Energy & Chemical Change

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1 & Chemical Change CHAPTER 7 Chemistry: The Molecular Nature of Matter, 6 th edition By Jesperson, Brady, & Hyslop 1 CHAPTER 7: & Chemical Change Learning Objectives! Potential vs Kinetic! Internal, Work, and Heat Calculations! System vs Surroundings! First Law of Thermodynamics! Units of! Endothermic vs Exothermic! Heat of Temperature Changes (Specific Heat and Heat Capacity)! Calorimetry: Constant Volume vs Constant Pressure! Heat Stoichiometry (Thermochemical Equations)! Hess s Law and Equation Summation! Heat of Reaction from Heats of Formation 2

2 Chemical Bonds Chemical bond Attractive forces that bind Atoms to each other in molecules, or Ions to each other in ionic compounds Give rise to compound s potential energy Chemical energy Potential energy stored in chemical bonds Chemical reactions Generally involve both breaking and making chemical bonds 3 Chemical Reactions Forming Bonds Atoms that are attracted to each other are moved closer together Decrease the potential energy of reacting system Releases energy Breaking Bonds Atoms that are attracted to each other are forced apart Increase the potential energy of reacting system Requires energy 4

3 Exothermic Reaction Reaction where products have less chemical energy than reactants Some chemical heat energy converted to kinetic energy Reaction releases heat energy to surroundings Heat leaves the system; q is negative ( ) Heat energy is a product Reaction gets warmer, temperature increases Example. CH 4 (g) + 2O 2 (g)! CO 2 (g) + 2H 2 O(g) + heat 5 Endothermic Reaction Reaction where products have more chemical energy than reactants Some kinetic energy converted to chemical energy Reaction absorbs heat from surroundings Heat added to system; q is positive (+) Heat energy is a reactant Reaction becomes colder, temperature decreases Example: Photosynthesis 6CO 2 (g) + 6H 2 O(g) + solar energy! C 6 H 12 O 6 (s) + 6O 2 (g) Michelle Molinari/Alamy 6

4 Bond Strength Measure of how much energy is needed to break bond or how much energy is released when bond is formed. Larger amount of energy equals a stronger bond Weak bonds require less energy to break than strong bonds Key to understanding reaction energies Example: If reaction has Weak bonds in reactants and Stronger bonds in products Heat released 7 Why Fuels Release Heat Methane and oxygen have weaker bonds Water and carbon dioxide have stronger bonds 8

5 Heat of Reaction Amount of heat absorbed or released in chemical reaction Determined by measuring temperature change they cause in surroundings Calorimeter Instrument used to measure temperature changes Container of known heat capacity Use results to calculate heat of reaction Calorimetry Science of using calorimeter to determine heats of reaction 9 Heats of Reaction Calorimeter design not standard Depends on Type of reaction Precision desired Usually measure heat of reaction under one of two sets of conditions Constant volume, q V Closed, rigid container Constant pressure, q P Open to atmosphere 10

6 What is Pressure? Amount of force acting on unit area Pressure = Atmospheric Pressure Pressure exerted by Earth s atmosphere by virtue of its weight. ~14.7 lb/in 2 force area Container open to atmosphere Under constant P conditions P ~ 14.7 lb/in 2 ~ 1 atm ~ 1 bar 11 Comparing q V and q P Difference between q V and q P can be significant Reactions involving large volume changes, Consumption or production of gas Consider gas phase reaction in cylinder immersed in bucket of water Reaction vessel is cylinder topped by piston Piston can be locked in place with pin Cylinder immersed in insulated bucket containing weighed amount of water Calorimeter consists of piston, cylinder, bucket, and water 12

7 Comparing q V and q P Heat capacity of calorimeter = kj/ C Reaction run twice, identical amounts of reactants Run 1: q V - Constant Volume Same reaction run once at constant volume and once at constant pressure Pin locked; t i = C; t f = C q Cal = C"t = J/ C # ( ) C = 39.8 kj q V = q Cal = 39.8 kj 13 Comparing q V and q P Run 2: q P Run at atmospheric pressure Pin unlocked t i = C; t f = C Heat absorbed by calorimeter is q Cal = C"t = J/ C # ( ) C = 34.2 kj q P = q Cal = 34.2 kj 14

