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PHALL not integrated into presentation Chapter 6.1-6.2 Thermochemistry THERMODYNAMICS Courtesy of lab-initio.com Chapter 6 Table of Contents • • • • • • 6.1 6.2 6.3 6.4 6.5 6.6 The Nature of Energy Enthalpy and Calorimetry Hess’s Law Standard Enthalpies of Formation Present Sources of Energy New Energy Sources Copyright © Cengage Learning. All rights reserved 3 Chapter 6 Table of Contents • Finishing ch. 5 test - on computer • Notes with computer for section 6.1-6.2 including problems (note to me: assigned as HW last year incorporated into notes this year) • HW: Pre-lab Calorimetry • Turn in part 2 ch. 5 take home Test • Turn in previous lab reports • Turn in ch. 5 Interactive Notes packet for grade. • Turn in any missing or late assignments - check Pinnacle for grades Copyright © Cengage Learning. All rights reserved 4 Section 6.1 The Nature of Energy Energy • • Capacity to do work or to produce heat. That which is needed to oppose natural attractions. • Law of conservation of energy – energy can be converted from one form to another but can be neither created nor destroyed. • The First Law of Thermodynamics The total energy content of the universe is constant. Return to TOC Copyright © Cengage Learning. All rights reserved 5 Section 6.1 The Nature of Energy Thermodynamics • Thermodynamics is the study of energy and its interconversions. Return to TOC Copyright © Cengage Learning. All rights reserved 6 Definitions #1 Energy: The capacity to do work or produce heat Potential Energy: Energy due to position or composition Kinetic Energy: Energy due to the motion of the object Section 6.1 The Nature of Energy Energy • • Potential energy – energy due to position or composition. Kinetic energy – energy due to motion of the object and depends on the mass of the object and its velocity. Return to TOC Copyright © Cengage Learning. All rights reserved 8 Section 6.1 The Nature of Energy Initial Position • In the initial position, ball A has a higher potential energy than ball B. Return to TOC Copyright © Cengage Learning. All rights reserved 9 Section 6.1 The Nature of Energy Final Position • After A has rolled down the hill, the potential energy lost by A has been converted to random motions of the components of the hill (frictional heating) and to the increase in the potential energy of B. Return to TOC Copyright © Cengage Learning. All rights reserved 10 Section 10.1 The Nature of Energy Energy Transformations Types of Energy 1. Kinetic energy 2. Potential energy 3. Chemical energy 4. Heat energy 5. Electric energy 6. Radiant energy Return to TOC Copyright © Cengage Learning. All rights reserved 11 Section 6.1 The Nature of Energy What are 2 ways energy is transferred? 1) By doing work 2) by Heat transfer - thermal energy is transferred Return to TOC Copyright © Cengage Learning. All rights reserved 12 Section 6.1 Energy Transfer by Doing work The Nature of Energy • Energy can be transferred from one object to another by doing work. When work is done on an object, it results in a change in the object's motion (more specifically, a change in the object's kinetic energy). • Energy is often defined as the ability to do work. Work equals force multiplied by distance. • An illustration of how doing work is an example of energy transfer • Suppose that a person exerts a force on the wheelbarrow that is initially at rest, causing it to move over a certain distance. Recall that the work done on the wheelbarrow by the person is equal to the product of the person's force multiplied by the distance traveled by the wheelbarrow. Notice that when the force is exerted on the wheelbarrow, there's a change in its motion. Its kinetic energy increases. But where did the wheelbarrow get its kinetic energy? It came from the person exerting the force, who used chemical energy stored in the food they ate to move the wheelbarrow. In other words, when the person did work on the wheelbarrow, they transferred a certain amount of chemical energy stored in the person was transferred to the wheelbarrow, causing its kinetic energy to increase. As a result, the person's store of chemical energy decreases and the