Thermochemistry A. Energy Changes energy is the capacity to do work, generate heat and/or generate electricity energy can be converted to other forms however the total energy of the system is conserved (1st Law of Thermodynamics) with every energy conversion, energy is always lost as heat (2nd 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 kJ 2. phase change: 10’s kJ 3. chemical change: 100-1000’s kJ 4. nuclear change: millions kJ 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 motion: rapid back and forth movement of bundled atoms with no change of location -- solid, liquid, gas 2. rotational motion: molecular rotation, no change in position --liquid, gas 3. translational motion: motion 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/gC Example Find the heat required to change 2.50 g of water from 10.0C to 27.0C . q = mcΔt = (2.50 g)(4.19 J/gC) (27.0C - 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 energy changes but do not involve a change in temperature are associated with potential energy and not kinetic energy energy from the surroundings separates the bonded molecules (intermolecular forces) thereby increasing their potential energy, or Ep Types of Phase Changes liquid evaporation melting condensation freezing sublimation solid 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 kJ (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 kJ/mol (H) molar enthalpy can be used to calculate the enthalpy change of a phase change: ΔH = nH where: ΔH = enthalpy change in J or kJ n = number of moles in mol H = molar enthalpy in J/mol or kJ/mol Example Find the energy required to melt 2.50 g of ice. H = nH = m H M = 2.50 g (6.01 kJ/mol) 18.02 g/mol = 0.8337957825 kJ = +0.834 kJ Your Assignment: pg 4 & 5 in workbook D. Total Energy Calculations a heating curve is a graph showing the phase and temperature changes as heat is added to a 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=nH during temperature changes the EK of the molecules change so you calculate heat using q = mct during phase changes there only a change in EP so you calculate heat using H = nH Heating Curve For Water H2O(l) H2O(g) 100C BP Temperature (C) 0C MP H2O(l) H2O(s) H2O(l) H2O(s) Time (min) H2O(g) Cooling Curve For Water H2O(g) 100C BP Temperature (C) H2O(l) H2O(g) H2O(l) H2O(s) H2O(l) 0C MP H2O(s) Time (min) Steps: 1. Always draw the heating curve first!!!! 2. On the curve, put a point where you begin and a 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=nH (vap or fus) 4. Perform the calculation. Example Find the total energy required to change 1.0 g of ice at -20C to steam at 110C. Heating Curve For Water 110C 100C Temperature (C) 0C -20C Time (min) ΔEtotal = + + + + = mcΔt + nHfus + mcΔt + nHvap + mcΔt = (1.0g)(2.00J/gC)(20C) + (1.0g/18.02g/mol) (6010J/mol) + (1.0g)(4.19J/gC)(100C) + (1.0g/18.02g/mol) (40650J/mol) + (1.0g)(2.02J/gC)(10C) = 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: reacting substances are placed in the bomb 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 10.0C to 52.0C. Calculate the heat released by this reaction. Give your answer in kJ. HL (rxn) = HG (water) q = mct q = (500 g)(4.19 J/gC)(52.0C – 10.0C ) q = 87 990 J = 87.990 kJ = 88.0 kJ Example 2 150 g of unknown metal X is at 100C. It is placed in a calorimeter with 200 mL of water at 23.0C. If the equilibrium temperature reached is 25.0C, what is the specific heat capacity of metal X? HL (metal) = HG (water) mct = mct (150g)c(100C – 25.0C) = (200g)(4.19 J/gC)(25.0C – 23.0C) 11250 c = 1676 c = 0.148977777 J/gC c = 0.149 J/gC Your Assignment: pg 6 in workbook Example 3 When 80.0 g of NaOH is added to 850 mL of water at 23.0C, the temperature of the water rose to 28.5C after the NaOH had dissolved. Calculate the molar enthalpy of dissolving. HL (dissolving) = HG (water) (m/M)H = mct (80.0g/40.00g/mol)H = (850g)(4.19J/gC)(28.5C – 23.0C) (2 mol)H = 19588.25 J H = 9794.125 J/mol H = – 9.79 x 103 J/mol or – 9.79 kJ/mol exothermic Example 4 When 52.5 g of LiNO3 is added to 150 mL of water in a calorimeter the initial temperature of the water was 18.0C and after the LiNO3 the temperature was 16.5C. Calculate the molar enthalpy of dissolving. HL (water) = HG (dissolving) mct = (m/M)H (150g)(4.19J/gC)(18.0C – 16.5C) = (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 kJ/mol Your Assignment: Molar Enthalpy Worksheet Example 5 15 g of ice at 5.0C is placed in a beaker of water at 30C. Calculate the mass of the water in the beaker if the final temperature at equilibrium is 10C. 30C temperature 5.0C time 10C HL (water) = HG (ice) mct = mct + (m/M)Hfus + mct m(4.19 J/gC)(30C – 10C ) = (15 g)(2.00 J/gC)(0C – (–5.0C)) +(15 g/18.02g/mol)(6010 J/mol) + (15 g)(4.19J/gC)(10C – 0C) (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 15C is placed in a calorimeter with 200 mL of water at 25C and stirred so that an equilibrium is reached, what is the final temperature of the mixture? 