energy  Thermal."— Presentation transcript:

1 How are they related?

2 Energy Encountered Daily

3 What is Energy?  Defined as the ability to do work or create heat.  Many types of energy  Thermal  Light  Gravitational  Kinetic  Potential

4 Light Energy Review  How is light energy produced?  Electrons release light energy when they fall from a high energy level to a lower energy.  We’re now going to talk about energy released or used in a chemical reaction. Heat energy

5 Thermochemistry  The study of heat used or released in a chemical reaction.  Let’s investigate heat as it compares to temperature using the Heat vs. Temperature Handout

6 Specific Heat Calculations  q = mCΔT q = heat (J or cal or Cal) 4.184 cal = 1 Joule 1000 cal = 1 Cal (dietary calorie) m = mass (g) o C = specific heat (J/g o C or cal/g o C) ΔT = change in temperature ( o C or K) = T f - T i

7 Specific Heat  Specific heat of water = 1 cal /g o C or = 4.184 J / g o C  Specific heat of most metals = < 1 J / g o C  Do metals heat slowly or quickly compared to water?  Do metals stay warm longer or shorter than water?

8 Practice Problem  How much energy is required to heat 120.0 g of water from 2.0 o C to 24.0 o C? q = mCΔT m= 120.0 g C = 4.184 J/g o C ΔT= (24.0 – 2.0) o C = 22.0 o C q = (120.0g)(4.184 J/g o C)(22.0 o C) =

9 Practice Problem  How much heat (in kJ) is given off when 85.0 g of lead cools from 200.0 o C to 10.0 o C? (Specific heat of lead = 0.129 J/g o C) q = mCΔT m = 85.0 g C = 0.129 J/g o C ΔT = (10.0 – 200.0) o C = - 190.0 o C q = (85.0 g)(0.129 J/g o C)(- 190.0 o C) = -

10 How Do Chemical Reactions Create Heat energy?  Consider the combustion of gasoline (octane) 2 C 8 H 18 +25 O 2  16 CO 2 +18 H 2 O  Potential Energy: Stored energy  Potential energy is stored in the bonds of the reactant s and the products  When bonds are broken, the energy is available  When produce bonds form, some energy is used in these bonds  The excess energy is released as heat

11 Kinetic Energy  Directly related to temperature

12 Is Heat Used or Released?  Endothermic reactions used heat from the surroundings  Sweating  Refrigeration  Exothermic heat releases heat to the surroundings  Hot hands  Combustion  Exercise

13 Endothermic Reactions  Decrease in kinetic energy  decrease in temperature  heat will transfer from the environment to the system resulting in a cooler environment  Absorbs heat from its surrounding.  The system gains heat  Positive value for q   H = q =  0  H products  H reactants

14 Exothermic Reactions  Increase in kinetic energy  increase in temperature of system  heat released to the environment resulting in a hotter environment  Releases heat to its surroundings  The system loses heat  Negative value for q   H = q =  0  H products  H reactants

15 Enthalpy  Heat content for systems at constant pressure  Symbol is H  Terms heat and enthalpy are used interchangeably for this course   H = q = m C  T  Heat moves from ________ to ___________.

16 Law of Conservation of Energy  Energy is not lost or gained in a chemical reaction  In a chemical reaction potential energy is transferred to kinetic energy

17 Thermochemical Equations  An equation that includes the heat change  Example: write the thermochemical equation for this reaction  CaO(s) + H 2 O(l)  Ca(OH) 2 (s)  H = -65.2 kJ CaO(s) + H 2 O(l)  Ca(OH) 2 (s) + 65.2 kJ

18 Stoichiometry and Thermochemistry Tin metal can be extracted from its oxide according to the following reaction: SnO 2 (s) + 4NO 2 (g) + 2H 2 O(l) + 192 kJ  Sn(s) + 4HNO 3 (aq) How much energy will be required to extract 59.5 grams of tin?

19 How to solve 1. Use your stoichiometry 2. Treat heat as a reactant or product SnO 2 (s) + 4NO 2 (g) + 2H 2 O(l) + 192 kJ  Sn(s) + 4HNO 3 (aq) 59.5 g Sn 1 mol Sn 192 kJ 1 g Sn 1 mol Sn

20 If an Object feels hot, it means it is undergoing a change with a  H that is: a. positive b. negative c. whether the object feels hot or not is unrelated to its  H d. I don’t know 

21 If the object feels hot, it means it is undergoing : a. an exothermic reaction b. an endothermic reaction c. whether it feels hot or not is unrelated to whether it is undergoing an exothermic or an endothermic change

22 How does ice melt?

23 Molar Heat of Fusion  Heat absorbed by one mole of a substance during melting  Constant temperature   H fus  H 2 O(s)  H 2 O(l)  H = 6.01 kJ/mol

24 Molar Heat of Solidification  Heat lost when 1 mole of a liquid solidifies  Temperature is constant   H solid   H fus = -  H solid  H 2 O(l)  H 2 O(s)  H = -6.01 kJ/mol

25 Molar Heat of Vaporization  Heat needed to vaporize 1 mole of a liquid   H vap  H 2 O(l)  H 2 O(g)  H vap = 40.7 kJ/mol

26 Molar Heat of Condensation  Heat released when 1 mole of vapor condenses   H cond  H 2 O(g)  H 2 O(s)  H cond = -40.7 kJ/mol   H vap = -  H cond

27 Phase Change Diagram for Water

28 Phase Change Diagram

29 The House that Heats Itself  http://www.sciencefriday.com/videos/watch/ 10007 http://www.sciencefriday.com/videos/watch/ 10007

30

31 Calorimetry  Method used to determine the heat involved in a physical or chemical change.  Relies on the law of conservation of energy

32 Calorimeter

33 Simple Calorimeter

34 Calorimetry Math  Heat gained by the water = q  Heat lost by the system = -q mC  T = q  T = T f –T i, m = mass, C = specific heat q gained by water = q lost by system  q water = - q system  mC  T = -mC  T (mass H 2 O)(spec. heat H 2 O)(  T H 2 O) = - (mass sys)(spec. heat sys)(  T sys)

35 Standard Heat of Reaction  Heat change for the equation as it is written  H =  H f (products) -  H f (reactants) Standard Heats of Formation (  H f )  Change in enthalpy when 1 mole of the compound is formed from its elements in their standard states at 25 o C and 101.3 kPa

36 Hess’s Law  A way to calculate the heat of a reaction that may be too slow or too fast to collect data from.  Add together several reactions that will result in the desired reaction. Add the ΔH for these reactions in the same way.   H total =  H products -  H reactants


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