wheelbarrow's kinetic energy increases. • Wherever you look, you can see examples of energy transfers. When you turn on a light, you see result of energy being transferred from the sun to the plants to the coal to electricity and finally to light you see. During each of these transfers, energy changes form. There are two main forms of energy, kinetic energyReturn to TOC (motion) and potential energy (position). To further classify energy, these forms are sometimes further Copyright © Cengage Learning. All rights reserved 13 Section 6.1 The Nature of Energy Energy • Heat involves the transfer of energy between two objects due to a temperature difference. • Work – force acting over a distance. • Energy is a state function; work and heat are not: State Function – property that does not depend in any way on the system’s past or future (only depends on present state). Return to TOC Copyright © Cengage Learning. All rights reserved 14 State Functions depend ONLY on the present state of the system ENERGY IS A STATE FUNCTION A person standing at the top of Mt. Everest has the same potential energy whether they got there by hiking up, or by falling down from a plane WORK IS NOT A STATE FUNCTION WHY NOT??? Section 6.1 The Nature of Energy Chemical Energy • • System – part of the universe on which we wish to focus attention. Surroundings – include everything else in the universe. Return to TOC Copyright © Cengage Learning. All rights reserved 16 Section 6.1 The Nature of Energy Chemical Energy • Exothermic Reaction: Energy flows out of the system. • Energy gained by the surroundings must be equal to the energy lost by the system. Return to TOC Copyright © Cengage Learning. All rights reserved 17 Section 6.1 The Nature of Energy Chemical Energy • Endothermic Reaction: Heat flow is into a system. Absorb energy from the surroundings. Return to TOC Copyright © Cengage Learning. All rights reserved 18 Section 6.1 The Nature of Energy Concept Check •Is the freezing of water an endothermic or exothermic process? Explain. Return to TOC Copyright © Cengage Learning. All rights reserved 19 Section 6.1 The Nature of Energy Concept Check •Classify each process as exothermic or endothermic. Explain. The system is underlined in each example. a) b) c) d) e) Your hand gets cold when you touch ice. The ice gets warmer when you touch it. Water boils in a kettle being heated on a stove. Water vapor condenses on a cold pipe. Ice cream melts. Copyright © Cengage Learning. All rights reserved 20 Return to TOC Section 6.1 The Nature of Energy Concept Check •Classify each process as exothermic or endothermic. Explain. The system is underlined in each example. Exo a) Endo Endo Exo Endo b) c) d) e) Your hand gets cold when you touch ice. The ice gets warmer when you touch it. Water boils in a kettle being heated on a stove. Water vapor condenses on a cold pipe. Ice cream melts. Copyright © Cengage Learning. All rights reserved 21 Return to TOC Section 6.1 The Nature of Energy Concept Check •For each of the following, define a system and its surroundings and give the direction of energy transfer. a) b) Methane is burning in a Bunsen burner in a laboratory. Water drops, sitting on your skin after swimming, evaporate. Return to TOC Copyright © Cengage Learning. All rights reserved 22 Section 6.1 The Nature of Energy Concept Check •Hydrogen gas and oxygen gas react violently to form water. Explain. Which is lower in energy: a mixture of hydrogen and oxygen gases, or water? Return to TOC Copyright © Cengage Learning. All rights reserved 23 Section 6.1 The Nature of Energy Internal Energy • Law of conservation of energy is often called the first law of thermodynamics. • Internal energy E of a system is the sum of the kinetic and potential energies of all the “particles” in the system. • To change the internal energy of a system: ΔE = q + w – q represents heat – w represents work Return to TOC Copyright © Cengage Learning. All rights reserved 24 Work vs. Energy Flow Click here to watch visualization. Copyright © Houghton Mifflin Company. All rights reserved. 