Chapter 6 Energy and Chemical Change Brady and Senese 5th Edition
Index 6.1 An object has energy if it is capable of doing work 6.2 Internal energy is the total energy of an object s molecules 6.3 Heat can be determined by measuring temperature changes 6.4 Energy is absorbed or released during most chemical reactions 6.5 Heats of reaction are measured at constant volume or constant pressure 6.6 Thermochemical equations are chemical equations that quantitatively include heat 6.7 Thermochemical equations can be combined because enthalpy is a state function 6.8 Tabulated standard heats of reaction can be used to predict any heat of reaction using Hess s law
Energy is the Ability to do Work Energy is the ability to do work (move mass over a distance) or transfer heat Types: kinetic and potential Kinetic - the energy of motion Potential - the stored energy in matter Internal energy (E) - the sum of the kinetic and potential energy for each particle in the system 6.1 An object has energy if it is capable of doing work 3
Kinetic Energy: The Energy of Motion KE = ½mv 2 Energy can be transferred by moving particles Collision of fast particles with slower particles causes the slow particle to speed up while the fast molecule slows This is why hot water cools in contact with cool air 6.1 An object has energy if it is capable of doing work 4
Potential Energy Depends on Position Potential energy increases when: Objects that attract move apart, or Objects that repel move toward each other Stored energy that can be converted to kinetic energy 6.1 An object has energy if it is capable of doing work 5
Your Turn! Which of the following is not a form of kinetic energy? A. A pencil rolls across a desk B. A pencil is sharpened C. A pencil is heated D. All are forms of kinetic energy E. None are forms of kinetic energy 6.1 An object has energy if it is capable of doing work 6
Law of Conservation of Energy Energy cannot be created or destroyed, but can be transformed from one form of energy to another Also known as the first law of thermodynamics How does water falling over a waterfall demonstrate this law? 6.1 An object has energy if it is capable of doing work 7
Heat and Temperature are Not the Same The temperature of an object is proportional to the average kinetic energy of its particles the higher the average kinetic energy, the higher the temperature Heat is energy (also called thermal energy) transferred between objects caused by differences in their temperatures until they reach thermal equilibrium 6.1 An object has energy if it is capable of doing work 8
Units of Energy SI unit is the Joule, J J = kg m 2 /s 2 If the calculated value is greater than 1000 J, use the kj Another unit is the calorie, cal cal = 4.184 J (exact) Nutritional unit is the Calorie (note capital C), which is one kilocalorie 1 Cal = 1 kcal = 4.184 kj 6.1 An object has energy if it is capable of doing work 9
Internal Energy is Conserved 1 st Law of Thermodynamics: For an isolated system the internal energy (E) is constant: Δ E = E f - E i = 0 Δ E = E product - E reactant = 0 We can t measure the internal energy of anything, so we measure the changes in energy E is a state function 6.2 Internal energy is the total energy of an object s molecules 10
What is Temperature? Temperature (T) is proportional to the average kinetic energy of all particle units: C, F, K KE average = ½ mv average 2 At a high temperature, most molecules are moving at higher average 6.2 Internal energy is the total energy of an object s molecules 11
State Function A property whose value depends only on the present state of the system, not on the method or mechanism used to arrive at that state Position is a state function: both train and car travel to the same locations although their paths vary The actual distance traveled does vary with path New York Los Angeles 6.2 Internal energy is the total energy of an object s molecules 12
