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Chapter 6 Thermochemistry. Deals with the energy changes Chemical reactions –Determines whether a reaction is exothermic or endothermic Physical Changes.

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Presentation on theme: "Chapter 6 Thermochemistry. Deals with the energy changes Chemical reactions –Determines whether a reaction is exothermic or endothermic Physical Changes."— Presentation transcript:

1 Chapter 6 Thermochemistry

2 Deals with the energy changes Chemical reactions –Determines whether a reaction is exothermic or endothermic Physical Changes –Energy changes

3 Vocabulary Energy - The ability to do work (E) in J Potential Energy – Energy of position (PE) Kinetic Energy – Energy of motion (KE) Temperature – Random motion of particles Heat – Transfer of energy (q) in J Work – Force x Distance (W) in J Enthalpy – Energy Change for reactions –(ΔH) in J

4 Vocabulary Pathway – How energy is transferred State Function – Properties that depend on current state. (Independent of pathway) Energy is a state function –Temperature, Volume, Pressure Heat and Work are not state functions

5 Example of Energy Change

6 Endothermic Reactions Energy is put into a reaction Product bonds are weaker than reactant bonds tesize/higher/img/chemistry/calculations_1/ pe_diags/fig10.gif

7 Exothermic Reactions Energy is released from a reaction Product bonds are stronger than reactant bonds /bitesize/higher/img/chemistry/calculatio ns_1/pe_diags/fig03.gif

8 Signs for Exo vs. Endo.

9 Calorimetry Measuring heat associated with a reaction Uses a calorimeter –Coffee Cup Calorimeter At constant pressure q = -ΔH q=m*s* ΔT q = heat, m = mass, ΔT = Change in Temp s=specific heat capacity

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11 Specific Heat Capacity Energy required to heat one gram of a substance 1 degree Celcius Units J/g*ºC Molar Heat Capacity = Just per mole Metals have low heat capacities

12 Example How much energy is required to heat 55.3 grams of Aluminum metal from 22.0ºC to ºC? Specific heat of water is 4.18 J/g*ºC the specific heat of aluminum is 0.89 J/g*ºC

13 Example #46 p. 282

14 Homework P. 281 #s 32, 37,40, 43, 48

15 Hesss Law In going from a particular set of reactants to a particular set of products, the change in enthalpy ( H) is the same whether the reaction takes place in one step or in a series of steps Energy is a state function

16 Comparison One step: N 2(g) + 2O 2(g) 2NO 2(g) H 1 = 68kJ Two step N 2(g) + O 2(g) 2NO (g) H 2 = 180kJ 2NO (g) + O 2(g) 2NO 2(g) H 3 = -112kJ N 2(g) + 2O 2(g) 2NO 2 (g) H 2 + H3 = 68kJ

17 Hesss Law Rules Start with the end reaction in mind and work backward rearrange your equations Work to cancel like terms When reversing the equation the sign of ΔH must be changed When multiplying an equation by a coefficient ΔH must be multiplied by the same coefficient

18 Example #54 p. 283 NH 3 ½ N 2 + 3/2 H 2 H = 46 kJ 2H 2 + O 2 2H 2 O H = -486 kJ 2N 2 + 6H 2 O 3O 2 + 4NH 3 H = ? Based on enthalpy is this a useful synthesis?

19 Example #52 p. 283 C 4 H 4 + 5O 2 4CO 2 + 2H 2 O H = -2341kJ C 4 H 8 + 6O 2 4CO 2 + 4H 2 O H = -2755kJ H 2 + ½ O 2 H 2 O H = -286kJ C 4 H 4 + 2H 2 C 4 H 8 H = ?

20 Homework P. 283 #s 51,53,55,57

21 Calculate how much energy (kJ) is required to heat 50.0 L of tap water from 10.0 ºC to 45.0 ºC. Assume the density of water is 1.00 g/mL and specific heat of water is 4.18 J/g* ºC –7320 kJ A McDonalds hamburger contains 250. Calories (thats 2.50x10 5 calories). How many burgers would it take to supply the same amount of energy? 1 calorie = 4.18 J –7.00 burgers

22 Enthalpy of Formation The change in enthalpy that accompanies the formation of one mole of a compound from its elements with all elements in their standard state Symbol = ΔH f º –f is formation –Degree symbol means standard states

23 Standard States –Temperature is 25 ºC –Pressure = 1.00 atm –Concentration = 1.00 M –The standard state is how something exists at these conditions –O 2(g), Na (s), Br 2(l)

24 Reactions 4C (s) + 2H 2(g) CH 4(g) ΔH f º = -75 kJ/mol H 2(g) + ½ O 2(g) H 2 O (l) ΔH f º = -286 kJ/mol Write the enthalpy of formation reaction for gaseous carbon dioxide. Dont worry about the value C (s) + O 2(g) CO 2(g) Values are in Appendix 4 (A21) ΔH f º = kJ/mol

25 Enthalpy of Formation Values Most values are negative –Compounds are more stable than element All elements are zero Even diatomic elements! States matter –Liquid water is -286 –Gaseous water is -242 Use this information to calculate enthalpy of the reaction!

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27 Standard Enthalpy of Reaction Standard Enthalpy of Reaction is the energy change at standard conditions Symbol = ΔHº ΔHº = Σ ΔH f º (Products) - Σ ΔH f º (Reactants) Remember to include coefficients Remember elements are zero!

28 Calculate the standard enthalpy of reaction for the following Al (s) + Fe 2 O 3(s) Al 2 O 3(s) + Fe (s) Al = 0, Fe = 0, Fe 2 O 3 = -826, Al 2 O 3 = -1676

29 Homework P. 284 #63


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