Energy Thermodynamics-study of energy and its interconversions Labs #22 Calorimetry 3/25/2017

Energy Work Heat Capacity to do work (or produce heat) Energy used to cause object with mass to move against a force Force x distance (force acting over distance) Heat Involves transfer of energy between two objects Energy used to cause temperature of object to increase Chemicals may store potential energy in bonds that can be released as heat energy 3/25/2017

Potential energy- energy due to position or composition Depends on position relative to pull of gravity Stored energy released when process occurs that changes its position When substances undergo chemical reactions, positions of atoms change, and energy is released or absorbed Chemical energy stored within molecules as consequence of attractions between electrons and atomic nuclei (electrostatic potential energy, Eel) Proportional to electrical charge on two interacting objects, Q1Q2, and inversely proportional to distance, d, separating them Based on magnitude of electron charge (1.60 x 10-19 C) If both have same sign, Eel is positive If opposite signs, Eel is negative Eel = KQ1Q2/d K = constant of proportionality, 8.99 x 109 J-m/C2 Lower energy of system/more strongly opposite charges interact, more stable the of system 3/25/2017

Kinetic energy- energy due to the motion of an object KE= ½ mv2 According to kinetic-molecular theory, all matter has thermal energy, because atoms and molecules are in constant motion Vibrate/rotate/move from one point to another More energy of molecular motion, higher temperature 3/25/2017

Temperature vs. Heat Temperature Heat Random motion of particles in substance Indicates direction heat energy will flow Heat Measure of energy content What is transferred (due to temperature difference between system and surroundings) Always flows from hotter to colder body 3/25/2017

State Functions (state property) property of system that depends only on its present state Independent of pathway (energy, temperature, pressure, enthalpy, entropy) Does not matter how you get there/how process carried out Liter of water behind dam has same potential energy for work regardless of whether it flowed downhill to dam, or was taken uphill to dam in bucket (PE state function dependent only on current position of water, not how it got there) If you climb mountain, you can take many paths to get to top, but once there, height is independent of path If NaCl is produced by one equation or many equations, change in heat still the same Work and heat are not state functions Travel to top of mountain by many paths, so distance traveled is dependent on path taken 3/25/2017

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Exothermic Reactions Give off energy (PE in chemical bonds) Products generally more stable (stronger bonds) than reactants Total chemical energy of products less than total chemical energy of reactants Heat released during chemical reaction comes from decrease in chemical potential energy as reactants are converted to products 3/25/2017

Endothermic Reactions Energy absorbed from surroundings Energy (heat) flows into system, increasing its PE Products generally less stable (weaker bonds) than reactants Products have greater chemical energy than reactants Difference in energy between reactants/products is equal to heat absorbed by system 3/25/2017

Thermodynamics (Euniverse is constant) First law of thermodynamics (law of conservation of energy) energy can be converted from one form to another, but cannot be created or destroyed (Euniverse is constant) 3/25/2017

Thermodynamic quantities ΔE = Efinal -Einitial Number/unit-gives magnitude of change Sign-indicates direction of flow +q (ΔE) Efinal > Einitial Endothermic reactions Energy flows into system (gained energy) System’s energy is increasing -q (ΔE) Efinal < Einitial Exothermic reactions Energy flows out of system (lost energy) System’s energy is decreasing 3/25/2017

System Energy Internal energy, E, of system ΔE = q + w Sum of KE/PE of all particles in system Change energy of system by flow of work, heat or both Heat gained/word done on system both positive quantities Both increase internal energy of system, causing ΔE to be positive quantity ΔE = q + w ΔE = change in system’s internal energy q = heat w = work 3/25/2017

Sign conventions for q, w and ΔE + means system gains heat - Means system loses heat For w + means work done on system - Means work done by system For ΔE + means net gain of energy by system - Means net loss of energy by system 3/25/2017