25C temperature 15C time tf HL (water) = HG (ice) mct = mct + (m/M)Hfus + mct (200 g)(4.19 J/gC)(25C – tf)= (10 g)(2.00 J/gC)(0C – (–15.0C)) +(10 g/18.02g/mol)(6010 J/mol) + (10 g)(4.19J/gC)(tf – 0C) 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 20C = tf Assignment: Enthalpy and Phase Change G. Chemical Change a chemical change is a transformation involving an energy change in which one substance 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 (CH3OH) was burned in a calorimeter which contained 500 mL of water at 25.0C. If the final temperature of the water is 50.0C, what is the molar heat of combustion for methanol? heat lost (combustion)= heat gained (water) (m/M)H = mct (2.65g/32.05g/mol)H = (500g)(4.19J/gC)(50.0C – 25.0C) (0.0826… mol )H = 52375 J H = 633441.038 J/mol H = – 6.33 x 105 J/mol or –633 kJ/mol Example 2 An 8.40 g sample of N2(g) is reacted with pure oxygen in a bomb calorimeter containing 1.00 kg of water to produce N2O. The temperature of the water dropped by 5.82C. What is the molar heat of reaction of N2(g)? heat lost (water) = heat gained (formation) mct = (m/M)Hf (1000 g)(4.19 J/gC)(5.82C) = (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 kJ/mol Your Assignment: pg 7 in workbook H. Industrial Bomb Calorimeters industrial calorimeters are used in research to measure 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 = mct (H2O) + mct (stirrer) + mct (bomb) + mct (thermometer) all of the “mc” parts are constant so they are replaced by one constant C, the heat capacity of the entire system in kJ/C Example 1 A 1.50 g sample of methane is completely burned in a calorimeter with a heat capacity of 11.3 kJ/C. The temperature increased from 20.15C to 27.45C. Calculate the molar enthalpy of combustion for methane. heat lost (combustion) = heat gained (calorimeter) (m/M)H = Ct (1.50 g/16.05 g/mol)H = (11.3 kJ/C)(27.45C – 20.15C) (0.0934 … mol )H= 82.49 kJ H = 882.6430002 kJ/mol H = – 883 kJ/mol Example 2 When 3.00 g of butter is burned in a bomb calorimeter with a heat capacity of 9.22 kJ/C the temperature changes from 19.62C to 31.89C. Calculate the specific enthalpy of combustion in kJ/g. ***note that in this question we are asked for enthalpy of combustion in kJ/g not kJ/mol. We substitute mass in for moles in the formula Hcomb = nHcomb heat lost (combustion) = heat gained (calorimeter) *** mH = Ct (3.00g)H = (9.22 kJ/C)(31.89C – 19.62C) (3.00 g)H= 113.1294 kJ H = 37.7098 kJ/g H = – 37.7 kJ/g Your Assignment: pg 8 in workbook I. Reaction Enthalpies the enthalpy change of a reaction refers to changes in EP and is called the heat of reaction the heat of reaction, H , 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 SO2(g) + O2(g) 2 SO3(g) H = –197.8 kJ Example Calculate the molar enthalpy of reaction (H) for sulphur dioxide using the following information: 2 SO2(g) + O2(g) 2 SO3(g) H = –197.8 kJ n = 2 mol H = –197.8 kJ H = nH H = H n = –197.8 kJ 2 mol = –98.9 kJ/mol 2. Inside Equation the heat of reaction can be written in the equation endothermic reaction… heat is on reactant side eg) H2O(l) + 285.8 kJ H2(g) + ½ O2(g) exothermic reaction… heat is on product side eg) Mg(s) + ½ O2(g) MgO(s) + 601.6 kJ Example Calculate the molar enthalpy (H) for oxygen in the decomposition of water using the following information: H2O(l) + 285.8 kJ H2(g) + ½ O2(g) H = +285.8 kJ n = ½ mol H = nH H = H n = +285.8 kJ ½ mol = +571.6 kJ/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 kJ of energy. H = –850 kJ m = 5.0 g M = 58.14 g/mol H = (m/M)H H = H (m/M) = –850 kJ (5.0 g/58.14 g/mol) = –9883.8 kJ/mol = –9.9 x 103 kJ/mol 4. Potential Energy Diagram reactants are separated from the products shape indicates whether the reaction is endothermic or exothermic Endothermic Exothermic products EP (kJ) H > 0 positive reactants Reaction Progress reactants EP (kJ) 