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Thermochemistry
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Energy definitions
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1. ENERGY The ability to do work or produce heat Energy is NOT matter (no mass, no space) 2 forms: potential & kinetic 2. POTENTIAL ENERGY Energy due to the position (ex: water stored behind a dam) or composition (ex: gasoline) of an object
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4. CHEMICAL POTENTIAL ENERGY Energy stored in chemical bonds
3. KINETIC ENERGY Energy of motion Ex: the PE of the dammed water is converted to KE as the dam gates are opened 4. CHEMICAL POTENTIAL ENERGY Energy stored in chemical bonds Ex: sugar (candy) – bonds are broken and energy is released
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5. THERMAL KINETIC ENERGY
5. THERMAL KINETIC ENERGY Total kinetic energy of all particles in a material Ex: big iceberg > pot of boiling water (more molecules => more energy) 6. LAW OF CONSERVATION OF ENERGY In any chemical reaction or physical process, energy can be converted from one form to another, but is neither created nor destroyed
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7. HEAT Form of energy that flows from the hotter object to the colder object TEMPERATURE Measure of heat intensity on a scale (K, oC, oF) 8.UNITS OF ENERGY & HEAT calorie - the amount of energy required to raise the temperature of one gram of pure water by 1 oC Calorie (nutritional) Joule (J) - unit for heat and energy
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Given: 230 nutritional Calories Wanted: joules 1 cal = 4.184 J
9. CONVERSIONS Given: 230 nutritional Calories Wanted: joules 1 cal = J 1000 cal = 1 Cal = 1 kcal 1 kJ = 1000 J 230 Cal cal J 1 Cal = 962,320 J 1 cal
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Measuring Energy
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Specific Heat Energy Specific Heat – The amount of ___________ required to ________ the temperature of __________ of a substance by ______. raise 1 gram 1oC
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Specific Heats of Common Substances
J/gC Air (g) 1.00 Aluminum (s) 0.897 Carbon (diamond) (s) 0.502 Carbon (graphite) (s) 0.711 Carbon dioxide 0.832 Copper (s) 0.387 Ethyl alcohol (l) 2.44 Gold (s) 0.129 Granite (s) 0.803 Iron (s) 0.449 Lead (s) 0.128 Paraffin (s) 2.9 Silver (s) 0.235 Stainless steel (s) 0.51 Water (l) 4.18 Water (s) 2.03 Water (g) 2.01 Differences in specific heat are due to: _________________________________________ What is the unit for specific heat?_________ Which substance has the highest specific heat capacity? _____________________ It means that water takes a lot of ______ to heat and takes a ____ time to lose the heat. Different densities J/goC Liquid water (H2O) energy long
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Calculating Heat Transfer
1. The heat absorbed or released by a substance during a change in temperature depends on the ______________of a substance, the ____________ of the substance, and the amount by which the _________________ changes. Formula: Q = mc T Q = ___________________________(unit) ___ m = ___________________ (unit) _________ c = ____________________ (unit) _________ Δ T = ________________________ (unit) ______ Is it intensive or extensive property? (Think back to the beginning of the school year) specific heat mass temperature Heat Energy J Mass g Specific Heat J/goC Change in Temperature oC Δ T = Tfinal - Tinitial This is a physical intensive property, it does not change based on size
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Substance Specific heat J/gC Air (g) 1.00 Aluminum (s) 0.897 Carbon (diamond) (s) 0.502 Carbon (graphite) (s) 0.711 Carbon dioxide 0.832 Copper (s) 0.387 Ethyl alcohol (l) 2.44 Gold (s) 0.129 Granite (s) 0.803 Iron (s) 0.449 Lead (s) 0.128 Paraffin (s) 2.9 Silver (s) 0.235 Stainless steel (s) 0.51 Water (l) 4.18 Water (s) 2.03 Water (g) 2.01 Example A 10-gram iron nail was heated to a final temperature of 52.8oC. If the original temperature was 25oC, how much heat did the nail absorb? Formula: Q= m= c= (iron) T= Q=mc T x = J 10g 0.449 J/goC = 27.8
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Substance Specific heat J/gC Air (g) 1.00 Aluminum (s) 0.897 Carbon (diamond) (s) 0.502 Carbon (graphite) (s) 0.711 Carbon dioxide 0.832 Copper (s) 0.387 Ethyl alcohol (l) 2.44 Gold (s) 0.129 Granite (s) 0.803 Iron (s) 0.449 Lead (s) 0.128 Paraffin (s) 2.9 Silver (s) 0.235 Stainless steel (s) 0.51 Water (l) 4.18 Water (s) 2.03 Water (g) 2.01 You try! An 18.0-g piece of an unidentified metal was heated from 21.5°C to 89.0°C. If 292 J of heat energy was absorbed by the metal in the heating process, what was the identity of the metal? Formula: Q= m= c= T= Q=mc T 292 J Silver 18g x = 67.5oC = 0.24 J/goC
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Calorimetry
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DEMO – What metal can it be?