6– Section 6.1 The Nature of Energy Internal Energy • • Sign reflects the system’s point of view. Endothermic Process: q is positive (heat enters the system) • Exothermic Process: q is negative (heat exits the system) Return to TOC Copyright © Cengage Learning. All rights reserved 26 Section 6.1 The Nature of Energy Internal Energy • • Sign reflects the system’s point of view. Surroundings do work on the system: w is positive • System does work on surroundings: w is negative Return to TOC Copyright © Cengage Learning. All rights reserved 27 Section 6.1 The Nature of Energy Work = F x d (from 9th grade) - applying to a gas • Work = P × A × Δh = PΔV P is pressure. A is area. Δh is the piston moving a distance. ΔV is the change in volume. Return to TOC Copyright © Cengage Learning. All rights reserved 28 Section 6.1 The Nature of Energy Work • • • For an expanding gas, ΔV is a positive quantity because the volume is increasing. Thus ΔV and w must have opposite signs: w = –PΔV To convert between L·atm and Joules, use 1 L·atm = 101.3 J. Return to TOC Copyright © Cengage Learning. All rights reserved 29 Work, Pressure, and Volume Expansion +V (increase) -w results Esystem decreases Work has been done by the system on the surroundings Compression -V (decrease) +w results Esystem increases Work has been done on the system by the surroundings Figure 6.4 The Piston, Moving a Distance Against a Pressure P, Does Work On the Surrounding s Copyright © Houghton Mifflin Company. All rights reserved. 6– Section 6.1 The Nature of Energy Exercise w = –PΔV •Which of the following performs more work? •a) A gas expanding against a pressure of 2 atm from 1.0 L to 4.0 L. b) A gas expanding against a pressure of 3 atm from 1.0 L to 3.0 L. Return to TOC Copyright © Cengage Learning. All rights reserved 32 Section 6.1 The Nature of Energy Exercise w = –PΔV •Which of the following performs more work? •a) A gas expanding against a pressure of 2 atm from 1.0 L to 4.0 L. b) A gas expanding against a pressure of 3 atm from 1.0 L to 3.0 L. • They perform the same amount of work. (-6 AllL*atm) Copyright © Cengage•Learning. rights reserved Return to TOC 33 E = q + w E = change in internal energy of a system q = heat flowing into or out of the system -q if energy is leaving to the surroundings +q if energy is entering from the surroundings w = work done by, or on, the system -w if work is done by the system on the surroundings +w if work is done on the system by the surroundings Section 6.1 The Nature of Energy Concept Check • ΔE = q + w w = –PΔV •Determine the sign of E for each of the following with the listed conditions: •a) An endothermic process that performs work on surrounding. |work| > |heat| Δ E = negative wk (-) heat + |work| < |heat| Δ E = positive •b) Work is done on a gas and the process is exothermic. |work| > |heat| Δ E = positive wk + and heat |work| < |heat| Δ E = negative Return to TOC Copyright © Cengage Learning. All rights reserved 35 Section 6.1 The Nature of Energy Thermodynamic Quantities Return to TOC Copyright © Cengage Learning. All rights reserved 36 Energy Change in Chemical Processes Endothermic: Reactions in which energy flows into the system as the reaction proceeds. + qsystem - qsurroundings Energy Change in Chemical Processes Exothermic: Reactions in which energy flows out of the system as the reaction proceeds. - qsystem + qsurroundings Section 6.1 The Nature of Energy Exercise 1 • Exercise 1 Internal Energy • Calculate ΔE for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system. Return to TOC Copyright © Cengage Learning. All rights reserved 39 Section 6.1 The Nature of Energy Exercise 1 • Exercise 1 Internal Energy • Calculate ΔE for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system. • ANSWER: 17.0 kJ Return to TOC Copyright © Cengage Learning. All rights reserved 40 Section 6.1 The Nature of Energy Exercise 2 •Exercise 2 PV Work •Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm. Return to TOC Copyright © Cengage Learning. All rights reserved 41 Section 6.1 The Nature of Energy Exercise 2 •Exercise 2 PV Work •Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm. •ANSWER: ‒270 L•atm Return to TOC Copyright © Cengage Learning. All rights reserved 42 Section 6.1 The Nature of Energy Exercise 3 Internal Energy, Heat, and