Presentation on theme: "Principles of Reactivity: Energy and Chemical Reactions"— Presentation transcript:
1 Principles of Reactivity: Energy and Chemical Reactions AP Notes Chapter 5Principles of Reactivity:Energy and Chemical Reactions
2 study of energy transfer or heat flow Thermodynamicsstudy of energy transfer or heat flowEnergyKinetic EnergyThermal - HeatMechanicalElectricalSoundPotential EnergyChemicalGravitationalElectrostatic
3 Energy is... The ability to do work. Conserved. made of heat and work. Work is a force acting over a distance.W=F x dPower is work done over time.P=WtHeat is energy transferred between objects because of temperature difference.
4 First Law of Thermodynamics Law of conservation of energyTotal energy of the universe is constantTemperature and heatHeat is not temperatureThermal energy = particle motionTotal thermal energy is sum of all a materials individual energies
5 The Universe is divided into two halves. the system and the surroundings.The system is the part you are concerned with.The surroundings are the rest.q into system = -q from surroundings
6 System: region of space where process occurs Surroundings: region of space around system
7 Thermodynamic Properties 1. State functions-depend on nature of process NOT method-path independent-denoted by capital lettersE → state function
8 Thermodynamic Properties 2. Path functions-depends on how process is carried out-path dependent-denoted by lower caseq & w → path functions
9 Work →ww = - (PV)w = - P V - V PConstant P wP = - P VConstant V wV = - V P
10 Work →ww = - P Vassume ideal piston(i.e. no heat loss)w = E E = - P Vin isothermicsw = -q
11 PV = nRTR=0.0821L atm/mol KIn isothermic conditions-q = w = -nRTR=8.314 J/mol K
12 DU = q + wIn isothermics DU = 0 so q=-wDU = q - PDVIn constant V or constant Pw=0 so DU = q
13 Calorie - Amount of heat needed to raise the temperature of 1 gram H20, 1 degree centigrade Nutritional calorie [Calorie]= 1 KcalSI unit → Joule1 cal = J1000cal = 1Kcal=4.184KJ
14 Direction of energy flow Every energy measurement has three parts.A unit ( Joules or calories).A number how many.and a sign to tell direction.negative - exothermicpositive- endothermicNo change in energy - isothermic
15 Factors Determining Amount of Heat Amount of materialTemperature changeHeat capacity
16 Specific HeatAmount of heat needed to raise the temperature of 1 gram of material 1 degree centigradeHeat = qq = m Cp Twhere Cp = Heat capacityT = T2 – T1
30 3. How much total heat is used to raise the temperature of 100 3. How much total heat is used to raise the temperature of g H2O from-15o to 125oC?
31 Some rules for heat and work Heat given off is negative.Heat absorbed is positive.Work done by system on surroundings is positive.Work done on system by surroundings is negative.Thermodynamics- The study of energy and the changes it undergoes.
32 First Law of Thermodynamics The energy of the universe is constant.Law of conservation of energy.q = heatw = workDE = q + wTake the systems point of view to decide signs.
33 What is work? Work is a force acting over a distance. w= F x Dd P = F/ aread = V/areaw= (P x area) x D (V/area)= PDVWork can be calculated by multiplying pressure by the change in volume at constant pressure.units of liter - atm L-atm
34 Work needs a signIf the volume of a gas increases, the system has done work on the surroundings.work is negativew = - PDVExpanding work is negative.Contracting, surroundings do work on the system w is positive.1 L atm = J
35 ExamplesWhat amount of work is done when 15 L of gas is expanded to 25 L at 2.4 atm pressure?If J of heat are absorbed by the gas above. what is the change in energy?How much heat would it take to change the gas without changing the internal energy of the gas?
36 Calorimetry Measuring heat. Use a calorimeter. Two kinds Constant pressure calorimeter (called a coffee cup calorimeter)heat capacity for a material, C is calculatedC= heat absorbed/ DT = DH/ DTspecific heat capacity = C/mass
37 Calorimetry molar heat capacity = C/moles heat = specific heat x m x DTheat = molar heat x moles x DTMake the units work and you’ve done the problem right.A coffee cup calorimeter measures DH.An insulated cup, full of water.The specific heat of water is 1 cal/gºCHeat of reaction= DH = sh x mass x DT
40 Calorimeter Constant heat capacity that is constant over a temperature rangewhereq(cal) = CC x TC(CC = calorimeter constant)
41 4. Determine the calorimeter constant for a bomb calorimeter by mixing 50.0 mL of water at 25oC with 50.0 mL of water at 60.0 oC. The final temp. attained is 40oC.
42 5. What is the molar heat of combustion of sulfur, if 2 5. What is the molar heat of combustion of sulfur, if 2.56 grams of solid flowers of sulfur are burned in excess oxygen in the bomb calorimeter in #4?
43 6. Determine the molar heat of solution of LiOH 6. Determine the molar heat of solution of LiOH. HINT: Find the calorimeter constant first.
44 7. The heat was absorbed by 815 g of water in calorimeter which had an initial temp of C. After equilibrium was reached, the final temperature was C.
