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Pembahasan Soal2 termokimia Pembahasan Soal2 termokimia Presentation Transcript

  • CHEMISTRY TASK SOAL & PEMBAHASAN OLEH KEL 3 XI IA-6: •DEBBY SIAGIAN (10)•JOYCE ANASTASIA SETYAWAN (17) •NAFIAH RAFIQAH R (22)•SHABRINA NABILA MAHMUDI (27)
  • QUADRA
  • NO 11.According to the Law Conservation of Energy, energy…..a. Can not be created (correction)b. Can be transferredc. Cannot be convertedd. Cannot form enthalpye. Can be distroyed
  • Solution for no 1 Law Conservation of Energy• "Energy can neither be created nor it is destroyed, however energy can be converted from one form energy to any other form of energy"
  • NO 2• To answer number 2 and 3,look at the reaction below.• C2H4+3O2 2CO2 + 2H2OThe enthalpy change of reaction is 1560 kJ/mole.2. It is an….reaction,with the value of H=…..a. balance; -1560 kJ/moleb. Exothermic; -1560 kJ/molec. Exothermic; +1560 kJ/moled. Endothermic; -1560 kJ/molee. Endothermic; +1560 kJ/mole
  • Solution for no 2 Combustion=exotherm• When a fuel reacts with Oxygen in a combustion reaction, a large amount of heat is generally released. (Something like a candle or a match for example, will also release heat in smaller amounts). The Release of Heat Energy is an Exothermic reaction.• the answer is (B) because an exotherm release heat so -1560 kJ/mole
  • NO 33. The enthalpy change measured for the reaction is 634 kJ. How much oxygen needed for this reaction?a. 8.91 Lb. 11.28 Lc. 22.4 Ld. 26.88 Le. 32.4 L
  • Solution no 3 C2H4+3O2 2CO2 + 2H2O ΔH = 1560• ΔH= 624 kJ (ΔH 1560 : 2.5)• O2=? L• C2H4+3O2 2CO2 + 2H2O ΔH = 1560 : 2.5• 0.4 C2H4+ 1.2 O2 0.8 CO2 + 0.8 H2OΔH = 624 kJ• 1.2 mole O2 x 22.4 = 26.88 L
  • NO 4Look at the enthalpy cycle below4. According to Hess’ Law,the value ΔH 2 is…a. ΔH 1 + ΔH 3 + ΔH 4b. ΔH 1 - ΔH 2 - ΔH 4c. ΔH 2 + ΔH 4 - ΔH 1d. ΔH 2 - ΔH 1 - ΔH 4e. ΔH 1 - ΔH 3 - ΔH 4
  • ΔH 1CaO(s) + H2O(l) Ca(s) + H2O(l) Route 1ΔH Route 2 Route 4 ΔH 4 Route 3CaO(s) + H2O(l) Ca(s) + H2O(l) ΔH 3
  • Solution for no 4 With Hess’LawCaO(s) + H2O(l) Ca(s) + H2O(l) ΔH 1• Ca(s) + H2O(l) CaO(s) + H2O(l) -ΔH 3• Ca(s) + H2O(l) Ca(s) + H2O(l) -ΔH 4CaO(s) + H2O(l) CaO(s) + H2O(l) ΔH 2
  • C6H12O6 + 6O2 6CO2 + 6H2OΔH = -2820 kJ/moleC2H5OH + 3O2 2CO2 + 3H2OΔH = -1380 kJ/moleUsing data above,the enthalpy change for fermentation with the equation of5.C6H12O6 2C2H5OH + 6CO2 is…..a. -4200 kJ/moleb. + 1440 kJ/molec. -1440 kJ/moled. +60 kJ/molee. -60 kJ/mole
