helpsheet_10_103

1 CHEMISTRY 103 – Help Sheet #10 Thermochemistry (Part II) – Module 6 Do the topics appropriate for your course Prepared by Dr. Tony Jacob https://clc.chem.wisc.edu (Resource page) Nuggets: Coffee Cup Calorimetry; Phase Changes ( D H fusion and D H vaporization ); Hess’s Law; D H f o CALORIMETRY – a technique used to measure heat transfer and from this determine D H rxn o or T f . Constant Pressure Calorimetry : P is constant; a Coffee Cup calorimeter is often used to do these measurements; in this type of calorimetry: q sys = q p = D H Typical Calorimetry Questions I. Given D H rxn o ® find T f II. Given T f ® find D H rxn o Step 1: Find q sys (scaling) See HelpSheet #9 Step 1: Find q surr q surr = C soln m total (T f – T i ) Step 2: Find q surr q surr = - q sys Step 2: Find q sys q sys = - q surr Step 3: Find T f q surr = C soln m total (T f – T i ) Step 3: Find D H rxn o (scaling) See HelpSheet #9 COFFEE CUP CALORIMETRY (constant pressure calorimetry) Example 1 (given T f find D H rxn o ) : When 2.50g H 2 SO 4 (l) at 21.5˚C was added to 75.0g water at 21.5˚C in a perfect insulating calorimeter, the temperature of the solution increased to 29.0˚C. What is the value of D H rxn o for H 2 SO 4 (l) ® H 2 SO 4 (aq)? C soln = 4.184J/g˚C Answer 1 : D H rxn o = -95.4kJ Step 1 (find q surr ): q surr = C soln m(T f – T i ) = (4.184J/g˚C)(75.0 + 2.50)(29.0 – 21.5) = 2432J Step 2 (find q sys ): q sys = -q surr ; q sys = -2432J; Step 3 (find D H rxn o ) scaling: ; -95,422J/mol x (1kJ/1000J) = -95.4kJ/mol} can also be done as a proportionality: ; ; cross multiply: 0.02549x = -2,432; x = -95,422J = -95.4kJ/mol Example 2 (given D H rxn o find T f ) : 10.0g KOH(s) is dissolved in 75.0ml water in a perfectly insulating calorimeter. The water and KOH are initially at 25.0˚C. What is the final temperature? C soln = 4.184J/g˚C, D soln = 1.0g/ml. KOH(s) ® KOH(aq) Δ H rxn o = -57.6kJ/mol Answer 2 : 54.0˚C Step 1 (find q sys ) scaling: can also be done as a proportionality: ; ; cross multiply: x = -10.3kJ} Step 2 (find q surr ): q surr = -q sys ⇒ q surr = +10.3kJ Step 3 (find T f ): m water = (D soln )(V) = (1.0g/ml)(75.0ml) = 75.0g; m soln = m water + m KOH = 75.0 + 10.0 = 85.0g q surr = C soln m soln (T f – T i ) ⇒ 10.3kJ(1000J/1kJ) = 10,300J; 10,300J = (4.184J/g˚C)(85.0g)(T f – 25.0); 10,300 = 355.64T f – 8891.0; 19,191 = 355.64T f ⇒ T f = 53.96˚C Bomb Calorimetry: Constant Volume Calorimetry : V is constant; similar to coffee-cup calorimeter except with constant volume instead of constant pressure ; D E is determined (instead of D H) since the volume is constant; in this type of calorimetry: q sys = q v = D E. − 2,432J 2.50gH 2 SO 4 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 98.09gH 2 SO 4 1molH 2 SO 4 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − 95,422J / mol 2 . 50 g H 2 SO 4 − 2 , 432 J = 1 mol H 2 SO 4 xJ 0 . 02549 mol H 2 SO 4 − 2 , 432 J = 1 mol H 2 SO 4 xJ 10.0gKOH 1molKOH 55.1gKOH ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − 57.6kJ 1molKOH ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − 10.3kJ 10.0gKOH xkJ = 1molKOH − 57.6kJ 0.1782molKOH xkJ = 1molKOH − 57.6kJ