45 ExamplesThe specific heat of graphite is 0.71 J/gºC. Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K.A 46.2 g sample of copper is heated to 95.4ºC and then placed in a calorimeter containing 75.0 g of water at 19.6ºC. The final temperature of both the water and the copper is 21.8ºC. What is the specific heat of copper?
46 Calorimetry Constant volume calorimeter is called a bomb calorimeter. Material is put in a container with pure oxygen. Wires are used to start the combustion. The container is put into a container of water.The heat capacity of the calorimeter is known and tested.Since DV = 0, PDV = 0, DE = q
47 Bomb Calorimeter thermometer stirrer full of water ignition wire Steel bombsample
48 Propertiesintensive properties not related to the amount of substance.density, specific heat, temperature.Extensive property - does depend on the amount of stuff.Heat capacity, mass, heat from a reaction.
49 Enthalpy abbreviated H H = E + PV (that’s the definition) PV = w at constant pressure.DH = DE + PDVthe heat at constant pressure qp can be calculated fromDE = qp + w = qp - PDVqp = DE + P DV = DH
50 Hess’s Law Enthalpy is a state function. It is independent of the path.We can add equations to come up with the desired final product, and add the DHTwo rulesIf the reaction is reversed the sign of DH is changedIf the reaction is multiplied, so is DH
51 Enthalpy and DHfN2 + 2O2 2NO2N2 + 2O2 + 68KJ/mol 2NO2Which side is the heat written on? Why?Look on your Heat Sheets!Why is it only 34KJ/mol?
57 8. Calculate the heatneeded to decomposeone mole of calciumcarbonate.
58 Standard EnthalpyThe enthalpy change for a reaction at standard conditions (25ºC, 1 atm , 1 M solutions)Symbol DHºWhen using Hess’s Law, work by adding the equations up to make it look like the answer.The other parts will cancel out.
59 Example Given calculate DHº for this reaction DHº= -1300. kJ
60 Example Given +1300. kJ + 394 kJ + 286 kJ calculate DHº for this reaction
61 Example Given +1300. kJ + 394 kJ + 286 kJ 1300.KJ + 2CO2(g) + H20(l) C2H2 (g) + 5/2 O2 (g)+ 394 kJ+ 286 kJcalculate DHº for this reaction
62 Example Given + 2(394 kJ) + 286 kJ calculate DHº for this reaction 1300.KJ + 2CO2(g) + H20(l) C2H2 (g) + 5/2 O2 (g)222+ 2(394 kJ)+ 286 kJcalculate DHº for this reaction
63 Example Given + 788 kJ + 286 kJ calculate DHº for this reaction 1300.KJ + 2CO2(g) + H20(l) C2H2 (g) + 5/2 O2 (g)242+ 788 kJ2+ 286 kJcalculate DHº for this reaction
64 Example Given + 788 kJ + 286 kJ calculate DHº for this reaction 1300.KJ + 2CO2(g) + H20(l) C2H2 (g) + 5/2 O2 (g)242+ 788 kJ2+ 286 kJcalculate DHº for this reaction
65 Example Given calculate DHº for this reaction 1300.KJ C2H2 (g) 2C(s) 788KJH2(g) 286KJ1300.KJ 788kJ + 286KJ1300.KJ 1074KJcalculate DHº for this reaction
66 Example Given calculate DHº for this reaction 1300.KJ 1074KJ 226KJ DHf = 1300KJ – 1074KJDHf = 226KJ (+sign shows Endothermic)calculate DHº for this reaction226KJ +
67 Example Now, You try it! Given DHº= +77.9kJ DHº= +495 kJ DHº= +435.9kJ Calculate DHº for this reaction
68 Standard Enthalpies of Formation Hess’s Law is much more useful if you know lots of reactions.Made a table of standard heats of formation. The amount of heat needed to for 1 mole of a compound from its elements in their standard states.Standard states are 1 atm, 1M and 25ºCFor an element it is 0There is a table in Appendix 4 (pg A22)
69 Standard Enthalpies of Formation Need to be able to write the equations.What is the equation for the formation of NO2 ?½N2 (g) + O2 (g) ® NO2 (g)Have to make one mole to meet the definition.Write the equation for the formation of methanol CH3OH.
70 Since we can manipulate the equations We can use heats of formation to figure out the heat of reaction.Lets do it with this equation.C2H5OH +3O2(g) ® 2CO2 + 3H2Owhich leads us to this rule.
71 Since we can manipulate the equations We can use heats of formation to figure out the heat of reaction.Lets do it with this equation.C2H5OH +3O2(g) ® 2CO2 + 3H2Owhich leads us to this rule.
72 9. Calculate Hf of SO3(g) from the data given. Forms of Equations: 2SO2(g) + O2(g) 2SO3(g) H = kjor2SO2(g) + O2(g) 2SO3(g) kj
73 10. If the metabolism of glucose is combustion of C6H12O6(s), how much heat is produced by the metabolism of 1.00 g of glucose?
74 Additivity of Heats of Reaction H = EnthalpyH = E + PVDH = DE + DPDV
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