Work Energy lost/gained by system by mechanical means, rather than by heat conduction Systems do not "contain" work or heat, but energy SI unit for energy (work, heat)-joule (J) Non-SI unit for heat-calorie (cal)-amount of heat needed to raise temperature of 1 gram of water by 1°C from 14.5°C to 15.5°C Heat or energy in calories can be converted to joules 1 cal = 4.184 J and 1 kcal = 4.184 kJ 3/25/2017

Since pressure = force / area, work = pressure  volume wsystem = PV work = force  distance Since pressure = force / area, work = pressure  volume wsystem = PV Units of pressure P = F/A = kg m s2/m2 = kg m-1 s-2 = 1 Pascal Units of volume P x V = kg m-1s-2 x m3 = kg m2 s-2 = 1 Joule 3/25/2017

Work done by gases: Work = P∆V for ideal gases (pressure constant) From point of view of system, W = -P∆V By gas (through expansion) System expands, positive work done on surroundings, negative work done on system ΔV is positive/w is negative To gas (by compression) System contracts, surrounding have done work on system, positive work done on system ΔV is negative/w is positive 3/25/2017

Calculate the work (with proper sign) associated with the contraction of a gas from 75 L to 30 L (work is done “on the system”) at a constant external pressure of 6.0 atm in: L atm ∆V = Vfinal –Vinitial = 30 L – 75 L = -45 L W = -P∆V = -6.0 atm (-45 L) = +270 L atm Joules (1 L atm = 101.3 J) +270 L atm 101.3 J = +2.7 x 104 J 1 L atm 3/25/2017

Calculate the change in energy of the system if 38 Calculate the change in energy of the system if 38.9 J of work is done by the system with an associated heat loss of 16.2 J. Get the sign correct: Q = - because heat is lost W = - because work is done by the system Solve: ΔE = q + w = -16.2 J + (-38.9 J) = -55.1 J The system has lost 55.1 J of energy. 3/25/2017

A piston is compressed from a volume of 8. 3 L to 2 A piston is compressed from a volume of 8.3 L to 2.8 L against a constant pressure of 1.9 atm. In the process, there is a heat gain by the system of 350 J. Calculate the change in energy of the system. w = -P∆V w = -1.9 atm (-5.5 L) = 10.45 L atm 10.45 L atm 101.3 J = +1059 J (q) L atm ΔE = q + w = 1059 + 350 = 1409 J = 1400 J 3/25/2017

Homework: Read 6.1, pp. 241-248 Q pp. 280-281, #9, 10, 18, 22, 24, 28 3/25/2017

Enthalpy (from Greek “to warm”) Formerly called “heat content” so it is still symbolized as H 3/25/2017

Enthalpy H = E + PV E = internal energy of system P = pressure of system V = volume of system Since internal energy, pressure and volume all state functions, enthalpy also state function (change in H independent of pathway) 3/25/2017

Enthalpy = ∆H = ∆E + ∆PV E = qP + w Since w = -P∆V E = qP  PV (qP = heat at constant P) qP = H only at constant pressure So H = E + PV Therefore, at constant pressure, qP = E + PV 3/25/2017

Enthalpy of reaction (rxn) or enthalpy change of system Heat transferred to (absorbed) or from (evolved) system at constant pressure) Difference between enthalpies of products and reactants ΔH = enthalpy of system after reaction – enthalpy of system before reaction Difference between sum of enthalpies of products and sum of enthalpies of reactants ΔH = ∑Hproducts – ∑Hreactants 3/25/2017

Changes in enthalpy can be positive or negative Endothermic reaction H(products) > H(reactants): H is positive, ΔH>0 Heat added (absorbed by system) is converted into potential energy of system Exothermic reaction H(reactants) > H(products): H is negative, ΔH<0 Loss of chemical potential energy accounts for the amount of heat released (given off by system) 3/25/2017

