Thermochemistry A Energy Changes energy is the capacity
- Slides: 88
Thermochemistry A. Energy Changes ¬energy is the capacity to do work, generate heat and/or generate electricity ¬energy can be however the converted to other forms total energy of the system is conserved (1 st Law of Thermodynamics) ¬with every energy conversion, energy is always lost as heat (2 nd Law of Thermodynamics)
¬the primary sources of energy are: 1. chemical: fossil fuels, plants
2. nuclear: uranium, hydrogen
3. solar: radiant energy, wind, hydroelectric
4. geothermal: geysers, hotsprings
¬there are 4 types of energy changes: 1. temperature change: 10’s k. J 2. phase change: 10’s k. J 3. chemical change: 100 -1000’s k. J 4. nuclear change: millions k. J TYPICAL DIPLOMA QUESTION
B. Temperature Changes ¬temperature or kinetic energy (EK) is the energy of motion of particles… increase in temperature means an increase in EK ¬heat is the transfer of thermal energy ¬ΔEK is the change in kinetic energy (results in a change in temperature)
¬EK can be classified as: 1. vibrational rapid motion: back and forth movement of bundled atoms with no change of location -- solid, liquid, gas 2. rotational molecu motion: lar rotation, no change in position --liquid, gas 3. moti translational motion: on from one point to another -liquid, gas
¬heat capacity is the heat required to change the temperature of a "unit mass" of a substance by 1 C ΔEK = q = mcΔt where: ΔEK = change in kinetic energy in J q = heat energy in J m = mass in g Δt = change in temperature in C c = specific heat capacity in J/g C
Example Find the heat required to change 2. 50 g of water from 10. 0 C to 27. 0 C. q = mcΔt = (2. 50 g)(4. 19 J/g C) (27. 0 C - 10. 0 C) = 178. 075 J = 178 J
Your Assignment: pg 1
C. Phase Changes 1. The Basics ¬phase change is a change of state ¬phase changes always involve but do energy changes not involve a change in temperature associated with potential energy and not kinetic energy ¬energy from the surroundings separates the bonded thereby molecules (intermolecular forces) increasing their potential energy, or Ep
Types of Phase Changes liquid evaporation melting condensation freezing solid sublimation deposition gas
endothermic = melting, evaporation, sublimation exothermic = freezing, condensation, deposition
2. Enthalpy and Molar Enthalpy ¬enthalpy is the total kinetic and potential energy of a chemical system under constant pressure and temperature ¬unfortunately, the enthalpy of individual substances cannot be measured directly (EK can with a thermometer but how do you measure EP? ) ¬changes in enthalpy occur whenever heat is released or absorbed in a physical or chemical change… fortunately, this can be measured ¬change in enthalpy is measured in J or k. J ( H)
¬during a phase change EK remains constant ¬endothermic enthalpy changes are positive values ¬exothermic enthalpy changes are negative values
¬molar enthalpy is the enthalpy change per mole of a substance ¬molar enthalpy is measured in J/mol or k. J/mol (H) ¬molar enthalpy can be used to calculate the enthalpy change of a phase change: ΔH = n. H where: ΔH = enthalpy change in J or k. J n = number of moles in mol H = molar enthalpy in J/mol or k. J/mol
Example Find the energy required to melt 2. 50 g of ice. H = n. H = m H M = 2. 50 g (6. 01 k. J/mol) 18. 02 g/mol = 0. 8337957825 k. J = +0. 834 k. J
Your Assignment: pg 4 & 5 in workbook
D. Total Energy Calculations ¬a heating curve is a graph showing the and phase changes as heat is added to a temperature substance over time ¬the total energy change that a substance goes through can be determined using a heating curve and the formulas q=mcΔt and ΔH=n. H ¬during temperature changes the of the molecules EK change so you calculate heat using q = mc t ¬during phase changes there only a change in EP so you calculate heat using H = n. H
Heating Curve For Water H 2 O(l) H 2 O(g) BP 100 C Temperature ( C) 0 C MP H 2 O(l) H 2 O(s) Time (min) H 2 O(g)
Cooling Curve For Water BP 100 C Temperature ( C) 0 C H 2 O(g) H 2 O(l) H 2 O(s) H 2 O(l) MP H 2 O(s) Time (min)
Steps: 1. Always draw the heating curve first!!!! 2. On the curve, put a point where and a you begin point where you end (temperatures). 3. Determine which formulas are needed and which part of the curve they apply to. ¬each new line segment, you have a new formula ¬diagonal lines represent a change in temperature q=mcΔt ¬horizontal lines represent a phase change ΔH=n. H (vap or fus) 4. Perform the calculation.