2 PHASE CHANGE : There is no temperature change during a phase change Solid ® Liquid: D H fusion o = D H fus o ; Liquid ® Solid: - D H fus o (reaction is reversed so sign of D H changes) ; Liquid ® Gas: D H vaporization o = D H vap o ; Gas ® Liquid: - D H vap o (reaction is reversed so sign of D H changes) ; s ® l: q = D H fus o x mol (if units of D H fus o = J/mol) or q = D H fus o x grams (if D H fus o = J/g) l ® g: q = D H vap o x mol (if units of D H vap o = J/mol) or q = D H vap o x grams (if D H vap o = J/g) Example 3 : How much heat is required to heat 25.0g H 2 O(s) from -50.0˚C to 150˚C? (C H 2 O(s) = 2.11J/g˚C; C H 2 O(l) = 4.184J/g˚C; C H 2 O(g) = 2.00J/g˚C; D H o fus = 333J/g; D H vap o = 2256J/g) Answer 3 : 80.3kJ q total = (q heat ice to mp -50 ® 0 ) + (q melt ice ) + (q heat water to bp 0 ® 100 ) + (q boil water ) + (q heat H 2 O (g) 100 ® 150 ) Step 1: q -50 ® 0 = C solid m D T = (2.11J/g˚C)(25.0g)(0-(-50.0˚C)) = 2638J Step 2: q melt = D H fus o x m = (333J/g)(25.0g) = 8325J Step 3: q 0 ® 100 = C liquid m D T = (4.184J/g˚C)(25.0g)(100.0-0.00˚C) = 10460J Step 4: q boil = D H vap o x m = (2256J/g)(25.0g) = 56400J Step 5: q 100 ® 150 = C gas m D T = (2.00J/g˚C)(25.0)(150-100) = 2500J q total = q 1 + q 2 + q 3 + q 4 + q 5 = 2638 + 8325 + 10460 + 56400 + 2500 = 80323J = 80.3kJ Note 1 : all the signs of q should be the same; in this case, heating the ice is endothermic so positive (+); if the system is cooled: all the signs of q should be negative (-); when cooling a system remember to change the signs of D H fus o and D H vap o because you have reversed the process. Note 2 : At the start of the s ® l phase change there is only solid; at the end there is only liquid; as you move left to right more solid melts, the amount of solid decreases and liquid increases. Likewise, at the start of the l ® g phase change there is only liquid; at the end there is only gas; as you move left to right more liquid boils, the amount of liquid decreases and gas increases. PHASE CHANGE + HEAT TRANSFER Example 4: 65.5g of iridium (Ir) is heated to 85.5˚C and then placed on top of a block of ice at 0.00˚C. After thermal equilibrium had been established, the temperature of the metal and ice were 0.00˚C, and it was found that 2.24g of ice had melted. What is the specific heat capacity of Ir? C s (H 2 O(s)) = 2.11J/g˚C; C s (H 2 O(l)) = 4.184J/g˚C; D H fus o (H 2 O) = 333J/g Answer 4: 0.133J/g˚C This is a heat transfer and phase change problem; heat lost + heat gained = 0; for Ir: q = C s m(T f - T i ); for ice: q = D H fus o (m). Add the heats and plug in values: C s (65.5g)(0 - 85.5˚C) + (333J/g)(2.24g) = 0; Distribute: -5600.3C s + 745.92 = 0; Solve for C s : 5600.3C s = 745.92; C s = 0.1332J/g˚C heating ice melting ice heating water boiling water heat water gas 1 2 3 4 5 temperature heat added 100 150 0 -50 Heating ice from -50 ˚C to water gas at 150 ˚C Steps 1, 3, and 5 use: q = Cm ! T Steps 2 and 4 use: q = ! H x mass Total heat required = Steps 1 + 2 + 3 + 4 + 5 s " l l " g