Work •Exercise 3 A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from 4.00 • 106 L to 4.50 • 106 L by the addition of 1.3 • 108 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate ΔE for the process. (To convert between L ⋅ atm and J, use 1 L ⋅ atm = 101.3 J.) Return to TOC Copyright © Cengage Learning. All rights reserved 43 Section 6.1 The Nature of Energy Exercise 3 Internal Energy, Heat, and Work •Exercise 3 A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from 4.00 • 106 L to 4.50 • 106 L by the addition of 1.3 • 108 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate ΔE for the process. (To convert between L ⋅ atm and J, use 1 L ⋅ atm = 101.3 J.) •ANSWER: 8.0 • 107 J Return to TOC Copyright © Cengage Learning. All rights reserved 44 Stopped here 2-5-13 - need to make ch.3-4 test shorter. Section 6.2 Temperature and Heat Temperature - Check out this simulation regarding the effect temperature on molecular motion. • Click on the following link and • http://phet.colorado.edu/en/simulation/states-of-matterbasics • Click on Run Now. Open with Java. • Adjust the temperatures, elements, and states of matter to see what happens from the solid, liquid, gas tab. Return to TOC Copyright © Cengage Learning. All rights reserved 46 Section 6.2 Temperature and Heat Temperature • A measure of the random motions of the components of a substance. Temperature (T)—is proportional to the average kinetic energy of the molecules, KEave . “Heat ‘em up and speed ‘em up”. Return to TOC Copyright © Cengage Learning. All rights reserved 47 Section 6.2 Temperature and Heat Heat • A flow of energy between two objects due to a temperature difference between the objects. Heat is the way in which thermal energy is transferred from a hot object to a colder object. Return to TOC Copyright © Cengage Learning. All rights reserved 48 Section 6.2 Temperature and Heat Heat • Two systems with different temperatures that are in thermal contact will exchange thermal energy, the quantity of which is call heat. This transfer of energy in a process (flows from a warmer object to a cooler one, transfers heat because of temperature difference but, remember, temperature is not a measure of energy—it just reflects the motion of particles) Return to TOC Copyright © Cengage Learning. All rights reserved 49 Section 6.2 Enthalpy Enthalpy and Calorimetry •Enthalpy (H)– flow of energy (heat exchange) at constant pressure when two systems are in contact. • Enthalpy of reaction (ΔHrxn) – amount of heat released (negative values) or absorbed (positive values) by a chemical reaction at constant pressure in kJ/molrxn • Enthalpy of combustion (ΔHcomb)—heat absorbed or released by burning (usually with O2) in kJ/molrxn; note that combustion reactions yield oxides of that which is combusted • Enthalpy of formation (ΔHf) – heat absorbed or released when ONE mole of compound is formed from elements in their standard states in kJ/molrxn • Enthalpy of fusion (ΔHfus)—heat absorbed to melt (overcome IMFs) 1 mole of solid to liquid @ MP expressed in kJ/molrxn • Enthalpy of vaporization (ΔHvap)—heat absorbed to vaporize or Copyright Cengage Learning. All rights reserved 1 mole liquid to vapor @BP in kJ/molrxn boil ©(overcome IMFs) 50 Return to TOC Section 6.2 Enthalpy and Calorimetry Change in Enthalpy (ΔH ) • • • State function ΔH = q at constant pressure (ex. atmospheric pressure) ΔH = Hproducts – Hreactants • Measure only the change in enthalpy, ΔH ( the difference between the potential energies of the products and the reactants) • A system that releases heat to the surroundings, an exothermic reaction, creates a negative ΔH because the enthalpy of the products is lower than the enthalpy of the reactants of the system (Figure 1). • A system that absorbs heat from the surroundings, an endothermic reaction, creates a positive ΔH, because the enthalpy of the products is higher than the enthalpy of the reactants of the system. • A positive ΔH is endothermic rxn • A negative ΔH is exothermic rxn Copyright © Cengage Learning. All rights reserved Return to TOC 51 Section 6.2 Enthalpy and Calorimetry Enthalpy Calculations • Enthalpy can be calculated from several • sources