Example Find the total energy required to change 1. 0 g of ice at -20 C to steam at 110 C. Heating Curve For Water 110 C 100 C Temperature ( C) 0 C -20 C Time (min)
ΔEtotal = + + = mcΔt + n. Hfus + mcΔt + n. Hvap + mcΔt = (1. 0 g)(2. 00 J/g C)(20 C) + (1. 0 g/18. 02 g/mol) (6010 J/mol) + (1. 0 g)(4. 19 J/g C)(100 C) + (1. 0 g/18. 02 g/mol) (40650 J/mol) + (1. 0 g)(2. 02 J/g C)(10 C) = 3068. 545172 J = 3. 1 103 J
Total Energy Calculation Practice Question Find the total energy required to changed 4. 4 g of methanol from a solid at -102 ˚C to a gas at 89. 0 ˚C. Fusion of methanol occurs at -98. 0 ˚C while vaporization occurs at 65. 0 ˚C.
E. Calorimetry ¬calorimetry is a technological process of measuring energy changes using an isolated system ¬the isolated system used to determine the heat involved in a phase change or in a chemical reaction is called a calorimeter Bomb Calorimeter insulation water ENERGY enclosed system (bomb)
¬here’s how it works: u u u reacting substances are placed in the bomb is placed in the calorimeter initial temp of water is recorded reaction is initiated final (maximum) temp of water is recorded
¬it is assumed that no energy is gained or lost by the system except for the energy required or released by the reaction or phase change ¬calculations are based on the Principle of Heat Transfer: HEAT LOST = HEAT GAINED
Example 1 A chemical reaction in a bomb calorimeter causes the temperature of 500 g of water to increase in temperature from 10. 0 C to 52. 0 C. Calculate the heat released by this reaction. Give your answer in k. J. HL (rxn) = HG (water) q = mc t q = (500 g)(4. 19 J/g C)(52. 0 C – 10. 0 C ) q = 87 990 J = 87. 990 k. J = 88. 0 k. J
Example 2 150 g of unknown metal X is at 100 C. It is placed in a calorimeter with 200 m. L of water at 23. 0 C. If the equilibrium temperature reached is 25. 0 C, what is the specific heat capacity of metal X? HL (metal) = HG (water) mc t = mc t (150 g)c(100 C – 25. 0 C) = (200 g)(4. 19 J/g C)(25. 0 C – 23. 0 C) 11250 c = 1676 c = 0. 148977777 J/g C c = 0. 149 J/g C
Your Assignment: pg 6 in workbook
Example 3 When 80. 0 g of Na. OH is added to 850 m. L of water at 23. 0 C, the temperature of the water rose to 28. 5 C after the Na. OH had dissolved. Calculate the molar enthalpy of dissolving. HL (dissolving) = HG (water) (m/M)H = mc t (80. 0 g/40. 00 g/mol)H = (850 g)(4. 19 J/g C)(28. 5 C – 23. 0 C) (2 mol)H = 19588. 25 J H = 9794. 125 J/mol H = – 9. 79 x 103 J/mol or – 9. 79 k. J/mol exothermic
Example 4 When 52. 5 g of Li. NO 3 is added to 150 m. L of water in a calorimeter the initial temperature of the water was 18. 0 C and after the Li. NO 3 the temperature was 16. 5 C. Calculate the molar enthalpy of dissolving. HL (water) = HG (dissolving) mc t = (m/M)H (150 g)(4. 19 J/g C)(18. 0 C – 16. 5 C) = (52. 5 g/68. 95 g/mol)H 942. 75 J = (0. 761… mol)H 1238. 145 J/mol = H + 1. 24 x 103 J/mol = H endothermic or +1. 24 k. J/mol
Your Assignment: Molar Enthalpy Worksheet
Example 5 15 g of ice at 5. 0 C is placed in a beaker of water at 30 C. Calculate the mass of the water in the beaker if the final temperature at equilibrium is 10 C. 30 C temperature 5. 0 C time 10 C