  • Solution for no 5C6H12O6 + 6O2 6CO2 + 6H2O4CO2 + 6H2O 2C2H5OH + 6O2 (reversed & x2)C6H12O6 2C2H5OH + 2CO2 (corrected)ΔH = ΔH 1+ ΔH 2ΔH =-2820+ (+1380*2) = -2820 + 2760 = +60 (D)
  • No. 6• Given that: MO₂ + CO MO + CO₂ ∆H⁰ = -20 kJ/mole M₃O₄ + CO 3MO + CO₂ ∆H⁰ = +6 kJ/mole 3M₂O₃ + CO 2M₃O₄ + CO₂ ∆H = -12 kJ/mole• The value of ∆H for the reactions 2MO₂ + CO M₂O₃ +CO₂ is …. A. -40 kJ D. -18 kJ B. -28 kJ E. 40 kJ C. -26 kJ
  • solution• 2MO₂ + CO M₂O₃ +CO₂x2 MO₂ + CO MO + CO₂ ∆H = -20 kJReversed, x⅔ M₃O₄ + CO 3MO + CO₂ ∆H = +6 kJReversed, x⅓ 3M₂O₃ + CO 2M₃O₄ + CO₂ ∆H = -12 kJ 2MO₂ + 2CO 2MO + 2CO₂ ∆H = -40 kJ 2MO + ⅔ CO₂ ⅔ M₃O₄ + ⅔ CO ∆H = -4 kJ ⅔M₃O₄ + ⅓CO₂ M₂O₃ + ⅓CO ∆H = +4 kJ• 2MO₂ + CO M₂O₃ +CO₂ ∆H = -40
  • No. 7• Given that N₂ + 3H₂ 2NH₃ (g) ∆H = -92 kJ The heat needed to dissociate 5.1 g ammonia (Mr = 17 g/mole) is …. A. 9.2 kJ/mole B. 13.8 kJ/mole C. 18.8 kJ/mole D. 27.6 kJ/mole E. 65.6 kJ/mole
  • solution• A reaction with negative amount of ∆H: will be positive when the reaction is reversed N₂ + 3H₂ (g) 2NH₃ (g) ∆Hᵣ = kJ -92 2NH₃ (g) N₂ + 3H₂ (g) ∆ Hᵣ= 92 kJ• The initial enthalpy change apply when 2mole of ammonia is reacted. In 5.1 g NH₃ there is• mole NH₃ = mass of NH₃ = 5.1 = 0,3 mole Mr of NH₃ 17Therefor, the enthalpy change is∆H: = 92:2= 46 x 0.3 = 13.8
  • No. 8• The enthalpy change of combustion of Fe₂O is a kJ/mole and the enthalpy of its formation is b kJ/mole. Thus, the heat change of Fe₂O₃ formation could be represented as … kJ/mole. A. a + b B. a – b C. 2a + b D. a - 2b E. 2b - a
  • solution• Fe₂O + O₂ Fe₂O₃ ∆H = a ∆Hf=b ∆Hf=?Hreaction = ∑∆Hf right - ∑∆Hf left a = ∆Hf Fe₂O₃ - b a + b = ∆Hf Fe₂O₃
  • No. 9• Given that formation enthalpy of H₂O (l) and H₂O (g) are -286 kJ/mole and -242kJ/mole. In evaporation of 4.5 g water will … of energy. A. release 11 kJ B. absorb 11 kJ C. Absorb 44 kJ D. Release 44 kJ E. Absorb 1332 kJ
  • solution• ∆H: H₂O (l) is -286 kJ/mole• ∆H: H₂O (g) is -242 kJ/mole• H₂O (l) H₂O (g)• ∆H = Hp - Hr = -242 – (- 286) = 44n = m = 4.5 g = 0.25 Mr 18So, the enthalpy change is 44x0.25=11 kJ
  • No.10• H₂ + Br₂ 2HBr ∆H = -72 kJ/mole. The amount of heat to break down 11.2 L of hydrogen bromide into its elements is…. A. 18 kJ/mole B. 36 kJ/mole C. 72 kJ/mole D. 144 kJ/mole E. 288 kJ/mole
  • solution• A reaction with negative amount of ∆H: will be positive when the reaction is reversed H₂ + Br₂ 2HBr ∆Hᵣ = kJ -72 2HBr H₂ +Br₂ ∆Hᵣ = 72 kJ• The initial enthalpy change apply when 2mole HBr is reacted. In 11.2 L HBr is…• mole HBr = Volume of HBr = 11.2 = 0.5 mole 22.4 22.4Therefor, the enthalpy change is∆H: = 72 : 2 = 36 x 0.5 = 18 kJ/mole