including: • • • • • Stoichiometry Calorimetry From tables of standard values Hess’s Law Bond energies Return to TOC Copyright © Cengage Learning. All rights reserved 52 Section 6.2 Stoichiometrically: Enthalpy and Calorimetry • Exercise 4 • Upon adding solid potassium hydroxide pellets to water the following reaction takes place: • KOH(s) → KOH(aq) + 43 kJ/mol • Answer the following questions regarding the addition of 14.0 g of KOH to water: • a) Does the beaker get warmer or colder? How do you know? • b) Is the reaction endothermic or exothermic? • c) What is the enthalpy change for the dissolution of the • 14.0 grams of KOH? Return to TOC Copyright © Cengage Learning. All rights reserved 53 Section 6.2 Stoichiometrically: Enthalpy and Calorimetry • Exercise 4 • Upon adding solid potassium hydroxide pellets to water the following reaction takes place: • KOH(s) → KOH(aq) + 43 kJ/mol • Answer the following questions regarding the addition of 14.0 g of KOH to water: • a) Does the beaker get warmer or colder? • b) Is the reaction endothermic or exothermic? • c) What is the enthalpy change for the dissolution of the • 14.0 grams of KOH? • ANSWER: (a) warmer (the +43 kJ/mole is heat on the product side) (b) Return to TOC exothermic (c) −10.7 kJ/molrxn (take 14 g divide by the molar mass of KOH 54 - to get # Copyright © Cengage Learning. All rights reserved Section 6.2 Enthalpy and Calorimetry Exercise 5 • Consider the combustion of propane: • C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) • ΔH = – 2221 kJ •Assume that all of the heat comes from the combustion of propane. Calculate ΔH in which 5.00 g of propane is burned in excess oxygen at constant pressure. Return to TOC Copyright © Cengage Learning. All rights reserved 55 Section 6.2 Enthalpy and Calorimetry Exercise 5 • Consider the combustion of propane: • C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) • ΔH = – 2221 kJ •Assume that all of the heat comes from the combustion of propane. Calculate ΔH in which 5.00 g of propane is burned in excess oxygen at constant pressure. Return to TOC Copyright © Cengage Learning. All rights reserved 56 Section 6.2 Calorimetry Enthalpy and Calorimetry • Science of measuring heat • Heat capacity – energy required to raise temp. by 1 degree (Joules/ °C) – Specific heat capacity: (s) • The energy required to raise the temperature of one gram of a substance by one degree Celsius. – Molar heat capacity: (Cp) • The energy required to raise the temperature of one mole of substance by one degree Celsius. Copyright © Cengage Learning. All rights reserved 57 Return to TOC Section 6.2 Enthalpy and Calorimetry Calorimetry • If two reactants at the same temperature are mixed and the resulting solution gets warmer, this means the reaction taking place is exothermic. • An endothermic reaction cools the solution. Return to TOC Copyright © Cengage Learning. All rights reserved 58 Calorimetry The amount of heat absorbed or released during a physical or chemical change can be measured, usually by the change in temperature of a known quantity of water in a calorimeter. Section 6.2 Enthalpy and Calorimetry A Coffee–Cup Calorimeter Made of Two Styrofoam Cups Coffee-cup calorimetry – in the lab this is how we experiment to find energy of a particular system. We use a Styrofoam® cup, reactants that begin at the same temperature and look for change in temperature. After all data is collected (mass or volume; initial and final temperatures) we can use the specific formula to find the energy released or absorbed. We refer to this process as constant pressure calorimetry. ** q = ΔH @ these conditions** Return to TOC Copyright © Cengage Learning. All rights reserved 60 Section 6.2 Enthalpy and Calorimetry Calorimetry • • • • q = s × m × ΔT • q = Energy released (heat) (or gained) s = specific heat capacity (J/°C·g) m = mass (g) ΔT = change in temperature (°C) • (Q = Cp x m × ΔT) - some use Cp as specific heat variable. Copyright © Cengage Learning. All rights reserved Return to TOC 61 Units for Measuring Heat The Joule is the SI system unit for measuring heat: The calorie is the heat required to raise the temperature of 1 gram of water by 1 Celsius degree 1 