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Enthalpy Diagrams Each line represents set of reactants or products for balanced chemical reaction When going from one line to another, atoms must balance Enthalpy associated with condensation of water CO2(g) on both sides cancel to yield Relative distance of each line reflect relative enthalpy difference (ΔH) between reactants/products If enthalpy change in going from reactants to products is negative, line for products must be below reactants Length of distance must be proportional 3/25/2017

Upon adding solid sodium hydroxide pellets to water, the following reaction takes place: NaOH(s)  NaOH(aq). For this reaction at constant pressure, ∆H = -43 kJ/mol. Answer the following questions regarding the addition of 14 g of NaOH to water: Does the beaker get warmer or colder? If ∆H < 0, heat is given off by system-beaker gets warmer. Is the reaction exo- or endothermic? If heat is given off by system, reaction is exothermic. What is the enthalpy change for the dissolution? ∆H = -43 kJ 1 mol NaOH 14 g NaOH = -15kJ mol 40.0 g NaOH 3/25/2017

Calorimetry Measurement of heat flow (released or absorbed) in reaction by calorimeter Experimental technique used to determine heat exchange (q) associated with reaction All calorimetry is based on measuring temperature change (t) of medium such as water Techniques/equipment used depend on nature of process being studied 3/25/2017

At constant pressure, q = ∆H At constant volume, q = ∆E In both cases, heat gain or loss is determined Amount of heat exchanged in reaction depends upon Net temperature change during reaction Amount of substances (more you have, more heat can be exchanged) Heat capacity (C) of substance Three ways to express heat capacity 3/25/2017

Heat Capacity (C) Amount of heat (q) required to raise temperature of given mass of substance by 1oC Units: J/oC or J/K Ratio of heat absorbed to increase in T C = q/Δt (amt. of heat flowing in/out of substance) Δt is temperature change given by tfinal – tinitial 3/25/2017

Specific Heat Capacity (Specific heat) Heat capacity of one gram of substance Unit = J/oC g or J/K g Energy required to raise temp of 1 gram of substance by 1°C Can calculate amount of heat absorbed by given amount of water/by object of given mass 3/25/2017

Molar Heat Capacity Heat capacity of one mole of substance Unit = J/oC-mol or J/K mol Energy required to raise temp of 1 mole of substance by 1°C 3/25/2017

Constant Pressure Calorimeter (solutions) Determines changes in enthalpy (heats of reactions) for reactions occurring in solutions “coffee-cup” calorimeter Constant atmospheric pressure No physical boundary between system/surroundings Reactants/products are system Water in which they dissolve/calorimeter are surroundings Assume calorimeter prevents gain/loss of heat from solution (heat does not escape) 3/25/2017

Calculating Heat of Rxn, ΔH When reactants (each at same T) are mixed, T of mixed solution is observed to increase or decrease Calculating Heat of Rxn, ΔH Energy released by reaction = energy absorbed by solution Energy released = s x m x ΔT qsoln = (sp. Heat solution) x (g of solution) x ΔT = -qrxn Heat of rxn is extensive property-dependent on amount of substance For dilute aqueous solutions, specific heat of solution approximately that of water (4.18 J/g-K) 3/25/2017

10.0 g NaOH 1 mol NaOH -43 kJ 1000 J = -10,750 J 40.0 g NaOH mol 1kJ If 10.0 g of solid NaOH is added to 1.00 L of water (specific heat capacity = 4.18 J/oC) at 25.0oC in a constant pressure calorimeter, what will be the final temperature of the solution? (assume the density of the final solution is 1.05 g/mL). We will use info from previous problem where enthalpy change per mole of NaOH is -43 kJ/mol. We need to know three things: Mass of solution: 1.00 L x 1050 g/L = 1050 g Heat capacity of solution: 4.18 J/oC Enthalpy of reactant: 10.0 g NaOH 1 mol NaOH -43 kJ 1000 J = -10,750 J 40.0 g NaOH mol 1kJ We want to know the change in temperature: Energy released = s x m x ΔT -10,750 J = (4.18 J/oC)(1050 g)(ΔT) = -2.45 25.0 + 2.45 = 27.45oC 3/25/2017