4 HEAT of FORMATION, D H f o : The heat, D H f o , required to form 1 mol of substance from its elements in their natural state under standard conditions D H rxn o = S n p D H f o (products) - S n r D H f o (reactants) D H f o = 0 for all elements in their natural state n p = # of product moles for each chemical, n r = # of reactant moles for each chemical Natural States of Elements: Diatomic elements ( memorize diatomic elements and phases ) : F 2 (g), Cl 2 (g), Br 2 (l), I 2 (l), H 2 (g), N 2 (g), O 2 (g); Other elements: Hg(l); C ( memorize ) : C graphite (s); Noble gases: (He(g), Ne(g), Ar(g), Kr(g), Rn(g)); Remaining elements are solids and monoatomic; (Note: phosphorous is P 4 (s) and sulfur is S 8 (s); use these polyatomic elemental forms if presented in the course) Example 6 : Write D H f o reaction for NH 4 ClO 4 (s). Answer 6 : 1 / 2 N 2 (g) + 2H 2 (g) + 1 / 2 Cl 2 (g) + 2O 2 (g) ® NH 4 ClO 4 (s) must include phases; must form only 1mol of product; fractions ok Example 7 : Calculate D H rxn o for 2CO(g) + O 2 (g) ® 2CO 2 (g) given D H f o (CO) = -110.5kJ/mol and D H f o (CO 2 ) = -393.5kJ/mol Answer 7 : -566kJ/mol D H rxn o = [2( D H f o (CO 2 ) )] – [2( D H f o (CO) ) + 1( D H f o (O 2 ) )] = [2(-393.5)] – [2(-110.5) + 1(0)] = -566kJ (Note: D H f o (O 2 (g)) = 0 because it is an element in its natural state) Calorimetry questions (simpler calorimetry problems; no limiting reagents) 1. When 5.00g CsOH was dissolved in 55.0ml of water in a perfectly insulating coffee cup, the temperature increased from 25.0˚C to 34.5˚C. What is the D H rxn o for CsOH dissolving? D soln = 1.00g/ml and C soln = 4.184J/g˚C. CsOH(s) ® CsOH(aq) 2. 14.20g NaCH 3 COO . 3H 2 O (molar mass = 136g/mol) was dissolved in a perfectly insulating container into 45.5ml H 2 O(l) at 33.5˚C according to the reaction: NaCH 3 COO . 3H 2 O(s) ® NaCH 3 COO(aq) + 3H 2 O(l) D H rxn o = 19.66kJ/mol What is the final temperature of the solution? The D soln = 1.00g/ml and C soln = 4.184J/g˚C. (harder calorimetry problems; with limiting reagents) 3. When 100.0ml 0.30M Pb(NO 3 ) 2 (aq) was mixed with 50.0ml 2.0M KI(aq) in a perfectly insulating coffee cup, the temperature increased from 15.00˚C to 19.16˚C. What is the D H rxn o for the formation of PbI 2 (s)? D soln = 1.00g/ml; C soln = 4.184J/g˚C. Pb 2+ (aq) + 2I - (aq) ® PbI 2 (s) 4. 50.0ml of a 2.00M HCl solution at 0.00˚C is mixed with 100.ml of a 2.00 M NaOH solution also at 0.00˚C in a perfectly insulating coffee cup. If D soln = 1.00g/ml, C soln = 4.184J/g˚C, and D H rxn o = -56.0kJ/mol, what is T f of the solution ? HCl(aq) + NaOH(aq) ® H 2 O(l) + NaCl(aq) D H rxn o = -56.0kJ/mol

5 5. (Hint: This problem is done by looking at the mass of solutions (proportional to their volumes) and the moles of chemicals reacting (mol = M x L).) When 50.0ml of 2.00M silver(I) nitrate (AgNO 3 ) solution is mixed with 50.0ml of 2.00M sodium iodide (NaI) solution in a perfectly insulating calorimeter a precipitate forms. The solutions were initially at 20.0˚C and the temperature rises to 50.0˚C. AgNO 3 (aq) + NaI(aq) ® AgI(s) + NaNO 3 (aq) I. What is the final temperature when 100.0ml of 2.00M silver(I) nitrate (AgNO 3 ) solution is mixed with 100.0ml of 2.00M sodium iodide (NaI) solution in a perfectly insulating calorimeter? II. What is the final temperature when 50.0ml of 1.00M silver(I) nitrate (AgNO 3 ) solution is mixed with 50.0ml of 1.00M sodium iodide (NaI) solution in a perfectly insulating calorimeter? III. What is the final