  • Number 11The bond enthalpy in C2H5Cl is...( C-H = 414 ; C-C = 347 ; C-Cl = 339 )a.2840 kJ/moleb. 2756 kJ/molec. 2409 kJ/moled.2070 kJ/molee. 2015kJ/mole
  • H H | |H – C – C – Cl (5 x C-H) + C-C + C-Cl = | | (5x414) + 347 + 339 = 2756kJ/mole (B) H H
  • Number 12The formation of hydrochloric acid, HCl, releases 183 kJ/mole energy. Using the data above, the bond enthalpy of Cl-Cl is... (H-H=436 ; H-Cl=431)a.67 kJ/moleb. 124 kJ/molec. 201 kJ/moled. 268 kJ/mole3. 312 kJ/mole
  • ½ H2 + ½ Cl2 → HCl ∆H=183kJ/mole (releases)(½ H2 + ½ Cl2)-HCl = -183½ (436 + Cl2) – 431 = -183½ (436 + Cl2) = 248436 +Cl2 = 496Cl2 = 2Cl = 60kJ/mole
  • Number 13The amount of energy released in the formation of 5g HF = 20g/mole, E H-F=565kJ/mole ; E F- F=158 kJ/mole, and E E-H=436 kJ/mole is...a. 67b. 124c. 201d. 268e. 312
  • mHF=5gr MR HF= 20 HF=0,25 mole½ H2 + ½ F2 → HF218 + 79 – 565 = 268 (for 1 mole)For 0,25 mole = 268 x 0,25 = 67 (A)
  • Number 14Methane fuel CH4 is used to heat up 80g water. The temperature rises from 25oC to 56oC. The amount of used methane is 3,2 g. The value of enthalpy change of CH4 is...a. 1041,6 kJb. 10416 kJc. 2083,2 kJd. 20832 kJe. 5208 kJ
  • ∆H = -QQ = m.c.∆TQ= 80 . 4,2 . 31 = 10416∆H = -10416m CH4 = 3,2 Mr = 163,2 : 16 = 0,2 mole∆H for 0,2 mole = 0,2 x -10416 = -2083,2 kJ (C)
  • Number 15Given that :C + O2 → CO2 ∆H= -395,22kJ/mole2H2 + O2 → 2H2O ∆H= -573,72kJ/moleIf the enthalpy change of C2H4 is +27,30kJ, then the enthalpy change of combustion of C2H4 is...a. -654,78kJ/moleb. -663,18kJ/molec. -709,38kJ/moled. -887,45kJ/molee. -1.336,86kJ/mole
  • C + O2 → CO2 ∆H= -395,22kJ2H2 + O2 → 2H2O ∆H= -573,72kJ2C + 2H2 → C2H4 ∆H= 27,30kJC2H4 + 3 O2 → 2CO2 + 2H2O ∆H=...?2C + 2O2 → 2CO2 ∆H= -395,22 x 2 kJ2H2 + O2 → 2H2O ∆H= -573,72kJC2H4 → 2C + 2H2 ∆H= -27,30kJ______________________________________+C2H4 + 3 O2 → 2CO2 + 2H2O ∆H=-1.391,46kJ (E)
  • No 1616. The entalphy change of reaction could be measured with the following methods, excepts… (a) coefficient of a reaction (b) bomb calorimeter (c) simple calorimeter (d) hess law (e) bond entalphyAnswer : A
  • No 1716. The material needed for succesful combustion are….. (a) carbon dyoxide, fuel , heat (b) oxygen, Fuel, heat (c) oxygen, carbon dyoxide, heat (d) oxygen, water, heat (e) carbon dyoxide, water, heatAnswer : E