Heat Transfer is a State Function Transfer of heat during a reaction is a state function. The route taken to arrive at the products does not affect the amount of heat that is transferred. The number of steps does not affect the amount of heat that is transferred. 6.3 Heat can be determined by measuring temperature changes 13
Heat Transfer, q Heat (q) - the transfer of energy from regions of high temperature to regions of lower temperature Units: J, cal, kg m 2 /s 2 A calorie is the amount of energy needed to raise the temperature of 1.00 g water from 14.5 to 15.5 C A metal spoon at 25 C is placed in boiling water. What happens? 6.3 Heat can be determined by measuring temperature changes 14
Surroundings / System / Universe System - the reaction or area under study Surroundings - the rest of the universe Open systems can gain or lose mass and energy across their boundaries i.e. the human body Closed systems can absorb or release energy, but not mass, across the boundary i.e. a light bulb Isolated systems (adiabatic) cannot exchange matter or energy with their surroundings i.e. a closed Thermos bottle 6.3 Heat can be determined by measuring temperature changes 15
The Sign Convention Endothermic systems require energy to be added to the system, thus the q is (+) Exothermic reactions release energy to the surroundings. Their q is (-) Energy changes are measured from the point of view of the system 6.3 Heat can be determined by measuring temperature changes 16
Your Turn! A cast iron skillet is moved from a hot oven to a sink full of water. Which of the following is not true? A. The water heats B. The skillet cools C. The heat transfer for the skillet has a (-) sign D. The heat transfer for the skillet is the same as the heat transfer for the water E. None of these are untrue 6.3 Heat can be determined by measuring temperature changes 17
Heat Capacity and Transfer Heat capacity (C) - the (extensive) ability of an object with constant mass to absorb heat. Calorimeter constant Varies with the sample mass and the identity of the substance Units: J C -1 q = C t q = heat transferred C = heat capacity of object Δt = Change in Temperature (t final - t initial ) 6.3 Heat can be determined by measuring temperature changes 18
Learning Check A cup of water is used in an experiment. Its heat capacity is known to be 720 J C -1. How much heat will it absorb if the experimental temperature changed from 19.2 C to 23.5 C? q = C t q = 720 J C -1 (23.5-19.2 C) q = 3.1 10 3 J 6.3 Heat can be determined by measuring temperature changes 19
Heat Transfer and Specific Heat Specific heat (s) - The intensive ability of a substance to store heat. C = m s Units: J g -1 C -1 or J g -1 K -1 or J mol -1 K - 1 q = m t s q = heat transferred m = mass of object Δt = change in temperature (t final - t initial ) 6.3 Heat can be determined by measuring temperature changes 20
Specific Heats Substances with high specific heats resist temperature changes Note that water has a very high specific heat This is why coastal temperatures are different from inland temperatures. Substance Specific Heat J g -1 C -1 (25 C) Carbon (graphite) 0.711 Copper 0.387 Ethyl alcohol 2.45 Gold 0.129 Granite 0.803 Iron 0.4498 Lead 0.128 Olive oil 2.0 Silver 0.235 Water (liquid) 4.18 6.3 Heat can be determined by measuring temperature changes 21
Learning Check Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a 32.91 g sample by 2.53 C. q = m s t q 235 J J s = = = 2.82 m t 32.91 g 2.53 C g C 6.3 Heat can be determined by measuring temperature changes 22
The First Law of Thermodynamics Explains Heat Transfer If we monitor the heat transfers (q) of all materials involved and all work processes, we can predict that their sum will be zero By monitoring the surroundings, we can predict what is happening to our system Heat transfers until thermal equilibrium, thus the final temperature is the same for all materials 6.3 Heat can be determined by measuring temperature changes 23