HL (water) = HG (ice) mc t = mc t + (m/M)Hfus + mc t m(4. 19 J/g C)(30 C – 10 C ) = (15 g)(2. 00 J/g C)(0 C – (– 5. 0 C)) +(15 g/18. 02 g/mol)(6010 J/mol) + (15 g)(4. 19 J/g C)(10 C – 0 C) (83. 8 J/g)m = 150 J + 5002. 77 J… + 628. 5 J m = 68. 966……g = 69 g
Example 6 If 10 g of ice at 15 C is placed in a calorimeter with 200 m. L of water at 25 C and stirred so that an equilibrium is reached, what is the final temperature of the mixture? 25 C temperature 15 C time tf
HL (water) = HG (ice) mc t = mc t + (m/M)Hfus + mc t (200 g)(4. 19 J/g C)(25 C – tf)= (10 g)(2. 00 J/g C)(0 C – (– 15. 0 C)) +(10 g/18. 02 g/mol)(6010 J/mol) + (10 g)(4. 19 J/g C)(tf – 0 C) 20 950 J – (838 J/ C) tf = 300 J + 3335. 18313 J + (41. 9 J/ C) tf 17314. 81687 J = (879. 9 J/ C) tf 19. 678… C = tf 20 C = tf
Assignment: Enthalpy and Phase Change
G. Chemical Change ¬a is a transformation involving chemical change one substance an energy change in which is converted into another substance
¬uses of chemical energy (exothermic): 1. steam generators from burning fossil fuels 2. motor vehicles where fuel is burned
3. natural gas, propane, coal, wood burned for heating 4. batteries
5. living organisms, cellular respiration
¬a calorimeter can be used to quantify the amount of heat lost or gained by a chemical reaction (still sticking to the heat lost = heat gained principle!!!!)
Example 1 A 2. 65 g sample of methanol (CH 3 OH) was burned in a calorimeter which contained 500 m. L of water at 25. 0 C. If the final temperature of the water is 50. 0 C, what is the molar heat of combustion for methanol? heat lost (combustion)= heat gained (water) (m/M)H = mc t (2. 65 g/32. 05 g/mol)H = (500 g)(4. 19 J/g C)(50. 0 C – 25. 0 C) (0. 0826… mol )H = 52375 J H = 633441. 038 J/mol H = – 6. 33 x 105 J/mol or – 633 k. J/mol
Example 2 An 8. 40 g sample of N 2(g) is reacted with pure oxygen in a bomb calorimeter containing 1. 00 kg of water to produce N 2 O. The temperature of the water dropped by 5. 82 C. What is the molar heat of reaction of N 2(g)? heat lost (water) = heat gained (formation) mc t = (m/M)Hf (1000 g)(4. 19 J/g C)(5. 82 C) = (8. 40 g/28. 02 g/mol)H 24385. 8 J = (0. 299… mol )H H = 81344. 06143 J/mol H = + 8. 13 x 104 J/mol or +81. 3 k. J/mol
Your Assignment: pg 7 in workbook
H. Industrial Bomb Calorimeters ¬industrial calorimeters are used in to measure research the heat of combustion of food, fuel, oil, crops, and explosives ¬modern calorimeters have fixed components eg) volume of water used, container (bomb) material, stirrer and thermometer
¬in calculating the energy of combustion, you take all components of the calorimeter into account: Etotal = mc t (H 2 O) + mc t (stirrer) + mc t (bomb) + mc t (thermometer) ¬all of the “mc” parts are constant so they are replaced by one constant C, the heat capacity of the entire system in k. J/ C
Example 1 A 1. 50 g sample of methane is completely burned in a calorimeter with a heat capacity of 11. 3 k. J/ C. The temperature increased from 20. 15 C to 27. 45 C. Calculate the molar enthalpy of combustion for methane. heat lost (combustion) = heat gained (calorimeter) (m/M)H = C t (1. 50 g/16. 05 g/mol)H = (11. 3 k. J/ C)(27. 45 C – 20. 15 C) (0. 0934 … mol )H= 82. 49 k. J H = 882. 6430002 k. J/mol H = – 883 k. J/mol