8 Comparing q V and q P q V = kj q P = kj System (reacting mixture) expands, pushes against atmosphere, does work Uses up some energy that would otherwise be heat Work = ( 39.8 kj) ( 34.2 kj) = 5.6 kj Expansion work or pressure volume work Minus sign means energy leaving system 15 Work Convention Work = P! "V P = opposing pressure against which piston pushes "V = change in volume of gas during expansion "V = V final V initial For Expansion Since V final > V initial "V must be positive So expansion work is negative Work done by system 16

9 First Law of Thermodynamics In an isolated system, the change in internal energy ("E) is constant: "E = E f E i = 0 Can t measure internal energy of anything Can measure changes in energy E is state function "E = heat + work "E = q + w "E = heat input + work input 17 First Law of Thermodynamics of system may be transferred as heat or work, but not lost or gained. If we monitor heat transfers (q) of all materials involved and all work processes, can predict that their sum will be zero Some energy transfers will be positive, gain in energy Some energy transfers will be negative, a loss in energy By monitoring surroundings, we can predict what is happening to system 18

10 First Law of Thermodynamics "E = q + w q is (+) q is ( ) w is (+) w is ( ) Heat absorbed by system (IN) Heat released by system (OUT) Work done on system (IN) Work done by system (OUT) Endothermic reaction "E = + Exothermic reaction "E = 19 "E is Independent of Path q and w NOT path independent NOT state functions Depend on how change takes place 20

11 Discharge of Car Battery Path a Short out with wrench All energy converted to heat, no work "E = q (w = 0) Path b Run motor converted to work and little heat "E = w + q (w >> q) "E is same for each path Partitioning between two paths differs 21 Bomb Calorimeter (Constant V) Apparatus for measuring "E in reactions at constant volume Vessel in center with rigid walls Heavily insulated vat Water bath No heat escapes "E = q v 22

12 Calorimeter Problem: Ex 4 When g of olive oil is completely burned in pure oxygen in a bomb calorimeter, the temperature of the water bath increases from C to C. a) How many Calories are in olive oil, per gram? The heat capacity of the calorimeter is kj/ C. q oil "t = C C = C q absorbed by calorimeter = C"t = kj/ C! C = kj q released by oil = q calorimeter = kj " kj 1 kcal (in cal/g) =!! g kj Cal/g oil 1 Cal 1 kcal 23 Calorimeter Problem: Ex 4 b) Olive oil is almost pure glyceryl trioleate, C 57 H 104 O 6. The equation for its combustion is C 57 H 104 O 6 (l) + 80O 2 (g)! 57CO 2 (g) + 52H 2 O What is "E for the combustion of one mole of glyceryl trioleate (MM = g/mol)? Assume the olive oil burned in part a) was pure glyceryl trioleate. " kj g C g H O ! 57H104O6 1 mol C57H104O6 "E = q V = 3.238! 10 4 kj/mol oil C 24

13 Enthalpy (H) Heat of reaction at constant Pressure (q P ) H = E + PV Similar to E, but for systems at constant P Now have P"V work + heat transfer H = state function At constant pressure "H = "E + P"V = (q P + w) + P"V If only work is P V work, w = P "V "H = (q P + w) w = q P 25 Enthalpy Change ("H) H is a state function "H = H final H initial "H = H products H reactants Significance of sign of "H Endothermic reaction System absorbs energy from surroundings "H positive Exothermic reaction System loses energy to surroundings "H negative 26