  • No 1816. The value of ΔH° could be calculated with the following formulas, except… (a) ΔH° = - Ccalorimeter . Δt (b) ΔH° = E + P Δv (c) ΔH° = ∑ (nproduct x ΔHf product) - ∑ (nreactant x ΔHf reactant) (d) ΔH° = ∑ (bond energy of reactant)- ∑ ( bond energy of product) (e) ΔH° = ∑ (nreactant x ΔHf reactant) - ∑ (nproduct x ΔHf product)Answer = E
  • No 1919. the standart entalphy change of formation ΔH°f of CH4 using the below experiment result, with condition of temperature 25°C and 1 atm is….. H2 (g) + ½ O2 H2O ΔH° = -285 kJ C(s,graphite) + O2 CO2 ΔH° = -393,5 kJ CH4 + 2 O2 CO2 ΔH° = -890,4 kJ (a) +182,4 kJ/mole (b) + 211,1kJ/mole (c) - 211,1 kJ/mole (d) + 74,7 kJ/mole (e) - 74,7kJ/mole
  • No 2020. A glass of water with the volume of 200ml is heated in the microwave. The amount of absorbed heat by water if the temperature is raised up up to 60°C is….(Cwater = 4,18 J/g°C, pwater = 1000 g / l)(a) 50,16 kJ(b) 501,6 kJ(c) 5016 kJ(d) 50160 kJ(e) 50160 kJ
  • No 20m=p.v = 1000. 200 = 200.000Q = m . C . Δt = 200.000 4,18 60 = 50.160.000 J = 50.160 kJanswer = D
  • FOTOCOPY SHEET
  • 21.Data:2C2H2(g)+ 5O2(g) 4CO2(g) + 2H2O(l)ΔH= -2,372.4kJ• The correct statement for the combustion reaction of 5.6 L C2H2 (STP) is….a. Releasing 593.1 kJ of heatb. Requiring 593.1 kJ of heatc. The enthalpy of the system rises 296.55kJd. Absorbing 296.55 kJ of heate. Releasing 296.55 kJ of heat
  • Solution for no 21ΔH 5.6 L of C2H2 = ?2C2H2(g)+ 5O2(g) 4CO2(g) + 2H2O(l)ΔH= -2,372.4 kJΔH = -2,372.4 kJ : 2 = - 1,186.2 kJmole C2H2 = 5.6 L /22,4 L = 0.25moleΔH = 0.25 x (-1,186.2 ) = 295.55 kJ (E)
  • • Calculating ΔH Value from Simple Experiment22. Data equation:NaOH (aq) + HCl (aq) NaCl (aq) + H2O(l)ΔH = -56 kJ/moleA 100 cm3 of 0.25 M HCl solution is mixed with 200 cm3 of 0.15 M NaOH.The enthalpy change for the equation is….a. -0.56 kJb. -3.06 kJc. -1.68 kJd. -1.40 kJe. -2.80 kJ
  • Solution for no 22• Acid + base → Salt + water + heat (neutralisation reaction).HCl=100ml x 0.25 = 0.025 moleNaOH=200ml x 0.15 = 0.03 mole• HCl + NaOH → NaCl + H2O The balanced equation above states that 1 mole of HCl will react with 1 mole of NaOH to produce 1 mole of NaCl and 1 mole of H2O There are 0.025 moles of HCl and 0.03 moles of NaOH. The coefficients indicate the mole ratios Since the balanced equation states that 1 mole of HCl will react with 1 mole of NaOH, the mole ratio = 1 : 1. So 0.025 moles of HCl will react with 0.025 moles of NaOH, to produce 0.025 moles of NaCl and 0.025 moles of H2O, and (0.03 – 0.025) 0.05 moles of NaOH will not react.
