chapter 6 thermochemistry. deals with the energy changes chemical reactions –determines whether a...
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Chapter 6
Thermochemistry
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Thermochemistry
• Deals with the energy changes
• Chemical reactions
– Determines whether a reaction is exothermic or endothermic
• Physical Changes
– Energy changes
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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
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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
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Example of Energy Change
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Endothermic Reactions• Energy is put into a reaction
• Product bonds are weaker than reactant bonds
http://www.bbc.co.uk/scotland/education/bitesize/higher/img/chemistry/calculations_1/pe_diags/fig10.gif
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Exothermic Reactions• Energy is released from a reaction
• Product bonds are stronger than reactant bonds
http://www.bbc.co.uk/scotland/education/bitesize/higher/img/chemistry/calculations_1/pe_diags/fig03.gif
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Signs for Exo vs. Endo.
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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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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
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• Example
How much energy is required to heat 55.3 grams of Aluminum metal from 22.0ºC to 100.0 ºC? Specific heat of water is 4.18 J/g*ºC the specific heat of aluminum is 0.89 J/g*ºC
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• Example
#46 p. 282
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Homework
• P. 281 #’s 32, 37,40, 43, 48
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Hess’s 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
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Comparison
• One step:
N2(g) + 2O2(g) 2NO2(g) H1 = 68kJ
• Two step
N2(g) + O2(g) 2NO(g)H2 = 180kJ
2NO(g) + O2(g) 2NO2(g) H3 = -112kJ
N2(g) + 2O2(g) 2NO2 (g) H2 + H3 = 68kJ
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Hess’s 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
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• Example #54 p. 283
NH3 ½ N2 + 3/2 H2 H = 46 kJ
2H2 + O2 2H2O H = -486 kJ
2N2 + 6H2O 3O2 + 4NH3 H = ?
Based on enthalpy is this a useful synthesis?
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• Example #52 p. 283
C4H4 + 5O2 4CO2 + 2H2O H = -2341kJ
C4H8 + 6O2 4CO2 + 4H2O H = -2755kJ
H2 + ½ O2 H2O H = -286kJ
C4H4 + 2H2 C4H8 H = ?
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Homework
• P. 283 #’s 51,53,55,57
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• 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 (that’s 2.50x105 calories). How many burgers would it take to supply the same amount of energy? 1 calorie = 4.18 J
– 7.00 burgers
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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 = ΔHfº
– f is formation– Degree symbol means standard states
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Standard States
• Standard States
– Temperature is 25 ºC
– Pressure = 1.00 atm
– Concentration = 1.00 M
– The standard state is how something exists at these conditions
– O2(g), Na(s), Br2(l)
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Reactions
• 4C(s) + 2H2(g) CH4(g) ΔHfº = -75 kJ/mol
• H2(g) + ½ O2(g) H2O(l) ΔHfº = -286 kJ/mol
• Write the enthalpy of formation reaction for gaseous carbon dioxide. Don’t worry about the value
• C(s) + O2(g) CO2(g)
• Values are in Appendix 4 (A21)
• ΔHfº = -393.5 kJ/mol
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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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Standard Enthalpy of Reaction
• Standard Enthalpy of Reaction is the energy change at standard conditions
• Symbol = ΔHº
• ΔHº = Σ ΔHfº(Products) - Σ ΔHfº(Reactants)
• Remember to include coefficients
• Remember elements are zero!
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• Calculate the standard enthalpy of reaction for the following
Al(s) + Fe2O3(s) Al2O3(s) + Fe(s)
Al = 0, Fe = 0, Fe2O3 = -826, Al2O3 = -1676
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Homework
• P. 284 #63