Example 2 When 3. 00 g of butter is burned in a bomb calorimeter with a heat capacity of 9. 22 k. J/ C the temperature changes from 19. 62 C to 31. 89 C. Calculate the specific enthalpy of combustion in k. J/g. ***note that in this question we are asked for enthalpy of combustion in k. J/g not k. J/mol. We substitute mass in for moles in the formula Hcomb = n. Hcomb heat lost (combustion) = heat gained (calorimeter) *** m. H = C t (3. 00 g)H = (9. 22 k. J/ C)(31. 89 C – 19. 62 C) (3. 00 g)H= 113. 1294 k. J H = 37. 7098 k. J/g H = – 37. 7 k. J/g
Your Assignment: pg 8 in workbook
I. Reaction Enthalpies ¬the enthalpy change of a refers to changes in reaction EP and is called the heat of reaction H ¬the heat of reaction, , can be expressed in 4 ways (vocabulary): 1. Outside Equation ¬the heat of reaction can be given as a H value outside the equation 2 SO 2(g) + O 2(g) 2 SO 3(g) H = – 197. 8 k. J
Example Calculate the molar enthalpy of reaction (H) for sulphur dioxide using the following information: 2 SO 2(g) + O 2(g) 2 SO 3(g) H = – 197. 8 k. J n = 2 mol H = n. H H = H n = – 197. 8 k. J 2 mol = – 98. 9 k. J/mol
2. Inside Equation ¬the heat of reaction can be written in the equation ¬endothermic reaction… heat is on reactant side eg) H 2 O(l) + 285. 8 k. J H 2(g) + ½ O 2(g) ¬exothermic reaction… heat is on product side eg) Mg(s) + ½ O 2(g) Mg. O(s) + 601. 6 k. J
Example Calculate the molar enthalpy (H) for oxygen in the decomposition of water using the following information: H 2 O(l) + 285. 8 k. J H 2(g) + ½ O 2(g) H = +285. 8 k. J n = ½ mol H = n. H H = H n = +285. 8 k. J ½ mol = +571. 6 k. J/mol
3. Molar Enthalpy, H ¬use the formula H = (m/M)H to calculate H Example Find the molar enthalpy when 5. 0 g of butane produces 850 k. J of energy. H = – 850 k. J m = 5. 0 g M = 58. 14 g/mol H = (m/M)H H = H (m/M) = – 850 k. J (5. 0 g/58. 14 g/mol) = – 9883. 8 k. J/mol = – 9. 9 x 103 k. J/mol
4. Potential Energy Diagram ¬reactants are separated from the products ¬shape indicates whether the reaction is endothermic or exothermic Endothermic Exothermic products EP (k. J) H > 0 positive reactants Reaction Progress reactants EP (k. J) H < 0 negative products Reaction Progress
Your Assignment: 1. pg 2 in workbook 2. pg 3 in workbook
J. Bond Energy ¬bond energy is the energy required to break a chemical bond or the energy released when a bond is formed ¬in both endothermic and exothermic reactions, the energy required to “pull apart” the atoms in the reactants is called the activation energy higher ¬the activation energy is always than the energy contained in the reactants and the products
Endothermic 2 H 2 O + energy 2 H 2 + O 2 H H O O Energy (k. J) activation energy needed to break bonds energy released when bonds form 2 H 2 + O 2 net energy for reaction 2 H 2 O Reaction Progress
Exothermic H 2 + Cl 2 2 HCl + energy activation H H Cl Cl energy needed to break bonds Energy (k. J) H 2 + Cl 2 energy released when bonds form net energy for reaction Reaction Progress 2 HCl
¬if a is used, it acts to catalyst lower the activation energy for the reaction (***typical diploma question)
Endothermic 2 H 2 O + energy 2 H 2 + O 2 = catalyzed reaction H H O O Energy (k. J) activation energy needed to break bonds energy released when bonds form 2 H 2 + O 2 net energy for reaction 2 H 2 O Reaction Progress
Exothermic H 2 + Cl 2 2 HCl + energy = catalyzed reaction activation H H Cl Cl energy needed to break bonds Energy (k. J) H 2 + Cl 2 energy released when bonds form net energy for reaction Reaction Progress 2 HCl