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 the activation energy is always higher than the energy contained in the reactants and the products Endothermic 2 H2O + energy 2 H2 + O2 HHHHOO Energy (kJ) activation energy needed to break bonds energy released when bonds form 2 H2 + O 2 net energy for reaction 2 H2O Reaction Progress Exothermic H2 + Cl2 2 HCl + energy activation energy needed to break bonds Energy (kJ) H2 + Cl2 H H Cl Cl energy released when bonds form net energy for reaction Reaction Progress 2 HCl if a catalyst is used, it acts to lower the activation energy for the reaction (***typical diploma question) Endothermic 2 H2O + energy 2 H2 + O2 = catalyzed reaction HHHHOO Energy (kJ) activation energy needed to break bonds energy released when bonds form 2 H2 + O2 net energy for reaction 2 H2 O Reaction Progress Exothermic H2 + Cl2 2 HCl + energy = catalyzed reaction activation energy needed to break bonds Energy (kJ) H2 + Cl2 H H Cl Cl 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 of reaction is the same 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 then 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 net reaction, if it is not given. 2. Manipulate the given equations so they will add to 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, flip the sign on H 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) + O2(g) CO2(g) H = –393.5 kJ H = –395.4 kJ C(s, di) + O2(g) CO2(g) CO2(g) C(s, gr) + O2(g) H = –395.4 kJ H = +393.5 kJ C(s, di) H = –1.9 kJ C(s, di) + O2(g) CO2(g) C(s, gr) Example 2 Find the heat of reaction for H2O2(l) H2O(l) + ½ O2(g) using the following reactions: flip H2(g) + O2(g) H2O2(l) H2(g) + ½ O2(g) H2O(l) H2(g) + ½ O2(g) H2O(l) H = –187.8 kJ H = –285.8 kJ H2O2(l) H2(g) + ½ O2(g) H = –285.8 kJ H = +187.8 kJ H2O2(l) H2O(l) + ½ O2(g) H = –98.0 kJ Example 3 Find the heat of reaction for C(s) + 2 H2(g) CH4(g) using the following reactions: C(s) + O2(g) CO2(g) 2 H2(g) + ½ O2(g) H2O(l) flip CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) C(s) + O2(g) CO2(g) H = –393.5 kJ H = –285.8 kJ H = –890.5 kJ H = –393.5 kJ CO2(g) + 2 H2O(l) CH4(s) + 2 O2(g) H = +890.5 kJ 2 H2(g) + 1 O2(g) 2 H2O(l) C(s) + 2 H2(g) CH4(g) H = –571.6 kJ H = –74.6 kJ Your Assignment: pgs 9-10 in workbook pg 11 in workbook 2. Using Standard Heats of Formation Hf 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 (Hf ) are the changes in EP that occur when compounds are formed from their elements Hf for elements cannot be directly measured therefore they are designated as zero…all other Hf values are in reference to this Hf for common compounds are listed on pages 6-7 in data booklet the Hf is an indirect measure of the stability of a compound the more exothermic the formation , the more 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 H2O(l) C2H4(g) N2O4(g) PCl3(l) Al2O3(s) Hf = Hf = Hf = Hf = Hf = –285.8 kJ/mol +52.4 kJ/mol +11.1 kJ/mol –319.7 kJ /mol –1675.7 kJ /mol Hess’s Law formula states that the Hr is the difference between the standard heats of formation of the reactants and the products Hr = nHf(products) nHf(reactants) Example 1 Calculate the standard heat of combustion for 2 CO(g) + O2(g) 2 CO2(g) and draw the EP diagram for this reaction. 2 CO(g) O2(g) + (2 mol)(-110.5 kJ/mol) + 0 kJ + 0 kJ -221.0 kJ 2 CO2(g) (2 mol)(-393.5 kJ/mol) -787.0 kJ Hc = nHf(products) nHf(reactants) = -787.0 kJ – (-221.0 kJ) = -566.0 kJ EP Diagram for 2 CO(g) + O2(g) 2 CO2(g) -221.0 2 CO(g) + O2(g) EP (kJ) -787.0 H = -566.0 kJ 2 CO2(g) Reaction Progress Example 2 Find the heat of combustion of ethane. The products of combustion are gases. 2 C2H6(g) (2mol)(-84.0kJ/mol) -168.0 kJ + 7 O2(g) 4 CO2(g) + 6 H2O(g) + 0 kJ + 0 kJ (4 mol)(-393.5 kJ/mol) + (6 mol)(-241.8 kJ/mol) -1574.0 kJ + -1450.8 kJ Hc = nHf(products) nHf(reactants) = (-1574.0 kJ + (-1450.8 kJ)) – (-168.0 kJ) = -2856.8 kJ Your Assignment: pgs 12-13 in workbook L. Energy Systems in Biological Processes Photosynthesis: energy + CO2(g) + H2O(l) C6H12O6(s) + O2(g) Cellular Respiration: C6H12O6(s) + O2(g) CO2(g) + H2O(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 H + 31 H 42 He + 10 n + 1.70 x 109 kJ 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 U 92 141 1 10 kJ + 10 n 92 Kr + Ba + 3 n + 1.9 x 10 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 fossils fuels are the most common source of energy 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