Learning Check A 43.29 g sample of solid is transferred from boiling water (t = 99.8 C) to 152 g water at 22.5 C in a coffee cup. The t water rose to 24.3 C. Calculate the specific heat of the solid. q sample + q water + q cup = 0 q cup is neglected in problem q sample = -q water q sample = m s t q sample = 43.29 g s (24.3-99.8 C) q water = 152 g 4.184 J g -1 C -1 (24.3 22.5 C) 43.29 g s (24.3-99.8 C) = -(152 g 4.184 J g -1 C -1 (24.3 22.5 C)) s (-3.27 10 3 ) g -1 C -1 = -1.14 10 3 J s = 0.349 J g -1 C -1 6.3 Heat can be determined by measuring temperature changes 24
Your Turn! What is the heat capacity of the container if 100. g of water (s = 4.184 J g -1 C -1 ) at 100. C are added to 100. g of water at 25 C in the container and the final temperature is 61 C? A. 870 J/ C B. 35 J/ C C. -35 J/ C D. -870 J/ C E. None of these 6.3 Heat can be determined by measuring temperature changes 25
Chemical Potential Energy Chemical bond - net attractive forces that bind atomic nuclei and electrons together Exothermic reactions form stronger bonds in the product than in the reactant and release energy Endothermic reactions break stronger bonds than they make and require energy 6.4 Energy is absorbed or released during most chemical reactions 26
Work and Pistons Pressure = force/area If the container volume changes, the pressure changes Work = -P ΔV Units: L atm 1 L atm = 101 J In expansion, V > 0, and is exothermic Work is done by the system in expansion 6.5 Heats of reaction are measured at constant volume or constant pressure 27
How does work relate to reactions? Work = Force Distance Is most often due to the expansion or contraction of a system due to changing moles of gas. Gases push against the atmospheric pressure, so P system = -P atm w = -P atm ΔV The deployment of an airbag is one example of this process. 1C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(g) 6 moles of gas 7 moles of gas 6.5 Heats of reaction are measured at constant volume or constant pressure 28
Learning Check: P-V work Ethyl chloride is prepared by reaction of ethylene with HCl. How much P-V work (in J) is done if 89.5 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -71.5 L? (1 L atm = 101 J) w = -1atm -71.5 L = 71.5 L atm w = 7.22 10 3 J Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 8.6 L to 4.3 L at a constant external pressure of 44 atm. w = -44 atm (4.3-8.6) L = 19 L atm w = 19 kj 6.5 Heats of reaction are measured at constant volume or constant pressure 29
Energy can be Transferred as Heat and Work E = q + w Internal energy changes are state functions 6.5 Heats of reaction are measured at constant volume or constant pressure 30
Your Turn! When TNT is combusted in air, it is according to the following reaction: 4C 6 H 2 (NO 2 ) 3 CH 3 (s) + 17O 2 (g) 24CO 2 (g) + 10H 2 O(l) + 6N 2 (g) The reaction will do work for all of these reasons except: A. The moles of gas increase B. The volume of gas increases C. The pressure of the gas increases D. None of these 6.5 Heats of reaction are measured at constant volume or constant pressure 31
Calorimetry is Used to Measure Heats of Reaction Heat of reaction - the amount of heat absorbed or released in a chemical reaction Calorimeter - an apparatus used to measure temperature changes in materials surrounding a reaction that result from a chemical reaction From the temperature changes we can calculate the heat of the reaction, q q v ; heat measured under constant volume conditions q p : heat measured under constant pressure conditions q p : termed enthalpy, H 6.5 Heats of reaction are measured at constant volume or constant pressure 32
Internal Energy is Measured with a Bomb Calorimeter Used for reactions in which there is change in the number of moles of gas present Measures q v Immovable walls mean that work is zero ΔE = q v 6.5 Heats of reaction are measured at constant volume or constant pressure 33