There are NO instruments that measure the specific heat of a substance directly. Heat transfer method is used to accomplish that task. Metals are good conductors of heat. Your chart should look like the one below: Metal Water Mass (m) Specific Heat (c) Initial temp (Ti) Final temp (Tf) T (Tf – Ti)
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Calorimetry Calculations
Remember the Law of Conservation of Energy Heat lost by a material = Heat gained by water Q released = Q absorbed, OR - mc ΔT = mc ΔT Calorimetry – A process of measuring heat ______________. 1. Calorimeter – an insulated (closed) device for measuring the amount of __________ absorbed or _____________ during a chemical or physical process. transfer energy released
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Calorimetry Example 1 100.0 grams of water at 22.4C is placed in a calorimeter. A gram sample of an unknown metal is heated in boiling water of 99.3C and then quickly placed in the calorimeter. If the temperature reaches equilibrium at 32.9C, what is the specific heat of the metal? What metal could it be? Metal Water Heat released = Heat absorbed - mc ΔT = mc ΔT - (75.25)(c)(-66.4) = (100)(4.18)(10.5) c = 4389 c = ___________ What metal is it? ____________ (Use the specific heat chart!) Mass (m) Specific Heat (c) Initial temp (Ti) Final temp (Tf) T (Tf – Ti) 75.25g g ? J/goC J/goC 99.3oC oC .88 J/goC 32.9oC oC -66.4oC oC Aluminum
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Calorimetry Example 2 A 25 gram sample of a metal at 75.0C is placed in a calorimeter containing 25 grams of water at 20.0C. The temperature stopped changing at 29.4C. What is the specific heat of the metal? Metal Water Heat released = Heat absorbed - mc ΔT = mc ΔT - (25)(c)(-45.6) = (25)(4.18)(9.4) 1140 c = 982.3 c = ___________ What metal is it? ____________ (Use the specific heat chart!) Mass (m) Specific Heat (c) Initial temp (Ti) Final temp (Tf) T (Tf – Ti) 25g g ? J/goC J/goC 75oC oC .86 J/goC 29.4oC oC -45.6oC oC Aluminum
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Phase Changes
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Phase Change Diagram
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Heating Curve for Water
Gas Phase Changes Condensation Heating water is an ________________________ process. Cooling water is an ________________________ process. Endothermic (absorbs energy) Evaporation Liquid Exothermic (releases energy) Freezing Melting Solid
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Equations Phase Change Equation: Q = mh
Substance Heat of Fusion (J/g) Heat of Vaporization (J/g) Copper 205 4726 Ethyl alcohol 109 879 Gold 64.5 1578 Lead 24.7 858 Silver 88 2300 Water 334 2260 Iron 267 6285 Equations Phase Change Equation: Q = mh (Q = _____________________________ ; m = ____________ ; h = __________________________________________________) Heat of Fusion- _____________________________________________________ Heat of Vaporization - _______________________________________________ ________ Heat Absorbed or Released (J) Mass (g) Heat of Fusion or Heat of Vaporization (Use the charts!) Heat required to melt 1g of a substance at its melting point Heat required to boil 1g of a substance at its boiling point
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Equations Cont’d Any time there is a temperature change (the points on the graph where there is not a phase change) you need to use Q = mc T
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Where on the chart would you use the formulas?