  • NaOH (aq) + HCl (aq) NaCl (aq) + H2O(l)ΔH = -56 kJ/mole0.025 NaOH (aq) + 0.025 HCl (aq) 0.025 NaCl (aq) + 0.025 H2O(l)1mole : 0.25 = 40ΔH = -56 : 40ΔH = -1.4 kJ
  • 23. When 100 mL of 1M NaOH is mixed with 100 mL of 1M HCl in a container,the temperature rises from 29C to37.5C (assume that specific heat is the same of that of pure water, 4.2 J/C). The enthalpy value is….a. -45.9 kJb. -54.6 kJc. +54.6 kJd. -71.4 kJe. -82.3 kJ
  • Solution for no 23• number of moles of each NaOH and HCl that reacted is = 100 x 1 = 100mmole = 1/10 mole• Δt = 37.5-29=8.5 C• m = 100 + 100 = 200 mL = 200 gram• Q = m x c x Δt = 200 x 4,2 x 8.5 = 7140 kJ = 7.14 kJ• If each NaOH and HCl reacted one mole,so• Q = 7.14 x 10 = 71.4 kJ/mole ΔH =-71.4 kJ/mole
  • • Calculating H Value According to Hess’ Law and Standard Enthalpy24. Look at this following diagram.a. ΔH 1 + ΔH 2 = ΔH 3 + ΔH 4b. ΔH 2 + ΔH 3 = ΔH 1 + ΔH 4c. ΔH 1 + ΔH 3 = ΔH 2 + ΔH 4d. ΔH 1 = ΔH 2 + ΔH 3 + ΔH 4e. ΔH 4 = ΔH 1 + ΔH 2 + ΔH 3
  • ΔH 1 A+B C+DΔH 2 ΔH 4 P+Q R+S ΔH 3
  • • See direction of the arrows !• Find the equation that starts and ends with the same reaction• Here starts with A+B and ends with R+S• ΔH 1 + ΔH 4 = R+S• ΔH 2 + ΔH 3 = R+S• So ΔH 2 + ΔH 3 = ΔH 1 + ΔH 4 (B)
  • 25. According to the diagram,the value of ΔH 3 is…a. ΔH 1 + ΔH 2 - ΔH 4b. ΔH 2 + ΔH 4 - ΔH 1c. ΔH 1 - ΔH 2 + ΔH 4d. ΔH 1 - ΔH 2 - ΔH 4e. ΔH 1 + ΔH 4 - ΔH 2
  • Mg ΔH 2 MgOΔH 1 ΔH 3 Mg(OH)2 ΔH 4
  • See direction of the arrows !Because ΔH 1 = ΔH 2 + ΔH 3 + ΔH 4So ΔH 3 = ΔH 1 - ΔH 2 - ΔH 4 (D)
  • No. 26• Data: MO₂ + CO MO + CO₂ ∆H = -20 kJ M₃O₄ + CO 3MO + CO₂ ∆H = +6 kJ 3M₂O₃ + CO 2M₃O₄ + CO₂ ∆H = -12 kJ• The value of ∆H for the reactions 2MO₂ + CO M₂O₃ +CO₂ is …. A. -40 D. -18 B. -28 E. +18 C. -26
  • solution• 2MO₂ + CO M₂O₃ +CO₂x2 MO₂ + CO MO + CO₂ ∆H = -20 kJReserved, x⅔ M₃O₄ + CO 3MO + CO₂ ∆H = +6 kJReserved, x⅓ 3M₂O₃ + CO 2M₃O₄ + CO₂ ∆H = -12 kJ 2MO₂ + 2CO 2MO + 2CO₂ ∆H = -40 kJ 2MO + ⅔ CO₂ ⅔ M₃O₄ + ⅔ CO ∆H = -4 kJ ⅔M₃O₄ + ⅓CO₂ M₂O₃ + ⅓CO ∆H = +4 kJ• 2MO₂ + CO M₂O₃ +CO₂ ∆H = -40
  • NO. 27N₂ + 3H₂ (g) 2NH₃ (g) ∆H = -92 kJThe heat required to decompose 5.1 gammonia (Mr = 17 g/mole) is ….A. 4.6 kJB. 9.2 kJC. 13.8 kJD. 18.8 kJE. 27.6 kJ
  • solution• A reaction with negative amount of ∆H: will be positive when the reaction is reserved N₂ + 3H₂ (g) 2NH₃ (g) ∆Hᵣ = kJ -92 2NH₃ (g) N₂ + 3H₂ (g) ∆ Hᵣ= 92 kJ• The initial enthalpy change apply when 2mole of ammonia is reacted. In 5.1 g NH₃ there is• mole NH₃ = mass of NH₃ = 5.1 = 0,3 mole Mr of NH₃ 17Therefor, the enthalpy change is∆H: = 92:2= 46 x 0.3 = 13.8
