Energy Changes in Chemical Reactions Most reactions give off or absorb energy Energy is the capacity to do work or supply heat. Heat: transfer of thermal (kinetic) energy between two systems at different temperatures (from hot to cold) Metal bar in water Metal bar drilled 1 Work (w): energy transfer when forces are applied to a system Heat (q): energy transferred from a hot object to a cold one Radiant energy heat from the sun Thermal energy associated with motion of particles Potential energy energy associated with object s position or substance s chemical bonds Kinetic energy energy associated with object s motion Describe the difference between the two. SI unit of energy: J 1kg m 1J s 1 watt = 1 J/s, so a 100 Watt bulb uses 100 J each second We often use the unit of kj to refer to chemical heat exchanges in a reaction. 1 kj = 1000 J Energy is also reported in calories: Amount of energy needed to raise 1 gram of water by 1 o C 1 cal = 4.184 J; 1 Cal = 4184 J Cal (or kcal) is used on food labels Molecular heat transfer Heat: form of energy transferred from object at higher temperature to one at lower temperature (from hot object to cold object) Thermochemistry: study of heat changes in chemical reactions, in part to predict whether or not a reaction will occur Thermodynamics: study of heat and its transformations First Law of Thermodynamics: Energy can be converted from one form to another but cannot be created or destroyed System loses heat (negative); gains heat (positive) 1
Endothermic reaction: q is positive (q > 0) Reaction (system) absorbs heat Surroundings feel cooler Exothermic reaction: q is negative (q < 0) Reaction releases heat Surroundings feel warmer Determine if the following processes are or exothermic Combustion of methane Reacting Ba(OH) with NH 4 Cl Neutralization of HCl Melting CaCO 3 (s) CaO (s) + CO (g) 8 Combustion of methane Reacting Ba(OH) with NH 4 Cl Neutralization of HCl Melting CaCO 3 (s) CaO (s) + CO (g) exothermic exothermic Combustion, neutralization, and combination reactions tend to be exothermic Decomposition reactions tend to be Melting, boiling, and sublimation are 9 10 Specific heat (sp. ht.): amount of heat required to raise 1 gram of substance by 1 o C Use mass, specific heat, and DT to calculate the amount of heat gained or lost: q = msdt ms = C q = CDT Heat capacity (C): amount of heat required to raise the temperature of a given quantity of a substance by 1 o C; C = q / DT = J / o C Molar heat capacity (C m ): amount of heat that can be absorbed by 1 mole of material when temperature increases 1 o C; q = (C m ) x (moles of substance) x (DT) = J / mol o C
Calculate the amount of heat transferred when 50 g of H O (with a specific heat of 4.184 J/g oc) is heated from o C to 98 o C. q = msdt Is heat being put into the system or given off by the system? If a piece of hot metal is placed in cold water, what gains heat and what loses heat? Which one will have a positive q value and which will have a negative q value? 34.8 g of an unknown metal at 5. o C is mixed with 60.1 g of H O at 96. o C (sp. ht. = 4.184 J/g oc). The final temperature of the system comes to 88.4 o C. Identify the unknown metal. Specific heats of metals: Al 0.897 J/g oc Fe 0.449 J/g oc Cu 0.386 J/g oc Sn 0.8J/g oc 14 Heat changes in a reaction can be determined by measuring the heat flow at constant pressure Apparatus to do this is called a calorimeter. Heat evolved by a reaction is absorbed by water; heat capacity of calorimeter is the heat capacity of water. A 8. gram sample of nickel is heated to 99.8 o C and placed in a coffee cup calorimeter containing 150.0 grams of water at 3.5 o C. After the metal cools, the final temperature of the metal and water is 5.0 o C. q absorbed + q released = 0 Which substance absorbed heat? Which substance released heat? Calculate the heat absorbed by the substance you indicated above. 