14 Coffee Cup Calorimeter Simple Measures q P Open to atmosphere Constant P Let heat be exchanged between reaction and water, and measure change in temperature Very little heat lost Calculate heat of reaction q P = C"t 27 Coffee Cup Calorimetry: Ex 5 NaOH and HCl undergo rapid and exothermic reaction when you mix 50.0 ml of 1.00 M HCl and 50.0 ml of 1.00 M NaOH. The initial t = 25.5 C and final t = 32.2 C. What is "H in kj/ mole of HCl? Assume for these solutions s = J g 1 C 1. Density: 1.00 M HCl = 1.02 g ml 1 ; 1.00 M NaOH = 1.04 g ml 1 NaOH(aq) + HCl(aq)! NaCl(aq) + H 2 O(aq) q absorbed by solution = mass # s # "t mass HCl = 50.0 ml # 1.02 g/ml = 51.0 g mass NaOH = 50.0 ml # 1.04 g/ml = 52.0 g mass final solution = 51.0 g g = g "t = ( ) C = 6.7 C 28

15 Coffee Cup Calorimetry: Ex 5 q cal = g # J g 1 C 1 # 6.7 C = 2890 J Rounds to q cal = 2.9 # 10 3 J = 2.9 kj q rxn = q calorimeter = 2.9 kj 1 mol HCl L HCl soln! 1 L HCl soln = mol HCl Heat evolved per mol HCl =!H = -2.9 kj mol HCl = -58 kj/mol 29 Enthalpy Changes in Chemical Reactions Focus on systems Endothermic Reactants + heat $! products Exothermic Reactants $! products + heat Want convenient way to use enthalpies to calculate reaction enthalpies Need way to tabulate enthalpies of reactions 30

16 The Standard State A standard state specifies all the necessary parameters to describe a system. Generally this includes the pressure, temperature, and amount and state of the substances involved. Standard state in thermochemistry Pressure = 1 atmosphere Temperature = 25 C = 298 K Amount of substance = 1 mol (for formation reactions and phase transitions) Amount of substance = moles in an equation (balanced with the smallest whole number coefficients) 31 Thermodynamic Quantities E and H are state functions and are also extensive properties "E and "H are measurable changes but still extensive properties. Often used where n is not standard, or specified "E and "H are standard changes and intensive properties Units of kj /mol for formation reactions and phase changes (e.g. "H f or "H vap ) Units of kj for balanced chemical equations ("H reaction ) 32

17 "H in Chemical Reactions Standard Conditions for "H 's 25 C and 1 atm and 1 mole Standard Heat of Reaction ("H ) Enthalpy change for reaction at 1 atm and 25 C Example: N 2 (g) + 3H 2 (g) $! 2 NH 3 (g) mol mol mol When N 2 and H 2 react to form NH 3 at 25 C and 1 atm kj released "H= kj 33 Thermochemical Equation Write "H immediately after equation N 2 (g) + 3H 2 (g)! 2NH 3 (g) "H = kj Must give physical states of products and reactants "H different for different states CH 4 (g) + 2O 2 (g)! CO 2 (g) + 2H 2 O(l ) "H rxn = kj CH 4 (g) + 2O 2 (g)! CO 2 (g) + 2H 2 O(g) "H rxn = kj Difference is equal to the energy to vaporize water 34

18 Thermochemical Equation Write "H immediately after equation N 2 (g) + 3H 2 (g)! 2NH 3 (g) "H= kj Assumes coefficients is the number of moles kj released when 2 moles of NH 3 formed If 10 mole of NH 3 formed 5N 2 (g) + 15H 2 (g)! 10NH 3 (g) "H= kj "H = (5! kj) = kj Can have fractional coefficients Fraction of mole, NOT fraction of molecule "N 2 (g) + 3 /2H 2 (g)! NH 3 (g) "H rxn = kj 35 State Matters! C 3 H 8 (g) + 5O 2 (g) # 3 CO 2 (g) + 4 H 2 O(g) $H rxn = 2043 kj C 3 H 8 (g) + 5O 2 (g) # 3 CO 2 (g) + 4 H 2 O(l ) $H rxn = 2219 kj Note: there is difference in energy because states do not match If H 2 O(l ) # H 2 O(g) $H vap = 44 kj/mol 4H 2 O(l ) # 4H 2 O(g) $H vap = 176 kj/mol Or 2219 kj kj = 2043 kj 36