BTU (British Thermal Unit) is the heat required to raise the temperature of 1 pound of water by 1 F Section 6.2 Measuring Energy Changes • The common energy units for heat are the calorie and the joule. calorie – the amount of energy (heat) required to raise the temperature of one gram of water 1oC. Joule – 1 calorie = 4.184 joules Return to TOC Copyright © Cengage Learning. All rights reserved 63 Section 6.2 Measuring Energy Changes Exercise 6 • Convert 60.1 cal to joules and kJ 1000 J = 1 kiloJoule (kJ) 251 J = 0.251 kJ Return to TOC Copyright © Cengage Learning. All rights reserved 64 Section 6.2 Measuring Energy Changes • • • • • • Energy (q) released or gained at constant pressure: q = m x s x ΔT q = quantity of heat (Joules or calories) m = mass in grams ΔT = Tf −Ti (final – initial) s = specific heat capacity ( J/g°C) (or Cp with some) • Specific heat of water (liquid state) = 4.184 J/g°C (or 1.00 cal/g°C) • *Water has one of the highest specific heats known! This property makes life on earth possible and regulates earth’s temperature year round! Return to TOC Copyright © Cengage Learning. All rights reserved 65 Section 6.2 Measuring Energy Changes Exercise 7 • Calculate the amount of heat energy (in joules) needed to raise the temperature of 6.25 g of water from 21.0°C to 39.0°C. • • Where are we going? We want to determine the amount of energy needed to increase the temperature of 6.25 g of water from 21.0°C to 39.0°C. What do we know? The mass of water and the temperature increase. • • Return to TOC Copyright © Cengage Learning. All rights reserved 66 Section 6.2 Measuring Energy Changes q = s × m × ΔT Example 7 • Calculate the amount of heat energy (in joules) needed to raise the temperature of 6.25 g of water from 21.0°C to 39.0°C. • • What information do we need? We need the specific heat capacity of water. 4.184 J/g°C How do we get there? • Return to TOC Copyright © Cengage Learning. All rights reserved 67 Section 6.2 Measuring Energy Changes q = s × m × ΔT Exercise 8 •A sample of pure iron requires 142 cal of energy to raise its temperature from 23ºC to 92ºC. What is the mass of the sample? (The specific heat capacity of iron is 0.45 J/gºC.) – – – – a) b) c) d) 0.052 g 4.6 g 19 g 590 g Return to TOC Copyright © Cengage Learning. All rights reserved 68 Section 6.2 Measuring Energy Changes q = s × m × ΔT Exercise 8 •A sample of pure iron requires 142 cal of energy to raise its temperature from 23ºC to 92ºC. What is the mass of the sample? (The specific heat capacity of iron is 0.45 J/gºC.) – – – – a) b) c) d) 0.052 g 4.6 g 19 g 590 g Return to TOC Copyright © Cengage Learning. All rights reserved 69 Section 6.2 q = s × m × ΔT Enthalpy and Calorimetry Concept Check Example 9 •A 100.0 g sample of water at 90°C is added to a 100.0 g sample of water at 10°C. •The final temperature of the water is: – a) Between 50°C and 90°C – b) 50°C – c) Between 10°C and 50°C Return to TOC Copyright © Cengage Learning. All rights reserved 70 Section 6.2 q = s × m × ΔT Enthalpy and Calorimetry Concept Check Example 9 •A 100.0 g sample of water at 90°C is added to a 100.0 g sample of water at 10°C. •The final temperature of the water is: – a) Between 50°C and 90°C – b) 50°C - work on next screens – c) Between 10°C and 50°C Return to TOC Copyright © Cengage Learning. All rights reserved 71 Section 6.2 Enthalpy and Calorimetry Problem • We start by calling the final, ending temperature 'x.' Keep in mind that BOTH water samples will wind up at the temperature we are calling 'x.' Also, make sure you understand that the 'x' we are using IS NOT the Δt, but the FINAL temperature. This is what we are solving for. • The warmer water goes down from to 90 to x, so this means its Δt equals 90 minus x. The colder water goes up in temperature, so its Δt equals x minus 10. • That last paragraph may be a bit confusing, so let's compare it to a number line: • ---10-------------------------x-----------------------------------90---- • To compute the absolute distance, it's the larger value minus the smaller value, so 90 to x is 90 minus x and the distance from x to 10 is x - 10. • These two distances on the number line represent our two Δt values: • a) the