temperature when 50.0ml of 4.00M silver(I) nitrate (AgNO 3 ) solution is mixed with 50.0ml of 4.00M sodium iodide (NaI) solution in a perfectly insulating calorimeter? IV. What is the final temperature when 50.0ml of 1.00M silver(I) nitrate (AgNO 3 ) solution is mixed with 50.0ml of 2.00M sodium iodide (NaI) solution in a perfectly insulating calorimeter? Phase changes 6. How much heat is required to convert 10.0g ice at –25.0˚C into water at 50.0˚C? The specific heat capacity of ice is 2.11J/g˚C, the specific heat capacity of water is 4.18J/g˚C, and the heat of fusion for ice is 333J/g. 7. 45.0g of vanadium (V) is heated to 225.0˚C and placed in 45.00g water at 100.0˚C. After thermal equilibrium has been established, it is found that 43.78g of water remain. What is the specific heat capacity of the V? C s (H 2 O(l)) = 4.184J/g˚C, C s (H 2 O(g)) = 2.00J/g˚C, D H vap o = 2256J/g 8. 50.00g of titanium is cooled to 42.25˚C and placed on 12.75g ice at 0.00˚C. After thermal equilibrium had been established, the temperature of the titanium and ice were 0.00˚C, and 9.43g of ice remains. What is the specific heat capacity of titanium? C s (H 2 O(l)) = 4.184J/g˚C, C s (H 2 O(s)) = 2.11J/g˚C, D H fus o = 333J/g

7 11. Calculate the standard enthalpy change D H rxn o for N 2 H 4 (l) + CH 4 O(l) ® CH 2 O(g) + N 2 (g) + 3H 2 (g) given the individual reaction standard enthalpy changes Rxn I: N 2 H 4 (l) + H 2 (g) ® 2NH 3 (g) D H rxn o = –9.0 kJ/mol Rxn II: N 2 (g) + 3H 2 (g) ® 2NH 3 (g) D H rxn o = –23.0 kJ/mol Rxn III: CH 2 O(g) + H 2 (g) ® CH 4 O(l) D H rxn o = 32.5 kJ/mol 12. (harder version of Hess’s Law) Calculate the standard enthalpy change D H rxn o for 2H 2 O(l) ® 2H 2 (g) + O 2 (g) given the individual reaction standard enthalpy changes Rxn I: CH 3 COOH(l) + 2O 2 (g) ® 2CO 2 (g) + 2H 2 O(l) D H rxn o = –1742 kJ/mol Rxn II: CO 2 (g) ® C(graphite)(s) + O 2 (g) D H rxn o = 788.0 kJ/mol Rxn III: CH 3 COOH(l) ® 2C(graphite)(s) + 2H 2 (g) + O 2 (g) D H rxn o = 978.0 kJ/mol D H f o 13. Write the D H f o reaction that corresponds with each chemical listed below. a. CH 3 OH(l) b. NH 4 OH(s) c. Cu(NO 3 ) 2 (s) 14. The standard enthalpy change for the reaction below is –1125kJ. Calculate the standard enthalpy of formation for SO 2 (g). 2H 2 S(g) + 3O 2 (g) ® 2H 2 O(l) + 2SO 2 (g) D H f o for H 2 S(g) = -20.2kJ/mol, D H f o for H 2 O(l) = -285.8kJ/mol, 15. Given the information below, what is the standard enthalpy of formation for KClO 3 (s)? 3 / 2 KClO 3 (s) ® 3 / 2 K(s) + 3 / 4 Cl 2 (g) + 9 / 4 O 2 (g) D H rxn o = 597kJ/mol (Hint: Start by writing out D H f o for KClO 3 .) 16. The combustion of methanol takes place according to the reaction: 2CH 3 OH(l) + 3O 2 (g) ® 2CO 2 (g) + 4H 2 O(l) Compute D H rxn o for the combustion of two moles of methanol under standard conditions and select the correct answer, given the following standard heats of formation: D H f o for CH 3 OH(l) = -238.5kJ/mol D H f o for CH 4 (g) = -74.87kJ/mol D H f o for CO 2 (g) = -393.5kJ/mol D H f o for CO(g) = -110.5kJ/mol D H f o for H 2 O(l) = -285.6kJ/mol D H f o for H 2 O(g) = -241.8kJ/mol 17. a. Write an appropriate equation illustrating the standard enthalpy of formation for MnO 2 (s). b. Given the two reactions below, calculate D H f o for MnO 2 (s). MnO 2 (s) ® MnO(s) + 1 / 2 O 2 (g) D H rxn o = 136.0kJ/mol MnO 2 (s) + Mn(s) ® 2MnO(s) D H rxn o = -248.9kJ/mol