Learning Check: Bomb Calorimeter A sample of 500. mg naphthalene (C 10 H 8 ) is combusted in a bomb calorimeter containing 1000. g of water. The temperature of the water increases from 20.00 C to 24.37 C. The calorimeter constant is 420 J/ C. What is the change in internal energy for the reaction? s water = 4.184 J/g C q v reaction + q water + q cal = 0 by the first law q water = 1000. g (24.37-20.00) C 4.184 J/g C q cal = 420 J/g (24.37-20.00) C q v reaction = E = -2.0 10 4 J 6.5 Heats of reaction are measured at constant volume or constant pressure 34
Enthalpy of Combustion When one mole of a fuel substance is reacted with elemental oxygen, a combustion reaction can be written Is always negative Learning Check: What is the equation associated with the enthalpy of combustion of C 6 H 12 O 6 (s)? C 6 H 12 O 6 (s) + 9O 2 (g) 6CO 2 (g) + 6H 2 O(l) 6.5 Heats of reaction are measured at constant volume or constant pressure 35
Your Turn! A 252 mg sample of benzoic acid, C 6 H 5 CO 2 H, is combusted in a bomb calorimeter containing 814 g water at 20.00 C. The reaction increases the temperature of the water to 21.70 C. What is the internal energy released by the process? A. -711 J B. -2.85 J C. +711 J D. +2.85 J E. None of these q w + q cal + q v reaction = 0; q cal is ignored by the problem q v reaction = -q w = -814 g (21.70-20.00) C 4.184 J g -1 q v reaction = -5789 J -5.79 kj s water = 4.184 J/g C 6.5 Heats of reaction are measured at constant volume or constant pressure 36
Enthalpy Change (ΔH) Enthalpy is the heat transferred at constant pressure ΔH = q p ΔE = q p - PΔV = ΔH - PΔV ΔH = ΔH final - ΔH initial ΔH = ΔH product - ΔH reactant 6.5 Heats of reaction are measured at constant volume or constant pressure 37
Enthalpy Measured in a Coffee Cup Calorimeter When no change in moles of gas is expected, we may use a coffee cup calorimeter The open system allows the pressure to remain constant Thus we measure q p E = q p + w or E = H P V Since there is no change in the moles of gas present, there is no work Thus we also are measuring E 6.5 Heats of reaction are measured at constant volume or constant pressure 38
Learning Check: Coffee Cup Calorimetry When 50.0 ml of 0.987 M H 2 SO 4 is added to 50.0 ml of 1.00 M NaOH at 25.0 C in a coffee cup calorimeter, the temperature of the aqueous solution increases to 31.7 C. Calculate heat for the reaction per mole of limiting reactant. Assume that the specific heat of the solution is 4.18 J/g C, the density is 1.00 g/ml, and that the calorimeter itself absorbs a negligible amount of heat H 2 SO 4 (aq) + 2NaOH(aq) 2H 2 O(l) + Na 2 SO 4 (aq) mol H 2 SO 4 = 0.0494 mol mol NaOH = 0.0500 mol, is limiting q p rxn + q cal + q soln = 0, thus q p rxn = -q soln q soln = 100 g soln (31.7-25.0) C 4.18 J/g C q p rxn = -2.8 10 3 J q p rxn = -5.6 10 4 J 6.5 Heats of reaction are measured at constant volume or constant pressure 39
Your Turn! A sample of 50.00 ml of 0.125 M HCl at 22.36 C is added to a 50.00 ml of 0.125 M Ca(OH) 2 at 22.36 C. The calorimeter constant was 72 J g -1 C -1. The temperature of the solution (s = 4.184 J g -1 C -1, d = 1.00 g/ml) climbed to 23.30 C. Which of the following is not true? A. q cal = 67.7 J B. q solution = 393.3 J C. q rxn = 461.0 J D. q rxn = -461.0 J E. None of these 6.5 Heats of reaction are measured at constant volume or constant pressure 40
Calorimetry Overview The equipment used depends on the reaction type. If there will be no change in the moles of gas, we may use a coffee-cup calorimeter or a closed system. Under these circumstances, we measure q p. If there is a large change in the moles of gas, we use a bomb calorimeter to measure q v. 6.5 Heats of reaction are measured at constant volume or constant pressure 41
Thermochemical Equations Relate the energy of a reaction to the quantities involved Must be balanced, but may use fractional coefficients Quantities are presumed to be in moles Example: C(s) + O 2 (g) CO 2 (g) H = -393.5 kj 6.6 Thermochemical equations are chemical equations that quantitatively include heat 42
Learning Check 2C 2 H 2 (g) + 5O 2 (g) 4CO 2 (g) + 2H 2 O(g) ΔH = -2511 kj The reactants (acetylene and oxygen) have 2511 kj more energy than the products. How many kj are released for 1 mol C 2 H 2? 1256 kj 6.6 Thermochemical equations are chemical equations that quantitatively include heat 43