19 Thermochemical Equations in Reverse To get energy out when form products, must put energy in to go back to reactants Consequence of Law of Conservation of If you know "H for reaction, you also know "H for the reverse CH 4 (g) + 2O 2 (g)! CO 2 (g) + 2H 2 O(g) "H reaction = kj Reverse thermochemical equation Must change sign of "H CO 2 (g) + 2H 2 O(g)! CH 4 (g) + 2O 2 (g) "H reaction = kj 37 Multiple Paths; Same "H Can often get from reactants to products by several different paths Reactants Products Intermediate A Intermediate B " Should get same "H " Enthalpy is state function and path independent " Let s see if this is true 38

20 Multiple Paths; Same "H : Ex 6 Path a: Single step C(s) + O 2 (g)! CO 2 (g) "H rxn = kj Path b: Two step Step 1: C(s) + "O 2 (g)! CO(g) "H rxn = kj Step 2: CO(g) + "O 2 (g)! CO 2 (g) "H rxn = kj Net Rxn: C(s) + O 2 (g)! CO 2 (g) "H rxn = kj Chemically and thermochemically, identical results True for exothermic reaction or for endothermic reaction 39 Multiple Paths; Same "H rxn : Ex 7 Path a: N 2 (g) + 2O 2 (g)! 2NO 2 (g) "H rxn = 68 kj Path b: Step 1: N 2 (g) + O 2 (g)! 2NO(g) "H rxn = 180. kj Step 2: 2NO(g) + O 2 (g)! 2NO 2 (g) "H rxn = 112 kj Net rxn: N 2 (g) + 2O 2 (g)! 2NO 2 (g) "H rxn = 68 kj Hess s Law of Heat Summation For any reaction that can be written into steps, value of "H rxn for reactions = sum of "H rxn values of each individual step 40

21 Enthalpy Diagrams Graphical description of Hess Law: Vertical axis = enthalpy scale Horizontal line =various states of reactions Higher up = larger enthalpy Lower down = smaller enthalpy 41 Enthalpy Diagrams Use to measure "H rxn Arrow down "H rxn = negative Arrow up "H rxn = positive Calculate cycle One step process = sum of two step process Example: H 2 O 2 (l )! H 2 O(l ) + "O 2 (g) 286 kj = 188 kj + "H rxn "H rxn = 286 kj ( 188 kj ) "H rxn = 98 kj 42

22 Hess s Law Hess s Law of Heat Summation Going from reactants to products Enthalpy change is same whether reaction takes place in one step or many Chief Use Calculation of "H rxn for reaction that can t be measured directly Thermochemical equations for individual steps of reaction sequence may be combined to obtain thermochemical equation of overall reaction 43 Rules for Manipulating Thermochemical Equations 1. When equation is reversed, sign of "H rxn must also be reversed. 2. If all coefficients of equation are multiplied or divided by same factor, value of "H rxn must likewise be multiplied or divided by that factor 3. Formulas canceled from both sides of equation must be for substance in same physical states 44

23 Strategy for Adding Reactions Together: 1. Choose most complex compound in equation for one-step path 2. Choose equation in multi-step path that contains that compound 3. Write equation down so that compound " is on appropriate side of equation " has appropriate coefficient for our reaction 4. Repeat steps 1 3 for next most complex compound, etc. 5. Choose equation that allows you to " cancel intermediates " multiply by appropriate coefficient 6. Add reactions together and cancel like terms 7. Add energies together, modifying enthalpy values in same way equation modified " If reversed equation, change sign on enthalpy " If doubled equation, double energy 45 Calculate "H rxn : Ex 8 C (s, graphite) $! C (s, diamond) Given C (s, gr) + O 2 (g)! CO 2 (g) "H rxn = 394 kj C (s, dia) + O 2 (g)! CO 2 (g) "H rxn = 396 kj To get desired equation, must reverse second equation and add resulting equations 1#[ ] C(s, gr) + O 2 (g)! CO 2 (g) CO 2 (g)! C(s, dia) + O 2 (g) "H rxn = 394 kj "H rxn = ( 396 kj) C(s, gr) + O 2 (g) + CO 2 (g)! C(s, dia) + O 2 (g) + CO 2 (g) "H = 394 kj kj = + 2 kj 46