15 A hot piece of copper (at 98.7 o C, specific heat = 0.385 J/g o C) weighs 34.6486 g. When placed in room temperature water, it is calculated that 915.1 J of heat are released by the metal. What gains heat? What loses heat? What is the final temperature of the metal? Watch signs!!!! Enthalpy (H) describes heat flow into and out of a system under constant pressure Enthalpy (a measure of energy) is heat transferred per mole of substance. At constant pressure, q p DH = H products H reactants DH > 0 (net absorption of energy from environment; products have more internal energy) DH < 0 exothermic (net loss of energy to environment; reactants have more internal energy) 3
Why does T become constant during melting and evaporating? Melting, vaporization, and sublimation are We can calculate total heat needed to convert a 15 gram piece of ice at -0 o C to steam at 10 o C..080 J/g o C 50 J/g 4.184 J/g o C 334 J/g.09 J/g o C Heat of fusion (DH fus ): Amount of heat required to melt (solid liquid) Heat of vaporization (DH vap ): Amount of heat required to evaporate (liquid gas) Heat of sublimation (DH sub ): Amount of heat required to sublime (solid gas) Why are there no values for DH freezing, DH condendsation, or DH deposition? 19 Shows both mass and enthalpy relationships Al (s) + Fe O 3 (s) Fe (s) + Al O 3 (s) DH o = -85 kj Amount of heat given off depends on amount of material: 85 kj of heat are released for every mol Al, 1 mol Fe O 3, mol Fe, and 1 mol Al O 3 Al (s) + Fe O 3 (s) Fe (s) + Al O 3 (s) DH o = -85 kj How much heat is released if 10.0 grams of Fe O 3 reacts with excess Al? What if we reversed the reaction? Heat would have to be put in to make the reaction proceed: Fe (s) + Al O 3 (s) Al (s) + Fe O 3 (s) DH o = +85 kj If a compound cannot be directly synthesized from its elements, we can add the enthalpies of multiple reactions to calculate the enthalpy of reaction in question. Hess s Law: change in enthalpy is the same whether the reaction occurs in one step or in a series of steps Look at direction of reaction and amount of reactants/products Value changes sign with direction Figure 8.5 4 4
Values of enthalpy change For a reaction in the reverse direction, enthalpy is numerically equal but opposite in sign Reverse direction, heat flow changes; becomes exothermic (and vice versa); sign of DH changes Proportional to the amount of reactant consumed Twice as many moles = twice as much heat; half as many moles = half as much heat DH T = DH 1 + DH + DH 3 +. Thermochemical equation: H (g) + I (s) HI(g) DH = +53.00 kj Two possible changes: Reverse the equation: HI(g) H (g) + I (s) DH = -53.00 kj Double the amount of material: H (g) + I (s) 4HI(g) DH = +106.00 kj 5 6 Calculate DH o for NO (g) + O (g) N O 4 (g) DH o =? N O 4 (g) NO (g) DH o = 57. kj NO (g) + ½ O (g) NO (g) DH o = -57.0 kj We can use known values of DH o to calculate unknown values for other reactions P 4 (s) + 3 O (g) P 4 O 6 (s) DH = -1640.1 kj P 4 (s) + 5 O (g) P 4 O 10 (s) DH = -940.1 kj What is DH o for the following reaction? P 4 O 6 (s) + O (g) P 4 O 10 (s) DH =? 8 Given: NH 3 (g) N H 4 (l) + H (g) DH = 54 kj 1 N (g) + 3 H (g) NH 3 (g) DH = -69 kj CH 4 O(l) CH O(g) + H (g) DH = -195 kj Find the enthalpy for the following reaction: N H 4 (l) + CH 4 O(l) CH O (g) + N (g) + 3H (g) DH =? kj 5