Learning Check 6CO 2 (g) + 6H 2 O(l) C 6 H 12 O 6 (s) + 6O 2 (g) H = 2816 kj How many kj are required for 44 g CO 2 (molar mass = 44.01 g/mol)? 470 kj If 100. kj are provided, what mass of CO 2 can be converted to glucose? 9.38 g 6.6 Thermochemical equations are chemical equations that quantitatively include heat 44
Learning Check: Calorimetry of Chemical Reactions The meals-ready-to-eat (MRE) in the military can be heated on a flameless heater. Assume the reaction in the heater is Mg(s) + 2H 2 O(l) Mg(OH) 2 (s) + H 2 (g) ΔH = -353 kj What quantity of magnesium is needed to supply the heat required to warm 25 ml of water from 25 to 85 C? Specific heat of water = 4.184 J g -1 C -1. Assume the density of the solution is the same as for water at 25 C, 1.00 g ml -1 mass soln = 25 ml 1.00 g ml -1 = 25 g q soln = 25 g (85-25) C 4.184 J g -1 C -1 = 6.3 (6.3 10 kj)(1 3 J mol Mg/353 kj)(24.3 g mol -1 Mg) = 0.43 g 6.6 Thermochemical equations are chemical equations that quantitatively include heat 45
Your Turn! Consider the thermite reaction. The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is -848 kj mol -1 Fe 2 O 3 at 298 K. 2Al(s) + Fe 2 O 3 (s) 2Fe(s) + Al 2 O 3 (s) What mass of Fe (molar mass: 55.847 g mol -1 ) is made when 500 kj are released? A. 65.9 g B. 0.587 g C. 32.8 g D. None of these 6.6 Thermochemical equations are chemical equations that quantitatively include heat 46
Learning Check: Ethyl Chloride Reaction Revisited Ethyl chloride is prepared by reaction of ethylene with HCl: C 2 H 4 (g) + HCl(g) C 2 H 5 Cl(g) ΔH = -72.3 kj What is the value of ΔE if 89.5 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -71.5 L? mol HCl: 3.43 mol mol C 2 H 4 : 3.19 mol, is limiting ΔH rxn =3.19mol -72.3 w = -1 atm -71.5 L kj/mol = 71.5 L atm = 7.222 kj =-230.6 kjδe= -230.6kJ+7.2kJ= -223 kj Ethylene = 28.05 g/mol; HCl = 36.46 g/mol 6.6 Thermochemical equations are chemical equations that quantitatively include heat 47
Enthalpy Diagram 6.7 Thermochemical equations can be combined because enthalpy is a state function 48
Hess s Law The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction For example: 2Fe(s) + 3/2O 2 (g) Fe 2 O 3 (s) H = -822.2 kj Fe 2 O 3 (s) + 2Al(s) Al 2 O 3 (s) + 2Fe(s) H = -848 kj 3/2O 2 (g) + 2Al(s) Al 2 O 3 (s) H = -822.2 kj + -848 kj -1670 kj 6.7 Thermochemical equations can be combined because enthalpy is a state function 49
Rules for Adding Thermochemical Reactions 1. When an equation is reversed written in the opposite direction the sign of H must also be reversed. 2. Formulas canceled from both sides of an equation must be for the substance in identical physical states. 3. If all the coefficients of an equation are multiplied or divided by the same factor, the value of H must likewise be multiplied or divided by that factor. 6.7 Thermochemical equations can be combined because enthalpy is a state function 50
Strategy for Adding Reactions Together 1. Choose the most complex compound in the equation (1) 2. Choose the equation (2 or 3 or ) that contains the compound 3. Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction 4. Look for the next most complex compound 6.7 Thermochemical equations can be combined because enthalpy is a state function 51
Hess s Law (Cont.) 5. Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient 6. Add the reactions together and cancel like terms 7. Add the energies together, modifying the enthalpy values in the same way that you modified the equation If you reversed an equation, change the sign on the enthalpy If you doubled an equation, double the energy 6.7 Thermochemical equations can be combined because enthalpy is a state function 52