Constant Volume Calorimetry (Bomb Calorimetry) Coffee cup calorimeter is not suitable for some reactions Greater precision may be required, or a device that allows higher temperatures Bomb calorimeters are used to measure heat evolved in combustion reactions of foods and fuels Heat capacity of calorimeter must be known, generally in kJ/°C Measurements of water/calorimeter related to heat evolved by reaction q and to H Energy released by reaction = energy absorbed by solution =specific heat capacity x mass of solution x increase in temperature Energy released = s x m x ΔT 3/25/2017

Stainless steel "bomb“ loaded with small amount of combustible substance and O2 at 30 atm of pressure, and is immersed in known amount of water Heat evolved during combustion absorbed by water/calorimeter assembly 3/25/2017

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Calculate heat capacity of bomb using data for glucose: The heat of combustion of glucose is 2800 kJ/mol. A sample of glucose weighing 5.00 g was burned with excess oxygen in a bomb calorimeter. The temperature of the bomb rose 2.4oC. What is the heat capacity of the calorimeter? A 4.40 g sample of propane was then burned with excess oxygen in the same bomb calorimeter. The temperature of the bomb increased 6.85oC. Calculate ΔEcombustion of propane. Calculate heat capacity of bomb using data for glucose: Convert to moles: 5.00 g 1 mol = 2.78 x 10-2 moles glucose 180.0 g Heat capacity = 2800 kJ/mol 2.78 x 10-2 moles 1 = 32.4 kJ/oC (kJ/ oC) 2.4oC Use this heat capacity to determine heat of combustion of propane: (kJ/mol) 32.4 kJ 6.85oC 1 = -2200 kJ/mol oC 0.100 mol 3/25/2017

Homework: Read 6.2, pp. 248-256 Q pp. 280-282, #12, 32, 34, 36, 38, 42, 46 3/25/2017

Hess’s Law In going from particular set of reactants to particular set of products, change in enthalpy (ΔH) is same whether reaction takes place in one step or in series of steps 3/25/2017

Using Hess’s Law Work backward from final reaction Reverse reactions as needed, being sure to also reverse ΔH Identical substances on both sides of summed equation cancel each other Can multiply entire equation by factor (3, 2, ½, or 1/3)-this includes ΔH 3/25/2017

Example: Step One: N2(g) + 2O2(g)  2NO2(g) ΔH1 = 68 kJ Step Two: 2NO(g) + O2 (g)  2NO2 g) ΔH3 = -112 kJ N2(g) + 2O2(g)  2NO2(g) ΔH2 + ΔH3 = 68 kJ 3/25/2017

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Example: S(s) + O2(g)  SO2(g)           H°rxn = –296.1 kJ H2(g) + ½ O2(g)  H2O(l)            H°rxn = –285.8 kJ H2S(g) + 3/2O2(g)  SO2(g) + H2O(l)     H°rxn = –561.7 kJ These equations can be arranged so that their sum is the formation reaction of hydrogen sulfide. The first one is unchanged: S(s) + O2(g)  SO2(g) The second one is unchanged: H2(g) + ½ O2(g)  H2O(l) The third one is reversed: SO2(g) + H2O(l)  H2S(g) + 3/2 O2(g) Their sum is overall equation: S(s) + H2(g) H2S(g) We must reverse the sign of the enthalpy change of the third reaction before summing the enthalpy changes. S(s) + O2(g)  SO2(g)            H°rxn = –296.1 kJ H2(g) + O2(g)  H2O(l)           H°rxn = –285.8 kJ SO2(g) + H2O(l)  H2S(g) + 3/2O2(g) H°rxn = 561.7 kJ S(s) + H2(g)  H2S(g)           H°f = –20.2 kJ Summation of the three reaction steps yields the net reaction, and summation of the H° values yields the enthalpy change of the net reaction. 3/25/2017