24 Tabulating "H values Need to Tabulate "H values Major problem is vast number of reactions Define standard reaction and tabulate these Use Hess s Law to calculate "H for any other reaction Standard Enthalpy of Formation, "H f Amount of heat absorbed or evolved when one mole of substance is formed at 1 atm (1 bar) and 25 C (298 K) from elements in their standard states Standard Heat of Formation 47 Standard State Most stable form and physical state of element at 1 atm (1 bar) and 25 C (298 K) Element O C H Al Ne Standard state O 2 (g) C (s, gr) H 2 (g) Al(s) Ne(g) Note: All "H f of elements in their standard states = 0 Forming element from itself. " See Appendix C in back of textbook and Table

25 Uses of Standard Enthalpy (Heat) of Formation, "H f 1. From definition of "H f, can write balanced equations directly "H f of C 2 H 5 OH(l ) 2C(s, gr) + 3H 2 (g) + "O 2 (g)! C 2 H 5 OH(l ) "H f = kj/mol "H f of Fe 2 O 3 (s) 2Fe(s) + 3 /2O 2 (g)! Fe 2 O 3 (s) "H f = kj/mol 49 Using "H f 2. Way to apply Hess s Law wighout needing to manipulate thermochemical equations Sum of all "H f of all "H reaction = of the products Sum of all "H f of all of the reactants Consider the reaction: aa + bb! cc + dd "H reaction = c! "H f (C) + d! "H f (D) {a!"h f (A) + b!"h f (B)} "H rxn has units of kj because Coefficients # heats of formation have units of mol # kj/mol ( ) ( ) o!h rxn = # $!H o f products " moles of product % ' & ( #!H o f ( reactants) " ( moles of reactant % ' $ )& "H rxn has units of kj "H f has units of kj/mol 50

26 Calculate "H rxn Using "H f : Ex 9 Calculate "H rxn using "H f data for the reaction SO 3 (g) $! SO 2 (g) + "O 2 (g) 1. Multiply each "H f (in kj/mol) by the number of moles in the equation 2. Add the "H f (in kj/mol) multiplied by the number of moles in the equation of each product 3. Subtract the "H f (in kj/mol) multiplied by the number of moles in the equation of each reactant o!h rxn ( ) ( ) ( ) = # $!H o f products " moles of product % ' & ( # $!H o f ( reactants) " moles of reactant % ' & "H rxn has units of kj "H f has units of kj/mol! H! rxn!!h rxn! f! f =! H (SO g )) + 1! H (O g )) "! H (SO g )) 2 ( 2 = "297 kj/mol + 1 (0 kj/mol) " ( " 396 kj/mol) 2 "H rxn = 99 kj 2 (! f 3 ( 51 Other Calculations Don t always want to know "H rxn Can use Hess s Law and "H rxn to calculate "H f for compound where not known Example 10: Given the following data, what is the value of "H f (C 2 H 3 O 2, aq)? Na + (aq) + C 2 H 3 O 2 (aq) + 3H 2 O(l )! NaC 2 H 3 O 2 %3H 2 O(s) Na(aq) "H rxn = 19.7 kj/mol "H f = kj/mol NaC 2 H 3 O 2 3H 2 O(s) H 2 O(l) "H f = kj/mol "H f = kj/mol 52

27 Other Calculations: Ex 10 Continued "H rxn = "H f (NaC 2 H 3 O 2 %3H 2 O, s) "H f (Na +, aq) "H f (C 2 H 3 O 2, aq) 3"H f (H 2 O, l ) Rearranging "H f (C 2 H 3 O 2, aq) = "H f (NaC 2 H 3 O 2 %3H 2 O, s) "H f (Na +, aq) "H rxn 3"H f (H 2 O, l) "H f (C 2 H 3 O 2, aq) = kj/mol ( 239.7kJ/mol) ( 19.7 kj/mol) 3( kj/mol) = kj/mol 53

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