K. Predicting Enthalpy ( Hr) Changes 1. Using Hess’s Law ¬because of the law of conservation of energy, the heat same of reaction is the whether the reactants are converted to the products in a single reaction or in a series of reactions ¬G. H. Hess (1840) suggested that if two or more thermochemical equations are added to give a final equation the enthalpies can be added to give the enthalpy for the final equation
¬sometimes the heat of reaction for a chemical change is not easily measured due to time of reaction, cost, rarity of reactants etc. so we use Hess’s Law to calculate Hr
Steps: 1. Write the , if it is not given. net reaction the given equations so they will to add 2. Manipulate yield the net equation. ¬if you multiply or divide an equation, multiply or divide the H by the same factor ¬if you flip an equation, the sign on H flip 3. Cancel the reactants and products where possible to simplify (you should end up with your net equation!) 4. Add the component enthalpy changes to get the net enthalpy change.
Example 1 Find the heat of reaction for C(s, di) C(s, gr) using the following reactions: flip C(s, gr) + O 2(g) CO 2(g) C(s, di) + O 2(g) CO 2(g) H = – 393. 5 k. J H = – 395. 4 k. J C(s, di) + O 2(g) CO 2(g) C(s, gr) + O 2(g) H = – 395. 4 k. J H = +393. 5 k. J C(s, di) H = – 1. 9 k. J C(s, gr)
Example 2 Find the heat of reaction for H 2 O 2(l) H 2 O(l) + ½ O 2(g) using the following reactions: flip H 2(g) + O 2(g) H 2 O 2(l) H 2(g) + ½ O 2(g) H 2 O(l) H = – 187. 8 k. J H = – 285. 8 k. J H 2 O 2(l) H 2(g) + O ½ 2(g) H = – 285. 8 k. J H = +187. 8 k. J H 2 O 2(l) H 2 O(l) + ½ O 2(g) H = – 98. 0 k. J
Example 3 Find the heat of reaction for C(s) + 2 H 2(g) CH 4(g) using the following reactions: C(s) + O 2(g) CO 2(g) 2 H 2(g) + ½ O 2(g) H 2 O(l) flip CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O(l) C(s) + O 2(g) CO 2(g) H = – 393. 5 k. J H = – 285. 8 k. J H = – 890. 5 k. J H = – 393. 5 k. J CO 2(g) + 2 H 2 O(l) CH 4(s) + 2 O 2(g) H = +890. 5 k. J 2 H 2(g) + 1 O 2(g) 2 H 2 O(l) H = – 571. 6 k. J C(s) + 2 H 2(g) CH 4(g) H = – 74. 6 k. J
Your Assignment: pgs 9 -10 in workbook pg 11 in workbook
2. Using Standard Heats of Formation H f ¬sometimes it is not easy to measure the heat change for a reaction (too slow/expensive) ¬in this case, H can be determined using heats of formation ¬heats of formation ( H f ) are the changes in EP that occur when compounds are formed from their elements
¬ H f for elements cannot be directly measured therefore they are designated as …all other H zero f values are in reference to this ¬ H f for common compounds are listed on pages 6 -7 in data booklet ¬the H stability f is an indirect measure of the of a compound
¬the more , the more exothermic the formation more stable the compound (this means you have to add that energy to decompose it) eg) List the following compounds in order from most stable to least stable. 3 5 4 2 1 H 2 O(l) C 2 H 4(g) N 2 O 4(g) PCl 3(l) Al 2 O 3(s) H f = H f = – 285. 8 k. J/mol +52. 4 k. J/mol +11. 1 k. J/mol – 319. 7 k. J /mol – 1675. 7 k. J /mol
Hr ¬Hess’s Law formula states that the is the difference between the standard heats of formation of the and the reactants products H r = n. H f(products) n. H f(reactants)