8 ANSWERS 1. -71.5kJ/mol { Step 1 : q surr = C soln m soln (T f - T i ); q surr = (4.184J/gC)(5.00 + 55.0g)(34.5 - 25.0˚C) = 2385J; Step 2 : q surr = -q sys ; q sys = -2385J; Step 3 scaling: ; -71,532.3J/mol x 1kJ/1000J = -71.5kJ/mol = D H o } can also be done as a proportionality: ; change to the same units: ; cross multiply: 0.03336x = -2,385; x = -71,502J = -71.5kJ/mol} 2. 25.3˚C { Step 1 scaling: ; can also be done as a proportionality: ; change to the same units: ; cross multiply: x = 2.053kJ Step 2 : q sys = -q surr ; q surr = -2.053kJ; Step 3 : q surr = C soln m soln (T f - T i ); -2.053kJ(1000J/1kJ) = (4.184J/gC)[(45.5ml)(1.00g/ml) + 14.20g)(T f - 33.5˚C); -2053J = 249.78T f – 8367.8; 6314.8 = 249.78T f ; T f = 25.28˚C} 3. -87.0kJ/mol { Step 1 : q surr = C soln m soln (T f - T i ); q surr = (4.184J/gC)(100.00g + 50.00g)(19.16 - 15.00˚C) = -2,610.8J; Step 2 : q surr = -q sys ; q sys = -2,610.8J; Find limiting reagent: mol Pb 2+ = M x L = (0.30)(0.1000) = 0.030mol Pb 2+ ; mol I - = M x L = (2.0)(0.0500) = 0.10mol I - ; since the ratio is 1 : 2 for Pb 2+ : I - the Pb 2+ is the limiting reagent; total moles PbI 2 (s) formed = 0.030mol PbI 2 (s) (can also use 0.030mol Pb 2+ or 0.060mol I - ; cannot use the total amount of I - , 0.10mol I - , since only 0.060mol I - reacted) Step 3 scaling: = D H o } can also be done as a proportionality: ; cross multiply: 0.030x = -2,610.8; x = -87,027J = -87.0kJ} 4. a. 8.92˚C {First, write reaction : HCl(aq) + NaOH(aq) ® H 2 O(l) + NaCl(aq); now find moles that react : limiting reagent problem: M HCl x vol HCl = mol HCl = (2.0M)(0.05L) = 0.1mol HCl; M NaOH x vol NaOH = mol NaOH = (2.0M)(0.1L) = 0.2mol NaOH; ratio is 1:1 ® HCl is the limiting reagent ; Step 1 : scaling: how much heat evolved when 0.1mol HCl reacts: ; can also be done as a proportionality: ; cross multiply: x = -5.60kJ = -5600J Step 2 : q surr = -q sys ; q surr = 5600J; Step 3 : q surr = C soln m soln (T f - T i ); the heat goes into the 2 solutions which total 50ml + 100ml = 150ml; convert to mass with density: D = mass/vol ® mass = D x vol = 1.00g/ml x 150. ml = 150. g solution; q surr = C soln m soln (T f - T i ); 5600J = (4.184J/g˚C)(150.g)(T f - 0.00˚C); 5600 = 627.6T f ; T f = 8.923˚C} − 2386J 5.00gCsOH ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 149.9gCsOH 1molCsOH ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − 71,532.3J / mol 5 . 00 g CsOH − 2 , 385 kJ = 1 mol CsOH x J 0 . 03336 mol CsOH − 2 , 385 kJ = 1 mol CsOH x J 14.20gNaC 2 H 3 O 2 ⋅ 3H 2 O 1molNaC 2 H 3 O 2 ⋅ 3H 2 O 136gNaC 2 H 3 O 2 ⋅ 3H 2 O ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 19.66kJ 1molNaC 2 H 3 O 2 ⋅ 3H 2 O ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 2.053kJ 1 mol NaCH 3 COO . 3 H 2 O 19 . 66 kJ = 14 . 20 g NaCH 3 COO . 3 H 2 O x kJ 1 mol NaCH 3 COO . 3 H 2 O 19 . 66 kJ = 0 . 1044 mol NaCH 3 COO . 3 H 2 O x kJ − 2,610.8J 0.030molPbI 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1mol PbI 2 ( ) = − 87,027J 1kJ 1000J ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − 87.0kJ 0 . 030 mol PbI 2 − 2 , 610 . 8 J = 1 mol PbI 2 x J 0.10molHCl − 56.0kJ 1molHCl ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − 5.60kJ = − 5600J 1 mol HCl − 56 . 0 kJ = 0 . 10 mol HCl x kJ

