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# Energy - PowerPoint PPT Presentation

Energy. Energy transfer occurs between a system and its surroundings. System = piece of the universe. Energy. Two types: heat and work. Energy. Heat = energy transfer that is the result of contact between two substances of differing temperatures. Energy.

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Energy
• Energy transfer occurs between a system and its surroundings.
• System = piece of the universe.
Energy
• Two types: heat and work.
Energy
• Heat = energy transfer that is the result of contact between two substances of differing temperatures.
Energy
• Energy can also be transferred between a system and its surroundings.
Energy
• The system may do work on the surroundings, or the surroundings may do work on the system.
Energy
• Work = F*d (a force is used to move an object).
• Ex. Cu strip being hammered.
• Ex. Gas engine.
Energy
• The state of a system is defined by its composition, temperature, and pressure.
Energy
• State Property: quality that is dependent only on the state of the system and not how that system reached that state.
Energy
• Example: reactions under constant volume and pressure do not do expansive work - their q values are different!
Conversions
• The joule is the SI unit of energy (1J = 1N*m)
Conversions
• 1 calorie = 4.184 J
• Dietary calorie = C = 1000 c = 4184 J
• 1 kJ = 1000 J
Exo vs Endothermic
• Exothermic: reaction releases heat into the surroundings.
• Endothermic: reaction absorbs heat.
Exo vs Endothermic
• Exothermic: products formed are at lower energy than the reactants. (usually spontaneous reaction)
Exo vs Endothermic
• Endothermic: products formed are at higher energy than the reactants (usually non-spontaneous).
Exo vs Endothermic
• For a reaction at constant P and T: endothermic: q = delta H > 0
• Exothermic q = delta H < 0
Activation Energy
• A reaction requires a collision among reactants. Activation energy is the minimal energy required to make this happen.
Activation Energy
• The net energy transfer is what determines whether the reaction is exothermic or endothermic.
Calorimetry
• The study of heat transfer using a calorimeter (a device that isolates a system and measures its temperature change)
Calorimetry
• Specific heat: the heat needed to raise 1 gram of a substance 1 degree Celsius.
• Different substances change temperature differently.
Calorimetry
• Specific heat of water = 4.184 J/g*C
• q = mass*delta T*specific heat
Calorimetry
• Ex. Suppose 652 J of heat is added to 15.0 g of water originally at 20.0o C. What is the final t?
• t = 30.4o C
Calorimetry
• In a coffee-cup calorimeter, delta H reaction = -q water. Heat given off by the reaction is absorbed by the water in the cup.
Calorimetry
• Ex. Suppose heat is absorbed by 412 g of water, increasing its t from 20.12 to 29.86oC. What is delta H?
• delta H = -16.8 kJ
Calorimetry
• In bomb calorimeters, some heat is absorbed by the metal as well as the water.
Calorimetry
• q reaction = -q calorimeter = -(C calorimeter) x delta t
• C calorimeter = total heat capacity of the bomb + water.
Calorimetry
• Ex. Suppose combustion of 1.60 g of methane in a bomb calorimeter raises the t by 5.14o C. What is the q reaction (C = 17.2 kJ/oC)?
• q reaction = -88.4 kJ
Phase changes
• Heat lost = heat gained
• Process: melting to boiling
• Heat of fusion, Heat of vaporization
• Q = mass * heat of fusion
Phase changes
• Hfof water = 333 J/g
• Hv of water = 2260 J/g
• Specific heat of ice = 2.06 J/g*C
• Steam = 2.02 J/g*C
Phase Change

How much energy is required to raise the temperature of 2.50g of water from -3.00 C to 108.00 C?

qtot = 7.58 kJ

Thermochem
• Rule 1: H is directly proportional to the amount of reactants or products.
• (stoichiometry)
Thermochem
• Rule 2: H for a reaction is equal in magnitude but opposite in sign to H for the reverse reaction or process.
• A(s) --> A(l) kJ
• A(l) --> A(s) kJ
Thermochem
• Rule 3: Reactions occur in steps. Hess’ Law: If equation 1 + equation 2 = equation 3 (the steps result in the overall net equation), then H3 = H1 +H2 .
Thermochem

Ex: step 1 - C(s) + 1/2 O2(g) --> CO(g); step 2 - CO(g) + 1/2 O2(g) --> CO2(g) H = -283.0 kJ; Hnet = (-393.5 kJ). What is the net equation and the H of the first step?

H1 = -110.5 kJ, C(s) + O2(g) --> CO2(g)

Heats of Formation
• Hof of a compound = H when one mole of the compound is formed from its elements in their stable state.
Thermochem
• Ex. 2 Ag(s) + Cl2(g) --> 2 AgCl(s) Ho = -254.0 kJ
• What is Hof for silver chloride?
• Hof = -127.0 kJ
Thermochem
• Ex. HgO(s) --> Hg(l) + 1/2 O2(g) Ho = +90.8 kJ
• What is Hof for mercury(II) oxide?
• Hof = -90.8 kJ
Heats of Formation
• For any thermochemical equation:
• Ho = f products) - f reactants)
• Note: the heat of formation of an element in its stable state = 0
Thermochem
• Ex. (use Table 8.3)
• Calculate the standard enthalpy change (o) for the combustion of methane.
• Ho = -890.3 kJ
Heats of Formation
• Can also apply to ions if you take the standard heat of formation of the hydrogen ion to be zero. (see Table 8.3 bottom)
Thermochem
• Ex. Calculate the standard enthalpy change (o) for Zn(s) + 2H+(aq) --> Zn2+ (aq) + H2(g).
• Ho = -153.9 kJ
Bond Energies
• Defined as H when one mole of bonds is broken in the gaseous state.
• Ex. Cl2(g) --> 2Cl(g)
• H = B.E. Cl-Cl = 243 kJ
Bond Energies
• In general, since multiple bonds involve more bonding pairs, they tend to be stronger than single bonds.
Bond Energies
• Ex. Use bond energies to to determine if the following is an exothermic or endothermic reaction:
• 2 CO(g) + O2(g) --> 2CO2(g).
• exothermic
Thermodynamics
• First law: E = q + w where the change in energy of a system, q = heat flow into the system, and w = work done on the system.
Thermodynamics
• Esystem = -Esurroundings.
• At constant volume: w=0, qv=E
• At constant pressure: w= -PV; q = H = E + ngRT