Example 1 Calculate the standard heat of combustion for 2 CO(g) + O 2(g) 2 CO 2(g) and draw the EP diagram for this reaction. 2 CO(g) O 2(g) + (2 mol)(-110. 5 k. J/mol) + 0 k. J -221. 0 k. J 2 CO 2(g) (2 mol)(-393. 5 k. J/mol) -787. 0 k. J H c = n. H f(products) n. H f(reactants) = -787. 0 k. J – (-221. 0 k. J) = -566. 0 k. J
EP Diagram for 2 CO(g) + O 2(g) 2 CO 2(g) -221. 0 2 CO(g) + O 2(g) EP (k. J) -787. 0 H = -566. 0 k. J 2 CO 2(g) Reaction Progress
Example 2 Find the heat of combustion of ethane. The products of combustion are gases. 2 C 2 H 6(g) + 7 O 2(g) 4 CO 2(g) + 6 H 2 O(g) (2 mol)(-84. 0 k. J/mol) -168. 0 k. J + 0 k. J (4 mol)(-393. 5 k. J/mol) + (6 mol)(-241. 8 k. J/mol) -1574. 0 k. J + -1450. 8 k. J H c = n. H f(products) n. H f(reactants) = (-1574. 0 k. J + (-1450. 8 k. J)) – (-168. 0 k. J) = -2856. 8 k. J
Your Assignment: pgs 12 -13 in workbook
L. Energy Systems in Biological Processes Photosynthesis: energy + CO 2(g) + H 2 O(l) C 6 H 12 O 6(s) + O 2(g) Cellular Respiration: C 6 H 12 O 6(s) + O 2(g) CO 2(g) + H 2 O(l) + energy **Water vapor condenses into liquid
M. Nuclear Change ¬enthalpy changes in nuclear reactions are the result of EP changes as rearrangements among the subatomic particles (protons and neutrons) occur ie) intranuclear forces ¬there are two types of nuclear reactions:
1. Fusion (Joining) ¬fusion of hydrogen to helium occurs on the sun and other stars ¬these types of reaction produce the greatest amount of energy and are necessary for life on Earth ¬require a great deal of heat and pressure 2 1 3 4 1 9 k. J H + H He + n + 1. 70 x 10 1 2 0 Top number = mass number Bottom number = number of protons
2. Fission (Splitting) ¬basis for nuclear power plants ¬uranium atoms can be split into two smaller nuclei which produces large quantities of energy ¬was discovered in the late 1930's when uranium was bombarded by neutrons causing it to split 235 1 92 141 1 10 k. J U + n Kr + Ba + 3 n + 1. 9 x 10 92 0 36 56 0 ¬the neutrons produced by fission allow a chain reaction to occur to keep the reaction self sustaining
N. Society and Technological Connections ¬we must assess the risks and benefits of relying on fossil fuels and nuclear energy as energy sources ¬we are limited by our scientific knowledge and by the technology that has been developed to date ¬ are the most common source of energy fossils fuels ¬many aspects of our society are based on the price of fuels like gasoline
Advantages vs. Disadvantages of Fossil Fuels Advantages ¬ relatively low cost ¬ readily available (market) ¬ plant set-up, vehicle design, expertise affordable ¬ used all over the world ¬ deposits are large Disadvantages ¬ release of gases that contribute to the greenhouse effect and acid rain when burned ¬ mining is detrimental to the environment ¬ non-renewable
Advantages vs. Disadvantages of Nuclear Power Advantages ¬ low cost of fuel ¬ lots of energy from small amount of fuel Disadvantages ¬ high cost of plant set-up, expertise, decommission ¬ cause thermal pollution ¬ doesn’t add to greenhouse effect, acid rain ¬ difficult to dispose of nuclear fuel wastes ¬ Canada has lots of uranium ore ¬ possibility of catastrophic accidents ¬ preserves existing fossil fuels ¬ non-renewable
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