10 9. a. ~ -63˚C {see the purple arrow labeled “a”; looking at the graph read across to the y-axis from the lower horizontal line : read the T: this is the melting point; since this was done using a small graph, a reasonable range of answers is acceptable} b. ~805J {see the blue lines labeled “b”; looking at the graph, start at -80.0˚C (heat = ~75J) and go to 20.0˚C (heat = ~880J); take the difference: D heat = heat final – heat initial = 880-75J = 805J; this is the heat required to take 5.00g chloroform from -80.0˚C to 20.0˚C; since this was done using a small graph, a reasonable range of answers is acceptable} c. ~5280J {see the red lines labeled “c”; looking at the graph start at the beginning of the boiling line at 61˚C (heat = ~1070J) and go to the end of the boiling line still at 61˚C (heat = ~2390J); take the difference: D heat = heat final – heat initial = 2390 - 1070 = 1320J for 5.0g; this is the heat required to boil 5.00g chloroform; to boil 20.0g chloroform multiply the heat by 4: 4 x 1320J = 5280J for 20.0g; since this was done using a small graph, a reasonable range of answers is acceptable} d. ~ -53˚C {see the green lines labeled “d”; looking at the graph, look at the cold chloroform at -100.0˚C: heat = ~50J, and the hot chloroform at 40.0˚C: heat = ~975J; the cold chloroform will gain heat while the hot chloroform will lose heat ; find the midpoint of this heat range : from the line drawing it occurs at ~525J; can also be done: ; then either add this value to the lower value or subtract this value from the higher value: 463 + 50 = 513 or 975 – 463 = 512J yields the midpoint heat ; now using this midpoint value of 513J, read up until you reach the heating curve and then read across to the y-axis and this is the temperature at which the two samples (one hot and one cold) will reach once thermally equilibrated; T = ~ - 53˚C; since this was done using a small graph, a reasonable range of answers is acceptable} e. ~0.6J/g˚C {see the hot pink lines labeled “e”; use q = Cm D T; solve for C: ; note the heat change: 2530-2390 = 140J; note the T change: 110.0-61.2=48.8˚C; plug into C equation: ; since this was done using a small graph, a reasonable range of answers would be accepted} f. ~73J/g {see the orange lines labeled “f”; q = ( D H vap )(mass); ; ; since this was done using a small graph, a reasonable range of answers is acceptable} 975 − 50 ( ) 2 = 925 2 = 463J C = q m Δ T C = (2530 − 2390J) (5g)(110 − (61.2) ! C) = 140 244 = 0.574J / g˚C Δ H vap = q mass Δ H vap = (475 − 110J) (5g) = 73J / g -140.0 -120.0 -100.0 -80.0 -60.0 -40.0 -20.0 0.0 20.0 40.0 60.0 80.0 100.0 120.0 0 250 500 750 1000 1250 1500 1750 2000 2250 2500 T(˚C) heat added (J) Chloroform (5.0g) Heating Curve b: 880-75 = 805J c: 2390-1070 = 1320J for 5.0g d a e: 2530-2390 = 140J f: 475-110 = 365J melting point T at thermal equilibrium