Δt of the warmer water is 90 minus x • b) the Δt of the cooler water is x minus 10 Return to TOC Copyright © Cengage Learning. All rights reserved 72 Section 6.2 Problem Enthalpy and Calorimetry • q lost = q gain • q = (mass) (Δt) (s) • So: • (mass) (Δt) (s) = (mass) (Δt) (s) • With qlost on the left side and qgain on the right side. Since these are both liquid water the specific heat capacity (s) are equal and can be eliminated from the equation. Also, since the amount of water is the same, it too could be eliminated. Then we let x be final temperature • Substituting values into the above, we then have • • • 90 - x = x - 10 Then solve for x (the final temperature) 100 = 2x so x = 50 The final temperature is at 50 degrees C. Return to TOC Copyright © Cengage Learning. All rights reserved 73 Section 6.2 Concept Check Enthalpy and Calorimetry Exercise 10 •A 100.0 g sample of water at 90.°C is added • to a 500.0 g sample of water at 10.°C. q = s × m × ΔT •The final temperature of the water is: – a) Between 50°C and 90°C – b) 50°C – c) Between 10°C and 50°C •Calculate the final temperature of the water. --------•100 x 4.19 x (90-t) = 500 x 4.19 x (t - 10) = •9000 -100t = 500t - 5000 so 14000 = 600t •t = 23.333 = 23°C Copyright © Cengage Learning. All rights reserved Return to TOC 74 Section 6.2 Enthalpy and Calorimetry Concept Check Exercise 11 q = s × m × ΔT •You have a Styrofoam cup with 50.0 g of water at 10.C. You add a 50.0 g iron ball at 90.C to the water. (sH2O = 4.18 J/°C·g and sFe = 0.45 J/°C·g) •The final temperature of the water is: – a) Between 50°C and 90°C – b) 50°C – c) Between 10°C and 50°C •Calculate the final temperature of the water. •(4.18) (t-10) = (.45) (90-t) t = 17.77 = 18°C Copyright © Cengage Learning. All rights reserved Return to TOC 75 Table 6.1 The Specific Heat Capacities of Some Common Substances Copyright © Houghton Mifflin Company. All rights reserved. 6– Section 6.2 Enthalpy and Calorimetry Type 2 Problem Exercise 12 A 110.0 g sample of Cu (specific heat capacity = .20 J/°C·g is heated to 82.4 °C and then placed in a container of water at 22.3°C. The final temperature of the water and the copper is 24.9°C. What was the mass of the water in the original container, assuming that all of the heat lost by the copper is gained by the water? The heat lost by copper is equal to: 110.0g Cu x 0.20 J/°C·g x (82.4°C - 24.9°C)= 1265 J The heat lost (-value) by Cu is equal to the heat gained by water. mH2O = q/(s x Δt) so 1265 J / (4.184 J x (24.9-22.3)) = 116.3 g water. Return to TOC Copyright © Cengage Learning. All rights reserved 77 120 The Heat Curve for Water, going from -20 to 120 oC, 78 The liquid is boiling at 100o 120 C; no temperature change (use q = mol x ΔHvaporization ) vap.) The gas temperature is rising from 100 to 120 oC (use (use q = s x m x ΔT) The liquid temperature is rising from 0 to 100 oC (q = s x m x ΔT) The solid is melting at 0o C; no temperature change (use q = mol x ΔHfusion) The solid temperature is rising from -20 to 0 oC (use q = s x m x ΔT) 79 Section 6.2 Exercise 13 Enthalpy and Calorimetry • 20.0 mL of pure water at 285 K is mixed with 48.0 mL of water at 315 K. What is the final temperature of the mix? Requires 5 steps. Assume all heat from warm water transferred to cold water. q gain = q lost (Density pure water is 1.0 g/mL so 1 gm/L = m/20 mL • • • • • a) b) c) d) e) m × ΔT x s = m × ΔT x s (20 g) (tf-285) (s) = (48 g) (315 - tf) (s) 20tf - 5700 = 15120 - 48tf 68tf = 20820 tf = 306.17 = 306 K Return to TOC Copyright © Cengage Learning. All rights reserved 80 Section 6.2 Exercise 14 Enthalpy and Calorimetry • 20.0 g of water at 28.5 C is mixed with 32.5 g of water at C. What is the final temperature after complete mixing? (Assume no energy is lost to the surroundings.) • q cold = q warm • • • • • a) b) c) d) e) 77.0 m × ΔT x s = m × ΔT x s (20 g) (tf-28.5) (s) = (32.5 g) (77 - tf) (s) (liquids - s cancels) 20tf - 570 = 2502.5 - 32.5tf 52.5tf = 3072.5 tf = 58.5 C Return to TOC Copyright © Cengage Learning. All rights reserved 81 Section 6.2 Exercise 15 Enthalpy and Calorimetry • Exercise 15 In a coffee cup calorimeter, 100.0 mL of 1.0 M NaOH and 100.0 mL of 1.0 M HCl are