Given the following equations: CO (g) O (g) + CO (g) DH = 566.0 kj ½ N (g) + ½ O (g) NO (g) DH = 90.3 kj Calculate the enthalpy change for: CO (g) + NO (g) CO (g) + N (g) DH =? Standard heat of formation (DH o f): heat needed to make 1 mole of a substance from its stable elements in their standard states DH o f = 0 for a stable (naturally occurring) element Which of these have DH o f = 0? CO(g), Cu(s), Br (l), Cl(g), O (g), O 3 (g), O (s), P 4 (s) Do the following equations represent standard enthalpies of formation? Why or why not? Ag (l) + Cl (g) AgCl (s) Ca (s) + F (g) CaF (s) 31 3 Can use measured enthalpies of formation to determine the enthalpy of a reaction (use Appendix B in back of book) DH o rxn = SnDH o f (products) SnDH o f (reactants) S sum; n = number of moles (coefficients) Direct calculation of enthalpy of reaction if the reactants are all in elemental form Sr (s) + Cl (g) SrCl (g) DH o rxn = [DH o f (SrCl )] [DH o f (Sr) + DH o f (Cl )] = -88.4 kj/mol Some Common Substances (5 o C) DH o rxn = S DH o f,products - S DH o f,reactants Calculate values of DH o for the following rxns: 1) CaCO 3 (s) CaO (s) + CO (g) ) C 6 H 6 (l) + 15O (g) 1CO (g) + 6H O (l) DH o f values: CaCO 3 : -107.1 kj/mol; CaO: -635.5 kj/mol; CO : -393.5 kj/mol; C 6 H 6 : 49.0 kj/mol; H O(l): -85.8 kj/mol Use Standard Heat of Formation values to calculate the enthalpy of reaction for: C 6 H 1 O 6 (s) C H 5 OH(l) + CO (g) Hint: Is the equation balanced? DH o f (C 6 H 1 O 6 (s ) = -160.0 kj/mol DH o f (C H 5 OH(l ) = -77.7 kj/mol DH o f (CO (g ) = -393.5 kj/mol 35 6
Bond Dissociation Energy (or Bond Energy, BE): energy required to break a bond in 1 mole of a gaseous molecule Reactions generally proceed to form compounds with more stable (stronger) bonds (greater bond energy) H Bond Energy Bond energies vary somewhat from one mole- cule to another so we use average bond dissociation energy (D) H-OH 50 kj/mol Avg O-H = 453 H-O 47 kj/mol kj/mol H-OOH 431 kj/mol 37 38 DH o rxn = SBE (reactants) + - SBE (products) released energy input SBE(react) > SBE(prod) SBE(react) < SBE(prod) exothermic exothermic energy Use only when heats of formation are not available, since bond energies are average values for gaseous molecules 39 40 Use bond energies to calculate the enthalpy change for the following reaction: N (g) + 3H (g) NH 3 (g) DH rxn = [BE N N + 3BE H-H ] + [-6BE N-H ] DH rxn = [945 + 3(436)] [6(390)] = -87 kj measured value = -9. kj Why are the calculated and measured values different? Use bond energies to calculate the enthalpy change for the decomposition of nitrogen trichloride: NCl 3 (g) N (g) + Cl (g) How many distinct bond types are there in each molecule? How many of each bond type do we need to calculate DH rxn? BE (N-Cl) = 00 kj/mol BE (N N) = 945 kj/mol BE (Cl-Cl) = 43 kj/mol 41 4 7
6(N-Cl) + -1(N N) + -3(Cl-Cl) 6(00) + -(945) + -3(43) = -474 kj Use q = msdt (s = J/g oc) If given mass of reactant, convert to moles and multiply by enthalpy to find total heat transferred If given multiple equations with enthalpies, use Hess s Law If given DH o f values: products reactants If given bond energy (BE) values: +reactants + -products Identify how to set up the following problems: Calculate the DH o of reaction for: C 3 H 8 (g) + 5O (g) 3CO (g) + 4H O (l) DH o f C 3 H 8 (g): -103.95 kj/mol; DH o f CO (g): -393.5 kj/mol; DH o f H O(l): -85.8 kj/mol 8750 J of heat are applied to a 170 g sample of metal, causing a 56 o C increase in its temperature. What is the specific heat of the metal? Which metal is it? C H 4 (g ) + 6F (g) CF 4 (g) + 4HF(g) DH o =? H (g) + F (g) HF (g) C (s) + F (g) CF 4 (g) DH o = -537 kj DH o = -680 kj C (s) + H (g) C H 4 (g) DH o = 5.3 kj Use average bond energies to determine the enthalpy of the following reaction CH 4 (g) + Cl (g) CH 3 Cl (g) + HCl (g) (BE C-Cl = 38 kj/mol) 8