Learning Check How can we calculate the enthalpy change for the reaction 2 H 2 (g) + N 2 (g) N 2 H 4 (g) using these equations? N 2 H 4 (g) + H 2 (g) 2NH 3 (g) 3H 2 (g) + N 2 (g) 2NH 3 (g) ΔH = -187.8 kj ΔH = -92.4 kj Reverse the first reaction (and change sign) 2NH 3 (g) N 2 H 4 (g) + H 2 (g) ΔH = +187.8 kj Add the second reaction (and add the enthalpy) 3H 2 (g) + N 2 (g) 2NH 3 (g) ΔH = -92.4 kj 2NH 3 (g) + 23H 2 (g) + N 2 (g) N 2 H 4 (g) + H 2 (g) + 2NH 3 (g) 2H 2 (g) + N 2 (g) N 2 H 4 (g) (187.8-92.4) = +95.4 kj 6.7 Thermochemical equations can be combined because enthalpy is a state function 53
Learning Check Calculate ΔH for 2C(s) + H 2 (g) C 2 H 2 (g) using: 2C 2 H 2 (g) + 5O 2 (g) 4CO 2 (g) + 2H 2 O(l) ΔH = -2599.2 kj C(s) + O 2 (g) CO 2 (g) ΔH = -393.5 kj H 2 O(l) H 2 (g) + ½ O 2 (g) ΔH = +285.9 kj -½(2C 2 H 2 (g) + 5O 2 (g) 4CO 2 (g) + 2H 2 O(l) ΔH = -2599.2) 2CO 2 (g) + H 2 O(l) C 2 H 2 (g) + 5/2O 2 (g) ΔH = 1299.6 2(C(s) + O 2 (g) CO 2 (g) 2C(s)+ 2O 2 (g) 2CO 2 (g) ΔH = -393.5 kj) ΔH = -787.0 kj -1(H 2 O(l) H 2 (g) + ½ O 2 (g) ΔH = +285.9 kj) H 2 (g) + ½ O 2 (g) H 2 O(l) ΔH = -285.9 kj 2C(s) + H 2 (g) C 2 H 2 (g) ΔH = +226.7 kj 6.7 Thermochemical equations can be combined because enthalpy is a state function 54
Your Turn! What is the energy of the following process: 6A + 9B + 3D + F 2G Given that: C A + 2B H = 20.2 kj/mol 2C + D E + B H = 30.1 kj/mol 3E + F 2G H = -80.1 kj/mol A. 70.6 kj B. -29.8 kj C. -111.0 kj D. None of these 6.7 Thermochemical equations can be combined because enthalpy is a state function 55
State Matters! C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(g) ΔH = -2043 kj C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(l) H = -2219 kj Note that there is a difference in energy because the states do not match If H 2 O(l) H 2 O(g) H = 44 kj/mol 4H 2 O(l) 4H 2 O(g) H = 176 kj/mol -2219 + 176 kj = -2043 kj 6.8 Tabulated standard heats of reaction can be used to predict any heat of reaction using Hess s law 56
Standard State Most stable form of the pure substance at 1 atm pressure Stated temperature. If temperature is not specified, assume 25 C Solutions are 1 M in concentration. Measurements made under standard state conditions have the mark: ΔH Most ΔH values are given for the most stable form of the compound or element. 6.8 Tabulated standard heats of reaction can be used to predict any heat of reaction using Hess s law 57
Determining the Most Stable State The most stable form of a substance: below the melting point is solid above the boiling point is gas between these temperatures is liquid What is the standard state of GeH 4? mp -165 C bp -88.5 C gas What is the standard state of GeCl 4? mp -49.5 C bp 84 C liquid 6.8 Tabulated standard heats of reaction can be used to predict any heat of reaction using Hess s law 58
Allotropes Are substances that have more than one form in the same physical state You should know which form is the most stable C, P, O and S all have multiple allotropes. Which is the standard state for each? C solid, graphite P solid, white O gas, O 2 S solid, rhombic 6.8 Tabulated standard heats of reaction can be used to predict any heat of reaction using Hess s law 59
Enthalpy of Formation Enthalpy of formation is the enthalpy change ΔH f for the formation of 1 mole of a substance in its standard state from elements in their standard states Note: ΔH f = 0 for an element in its standard state Learning Check: What is the equation that describes the formation of CaCO 3 (s)? Ca(s) + C(graphite) + 3/2O 2 (g) CaCO 3 (s) 6.8 Tabulated standard heats of reaction can be used to predict any heat of reaction using Hess s law 60
Calculating ΔH for Reactions Using ΔH f ΔH rxn = [sum of ΔH f of all products] [sum of ΔH f of all reactants] 2Fe(s) + 6H 2 O(l) 2Fe(OH) 3 (s) + 3H 2 (g) 0-285.8-696.5 0 ΔH rxn = 321.8 kj CO 2 (g) + 2H 2 O(l) 2O 2 (g) + CH 4 (g) -393.5-285.8 0-74.8 ΔH rxn = 890.3 kj 6.8 Tabulated standard heats of reaction can be used to predict any heat of reaction using Hess s law 61
Your Turn! What is the enthalpy for the following reaction? 2H 2 CO 3 (aq) + 2OH - (aq) 2H 2 O(l) + 2HCO 3 - (aq) H f º -699.65 kj/mol A. 96.5 kj B. -96.5 kj C. 48.2 kj D. -48.2 kj E. None of these -230.0 kj/mol -285.9 kj/mol -691.99 kj/mol