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Homework: Read 6.3, pp. 256-260 Q pp. 280-283, #14, 52, 54, 56, 58 3/25/2017

Enthalpy of formation (heat of formation) Tables for Enthalpies of vaporization (ΔH for converting LG) Enthalpies of fusion (ΔH for melting solids) Enthalpies of combustion (ΔH for combusting substance in oxygen) Enthalpy (heat) of formation (ΔHf) associated with process of forming products from reactants Conditions influencing enthalpy changes Temperature Pressure State of reactants/products (s/l/g/aq) 3/25/2017

Standard enthalpy of formation Change in enthalpy that accompanies formation of one mole of compound from its elements with all substances in their standard states Form most stable at 298K/1 atm 3/25/2017

Standard Enthalpy of Formation (ΔHf°) “o” symbol indicates standard conditions. Standard state for a substance is precisely defined reference state Elements/diatomic gases = 0 For a compound Gaseous substance at pressure of exactly 1 atmosphere Condensed state (pure liquid/solid) Substance present in solution of exactly 1M concentration For an element Form in which the element exists under conditions of 1 atmosphere and 25oC [O2(g), Na(s), Hg(l), etc.]. 3/25/2017

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SO2(g) + ½ O2(g)  SO3(g) H° = –99 kJ The enthalpy change in kilojoules applies to the reaction of the amount of substance shown in the balanced equation. SO2(g) + ½ O2(g)  SO3(g)           H° = –99 kJ When 2 mol SO2 react with 1 mol O2, 198 kJ of heat will be released If the above equation was divided by 2 and written then the enthalpy change (–198 kJ) must be divide by 2 also Remember that the heat evolved or absorbed by a chemical reaction is an extensive property: that is, it depends on the amount of reactants 3/25/2017

The following reaction absorbs heat If we write the reverse of a chemical reaction, the magnitude of H is the same, but its sign changes. The following reaction absorbs heat CaCO3(s)  CaO(s) + CO2(g)           H° = 178 kJ But the reverse reaction must release the same amount of heat CaO(s) + CO2(g)  CaCO3(s)           H° = –178 kJ 3/25/2017

In the combustion of methane, water is a product We must always specify the physical states of all reactants and products, because they help determine the magnitude of the enthalpy change. In the combustion of methane, water is a product The enthalpy change is different when H2O is formed as a gas, than when H2O is a liquid CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)      H° = –802 kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)       H° = –890 kJ 3/25/2017

3C(graphite) + 3O2(g) -> 3CO2(g) DH�f = �1180.5 kJ 3C(graphite) + 4H2(g) -> C3H8(g) DH�f = -103.85 kJ (reverse equation/sign) 3C(graphite) + 3O2(g) -> 3CO2(g) DH�f = �1180.5 kJ C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l) ΔH = ∑Hproducts – ∑Hreactants [3(-393.5) + 4(-285.8)] – [(-103.85) + 0] = -2219.8 kJ 3/25/2017

Calculate ΔH° for the following reaction: 2C3H6(g) + 9O2(g)  6CO2 (g) + 6H2O ΔHreactionO = ΣΔHfO (Products) - ΣΔHfO (Reactants) Appendix 4 in the textbook lists them [6(-286 kJ/mol) + 6(-393.5 kJ/mol)] – [2(20.9 kJ/mol) + 9(0 kJ/mol)] –remember, elements are 0 ΔH° = -4119 kJ 3/25/2017

Homework: Read 6.4 –6.6, pp. 260-278 Q pg. 284, #60, 62, 64, 66, and try 72 (extra credit) Submit quizzes by email to me: http://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch06_ace1.xml http://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch06_ace2.xml http://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch06_ace3.xml 3/25/2017