mixed. Both solutions were originally at 24.6°C. After the reaction, the final temperature is 31.3°C. Assuming that all solutions have a density of 1.0 g/cm3 and a specific heat capacity of 4.184 J/g°C, calculate the enthalpy change for the neutralization of HCl by NaOH. Assume that no heat is lost to the surroundings or the calorimeter. Return to TOC Copyright © Cengage Learning. All rights reserved 82 Section 6.2 Exercise 15 Enthalpy and Calorimetry • Exercise 15 In a coffee cup calorimeter, 100.0 mL of 1.0 M NaOH and 100.0 mL of 1.0 M HCl are mixed. Both solutions were originally at 24.6°C. After the reaction, the final temperature is 31.3°C. Assuming that all solutions have a density of 1.0 g/cm3 and a specific heat capacity of 4.184 J/g°C, calculate the enthalpy change for the neutralization of HCl by NaOH. Assume that no heat is lost to the surroundings or the calorimeter. • ANSWER: ‒5.6 kJ/mol (m = mass = 200 g (because we have 100 mL NaOH plus 100 mL HCl each with a density of 1 g / mL) Then using the equation q = m * s * ΔT = (200 g) (4.184 J/g*C) * 6.7 C) = 5606.56 J = 5.6 kJ • Since this is an exothermic reaction, q is negative. Return to TOC Copyright © Cengage Learning. All rights reserved 83 Section 6.2 Exercise 16 Enthalpy and Calorimetry • When 1 mole of methane (CH4) is burned at constant pressure, 890 kJ/mol of energy is released as heat. Calculate ΔH for a process in which a 5.8 gram sample of methane is burned at constant pressure. Return to TOC Copyright © Cengage Learning. All rights reserved 84 Section 6.2 Exercise 16 Enthalpy and Calorimetry • When 1 mole of methane (CH4) is burned at constant pressure, 890 kJ/mol of energy is released as heat. Calculate ΔH for a process in which a 5.8 gram sample of methane is burned at constant • pressure. ANSWER: ΔH = heat flow = ‒320 kJ/mol 5.8 g / 16.0 g = .3625 moles so set up proportion 1 mole/890 kJ = 0.36 mol/x so x = (890)(.36)/(1) = 320.4 = 320 kJ/mol and is -320 kJ/mol since exothermic reaction Return to TOC Copyright © Cengage Learning. All rights reserved 85 Section 6.2 Exercise 17 Enthalpy and Calorimetry • Exercise 17 Constant-Pressure Calorimetry • When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0°C is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25C in a calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to 28.1C. Assuming that the calorimeter absorbs only a negligible quantity of heat, and that the specific heat capacity of the solution is 4.18 J/C ⋅ g, and that the density of the final solution is 1.0 g/mL, calculate the enthalpy change per mole of BaSO4 formed. Return to TOC Copyright © Cengage Learning. All rights reserved 86 Section 6.2 Exercise 17 Enthalpy and Calorimetry • Exercise 17 Constant-Pressure Calorimetry • When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0°C is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25°C in a calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to 28.1C. Assuming that the calorimeter absorbs only a negligible quantity of heat, and that the specific heat capacity of the solution is 4.18 J/C ⋅ g, and that the density of the final solution is 1.0 g/mL, calculate the enthalpy change per mole of BaSO4 formed. • ANSWER = ‒26 kJ/mol • 2 L total volume = 2000 mL x (1.0 g/mL) = 2000 grams of sol. • ΔH = q at constant pressure = q = s × m × ΔT • = (4.18 J/C*g) (2000 g) (28.1-25.0 C) = 25,916 Joules or 25.9 = 26 kJ per mole Return to TOC Copyright © Cengage Learning. All rights reserved 87 Specific Heat The amount of heat required to raise the temperature of one gram of substance by one degree Celsius. Substance Specific Heat (J/g·C) Water (liquid) 4.18 Ethanol (liquid) 2.44 Water (solid) 2.06 Water (vapor) 1.87 Aluminum (solid) 0.897 Carbon (graphite,solid) 0.709 Iron (solid) 0.449 Copper (solid) 0.385 Mercury (liquid) 0.140 Lead (solid) 0.129 Gold (solid) 0.129 Section 6.2 Enthalpy and Calorimetry END OF NOTES 6.1-6.2 • HW: Read Section 6.1-6.3 by next Tuesday. • HW: Pre-lab Calorimetry due on Friday for lab • Turn in assignments including take home test ch. 5 if not done already no later than Friday if not turned in today. Return to TOC Copyright © Cengage Learning. All rights reserved 89