11 10. 226 kJ/mol {multiply Rxn II by 2; no change to Rxn III; reverse Rxn I; then add modified reactions; D H rxn o = [(-1)(-1300)] + [2(-394)] + [(-286)] = 226kJ; Process: • C: only appears in Rxn II; need two on the left and have one on the left ® multiply Rxn II by 2; • H 2 : only appears in Rxn III; need one on the left and have one on the left ® no change needed for Rxn III; • C 2 H 2 : only appears in Rxn I; need one on the right and have one on the left ® reverse Rxn I} 11. –18.5kJ/mol {no change to Rxn I; reverse Rxn III; reverse Rxn II; then add modified reactions; D H rxn o = [(-9.0)] + [(-1)(-23.0)] + [(-1)(32.5)] = -18.5kJ; Process: • N 2 H 4 : only appears in Rxn I; need one on the left and have one on the left ® no change needed for Rxn I; • CH 4 O: only appears in Rxn III; need one on the right and have one on the left ® reverse Rxn III; • CH 2 O: only appears in Rxn III; this reaction is already done • N 2 : only appears in Rxn II; need one on the left and have one on the right ® reverse Rxn II} 12. 1144kJ/mol {Rxn I: reverse; Rxn III: no change; by inspection of Rxn I and Rxn III ® Rxn II: reverse and multiply by 2; then add modified reactions; D H rxn o = [(-1)(-1742)] + [(-1)(2)(788.0)] + [(978.0)] = 1144kJ; Process: • H 2 O: only appears in Rxn I; need two on the right and have two on the left ® reverse Rxn I; • H 2 : only appears in Rxn III; need two on the right and have two on the right ® no change to Rxn III; • O 2 : appears in Rxns I, II, and III; skip this chemical • Still need to figure out Rxn II: done by inspection; can focus on any chemical in Rxn II but it is easier if you choose a chemical in Rxn II that does not appear in the overall reaction; for example, it is easier if you choose CO 2 or C(graphite); I will choose CO 2 ; the CO 2 must cancel out since there is no CO 2 in the overall reaction; the only other source is in Rxn I; in that reaction there are two CO 2 on the left (don’t forget Rxn I was reversed!); to cancel those two CO 2 on the left using Rxn II, I will need two CO 2 on the right from Rxn II and to get this I will reverse Rxn II and multiply Rxn II by 2} 13. For the reactions below, the elements must be in their natural state and the balanced reaction must form only 1 mol of product a. C(graphite)(s) + 2H 2 (g) + 1 / 2 O 2 (g) ® CH 3 OH(l) b. 1 / 2 N 2 (g) + 1 / 2 O 2 (g) + 5 / 2 H 2 (g) ® NH 4 OH(s) c. Cu(s) + N 2 (g) + 3O 2 (g) ® Cu(NO 3 ) 2 (s) 14. -297kJ/mol {-1125kJ = [2mol(-285.8kJ/mol) + 2mol( D H f o SO 2 )] – [2mol(-20.2kJ/mol) + 3(0)]; solve for D H f o SO 2 ; -1125kJ = -571.6kJ + 2mol( D H f o SO 2 ) + 40.4kJ; -593.8kJ = 2mol( D H f o SO 2 ); D H f o SO 2 = -296.9kJ/mol} 15. -398kJ {write D H f o for KClO 3 : K(s) + 1 / 2 Cl 2 (g) + 3 / 2 O 2 (g) ® KClO 3 (s); take reaction given in problem and reverse and multiply by 2/3; do the same with D H rxn o and that is the value for D H f o ; D H f o = (-1)( 2 / 3 )(597kJ) = -398kJ} 16. -1452kJ { D H rxn o = [2mol(-393.5kJ/mol) + 4mol(-285.6kJ/mol)] – [2mol(-238.5kJ/mol) + 3(0)] = -1452.4kJ} 17. a. Mn(s) + O 2 (g) ® MnO 2 (s) b. -520.9 kJ/mol {Rxn II: no change; Rxn I: Reverse and x2; D H rxn o = (-1)(2)(136.0) + (-248.9) = -520.9kJ/mol; Process: • Mn: only appears in Rxn II; need one on the left and have one on the left ® no change to Rxn II; • O 2 : only appears in Rxn I; need one on the left and have one half on the right ® reverse Rxn I and multiply by 2}