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Explore energy changes in Thermochemistry, including types of energy, temperature changes, phase changes, and total energy calculations using helpful examples and formulas.
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Thermochemistry A. Energy Changes • energy is the capacity toand/or do work, generate heat generate electricity • energy can be however the converted to other forms 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
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 temperature) (results in a change in
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 = q = m = Δt = c = change in kinetic energy in J heat energy in J mass in g change in temperature in °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
C. Phase Changes 1. The Basics • phase change is a change of state • phase changes always involve but do not involve energy changes a change in temperature are associated withand not potential energy 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 gas solid deposition
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
per mole of a substance • molar enthalpy is the enthalpy change • 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 = n = H = enthalpy change in J or kJ number of moles in mol 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
D. Total Energy Calculations • a heating curve is ashowing the and changes as heat is added to a substance over time phase graph temperature • the total energy change that a substance goes through can be determined using aand the formulasand heating curve q=mcΔt ΔH=nH • during temperature changes the of the molecules change so you calculate heat using EK q = mct • during phase changes there only a change in so you calculate heat using EP H = nH
Heating Curve For Water H2O(g) H2O(l) H2O(g) BP 100C Temperature (C) H2O(l) H2O(s) H2O(l) MP 0C H2O(s) Time (min)
Cooling Curve For Water H2O(g) BP 100C H2O(l) H2O(g) Temperature (C) H2O(l) H2O(s) H2O(l) MP 0C H2O(s) Time (min)
Steps: 1. Always draw the heating curve first!!!! 2. On the curve, put a point where and a point you begin 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.
Bomb Calorimeter E. Calorimetry • calorimetry is a technological process of measuring energy changes using an isolated system • the isolated system used to determine theheat involved in a phase change or in a chemical reaction is called a 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 • is recorded initial temp of water • reaction is initiated • is recorded final (maximum) temp of water
it is assumed that no energy isby the systemexcept for the energy required or released by the gained or lost 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) 11250c=1676 c=0.148977777 J/gC c=0.149 J/gC
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
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 10C 5.0C time
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
25C temperature 15C time 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? 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
G. Chemical Change • a is a transformation involving an energy change in which chemical change 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
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 eg) volume of water used, container (bomb) material, stirrer and thermometer fixed components
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