Q=mc T Q = mhvap Q=mc T Q = mhfus Q=mc T Q=
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Example 1 Substance Specific heat J/gC Air (g) 1.00 Aluminum (s) 0.897 Carbon (diamond) (s) 0.502 Carbon (graphite) (s) 0.711 Carbon dioxide 0.832 Copper (s) 0.387 Ethyl alcohol (l) 2.44 Gold (s) 0.129 Granite (s) 0.803 Iron (s) 0.449 Lead (s) 0.128 Paraffin (s) 2.9 Silver (s) 0.235 Stainless steel (s) 0.51 Water (l) 4.18 Water (s) 2.03 Water (g) 2.01 How much energy is required to heat 30g of copper from a solid at 1000C to a liquid at 1500C? (Melting point = 1085C , Boiling point = 2570C) Substance Heat of Fusion (J/g) Heat of Vaporization (J/g) Copper 205 4726 Ethyl alcohol 109 879 Gold 64.5 1578 Lead 24.7 858 Silver 88 2300 Water 334 2260 Iron 267 6285
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Example 2 Substance Specific heat J/gC Air (g) 1.00 Aluminum (s) 0.897 Carbon (diamond) (s) 0.502 Carbon (graphite) (s) 0.711 Carbon dioxide 0.832 Copper (s) 0.387 Ethyl alcohol (l) 2.44 Gold (s) 0.129 Granite (s) 0.803 Iron (s) 0.449 Lead (s) 0.128 Paraffin (s) 2.9 Silver (s) 0.235 Stainless steel (s) 0.51 Water (l) 4.18 Water (s) 2.03 Water (g) 2.01 How much energy is required to heat 10.0g of water from -10C to steam at 150C? (Water is the only substance w/ different specific heats for s, l, and g.) Substance Heat of Fusion (J/g) Heat of Vaporization (J/g) Copper 205 4726 Ethyl alcohol 109 879 Gold 64.5 1578 Lead 24.7 858 Silver 88 2300 Water 334 2260 Iron 267 6285
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Example 3 Substance Specific heat J/gC Air (g) 1.00 Aluminum (s) 0.897 Carbon (diamond) (s) 0.502 Carbon (graphite) (s) 0.711 Carbon dioxide 0.832 Copper (s) 0.387 Ethyl alcohol (l) 2.44 Gold (s) 0.129 Granite (s) 0.803 Iron (s) 0.449 Lead (s) 0.128 Paraffin (s) 2.9 Silver (s) 0.235 Stainless steel (s) 0.51 Water (l) 4.18 Water (s) 2.03 Water (g) 2.01 How much energy is required to heat 2.4g of iron from 23C to liquid at 1536C? (Melting point = 1536C , Boiling point = 2860C) Substance Heat of Fusion (J/g) Heat of Vaporization (J/g) Copper 205 4726 Ethyl alcohol 109 879 Gold 64.5 1578 Lead 24.7 858 Silver 88 2300 Water 334 2260 Iron 267 6285
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Enthalpy
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What is Enthalpy? heat A. Thermochemistry – the study of ____________ changes that accompany _____________reactions and _______________ changes. B. Enthalpy (H) – The ____________ content of a system at constant pressure. (The total amount of energy a substance contains (enthalpy) depends on many factors and can’t be measured, but you can measure the ________________in enthalpy.) chemical physical HEAT change in
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Change in Enthalpy H 1. Change in enthalpy (H) – The amount of heat _________________ or ________________ by a system (reaction) during a process at constant __________________. a. The H that occurs in a chemical reactionn is due to the energy absorbed to ___________the chemical bonds in the reactants, and the energy released by _____________ the chemical bonds in the products. absorbed released pressure break forming
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Exothermic vs Endothermic Reactions
released 2. Exothermic reaction – Energy is ____________. (Energy (H) reactants > Energy (H)products) H___0 3. Endothermic reaction – Energy is ___________. (Energy (H) reactants < Energy (H)products) H___0 < absorbed >
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> < Energy Diagrams negative positive Activation Energy –
H reactants___H products H reactants __H products H is _____________ H is ____________ energy needed to start a reaction > < negative positive
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Calculating Enthalpy heat change enthalpy mole
Thermochemical equation – an equation that includes the _________________. Standard Heat of Formation (ΔH0f) of a compound is the change in _________ that accomponies the formation of one _________of a compound from its elements with all substances in their stadard states at 250C & 1 atm (standard conditions) ΔH0f of an element (& diatomic molecules) = ___ ΔH reaction = ΔH0f(products) – ΔH0f(reactants) heat change enthalpy mole ** Refer to the chart for ΔHf of different compounds **
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Calculating Enthalpy Example
What is the standard heat of reaction (ΔH0) for the reaction of gaseous carbon monoxide with oxygen to form gaseous carbon dioxide? Exo- or Endo- reaction? ___________ These values come from the ΔHf chart, and the 2’s come from the balanced equation Balanced Equation: 2 CO(g) + 1 O2(g) 2 CO2(g) ΔH = [ΔHf products] – [ΔHf reactants] ΔH = [2( kJ)] – [2( kJ)] ΔH = [ kJ] – [ kJ] Since the change in enthalpy is negative, the reaction is releasing heat. Any time a reaction releases heat, it is an exothermic reaction. ΔH = kJ Exothermic
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