  • No. 28• Data:• ∆Hс C₂H₅OH (g) is -728 kJ/mole• ∆Hf CO₂ (g) is -394 kJ/mole• ∆Hf H₂O (l) is -286 kJ/mole• The enthalpy change for the formation of C₂H₅OH (g) in kJ/mole is …. A. -238 D. -952 B. -478 E. -714 C. -918
  • solution• The chemical equation of C₂H₅OH formation is• C₂H₅OH + 3O₂ 2CO₂ + 3H₂O• Applying the formula of calculating the enthalpy change of formation from the enthalpy change combustion, we have:• ∆H = ∑∆Hf:product ‒∑∆Hf:reactan = (2xCO₂ + 3xH₂O) ─ (C₂H₅OH + 3O₂) = (2x-394 + 3x-286) ‒ (-728 + 3x0) = -788 + -858 + 728 = -918
  • No. 29• H₂ (g) + Br₂ (g) 2HBr (g) ∆H = -72 kJ• The heat required to decompose 11.2 L HBr (STP) to H₂ and Br₂ is …. A. 9 kJ B. 18 kJ C. 36 kJ D. 72 kJ E. 144 kJ
  • solution• A reaction with negative amount of ∆H: will be positive when the reaction is reserved H₂ (g) + Br₂ (g) 2HBr (g) ∆Hᵣ = kJ -72 2HBr (g) H₂ (g) +Br₂ (g) ∆Hᵣ = 72 kJ• The initial enthalpy change apply when 2mole HBr is reacted. In 11.2 L HBr is…• mole HBr = Volume of HBr = 11.2 = 0.5 mole 22.4 22.4Therefor, the enthalpy change is∆H: = 72 : 2 = 36 x 0.5 = 18 kJ
  • No. 30• The enthalpy for formation of H₂O and NH₃ are a kcal/mole and b kcal/mole respectively. The enthalpy of combustion of 4NH₃ + 7O₂ 4NO₂ + 6H₂O is c. The enthalpy change of formation of NO₂ is ….A. a – 3b + ½ cB. c + b – aC. c + b -1½ aD. 1½ a – b ─ ½ cE. 1½ a + b + c
  • solution• ∆Hf H₂O = a kcal• ∆Hf NH₃ = b kcal• ∆Hf NO₂ = ?• 4NH₃ + 7O₂ 4NO₂ + 6H₂O ∆Hc = c• ∆H = ∑∆Hf:product ‒∑∆Hf:reactan c = (4x + 6a) ─ (4b + 0 ) c = 4x + 6a -4b 4x = -c +6a -4b x = -c + 3a –b 4 2
  • Number 31When the amount of magnesium is burned to produce 1 gram of MgO, 14,4kJ of heat is released. The heat of formation of MgO is .... kJa. 14,4b. -14,4c. 288d. -288e. 576
  • Mg + ½ O2 → MgOm MgO = 1 gram Mr MgO = 40 gramSo 1 gram of MgO = moleIf ∆H for mole MgO is 14,4 kJ then the ∆H0f is : 40 x 14,4 = 576kJ (E)
  • Number 32The heat of the combustion of Fe2O is a kcal/moleThe heat of the formation of Fe2O is b kcal/moleThe heat of the formation of Fe2O3 is...kcala. (a+b)b. (a-b)c. (a+2b)d. (2a+b)e. (2a+2b)
  • Fe2O + O2 → Fe2O3 a kcal/moleFe2 + ½ O2 → Fe2O b kcal/mole_______________________________ +Fe2 + O2 → Fe2O3 a+b kcal (A)
  • Number 33The enthalpy of formation of H2O(l) and H2O(g) is - 286kJ/mole and -242kJ/mole respectively. When we vaporize 4,5 grams of water, the process will....kJ of heata. Release 11b. Absorb 11c. Release 44d. Absorb 132e. Absorb 198
  • H2 + O2 → H2O(l) ∆H = -286kJ/moleH2 + O2 → H2O(g) ∆H = -242kJ/moleH2O(l) → H2 + O2 ∆H = -286kJ/moleH2 + O2 → H2O(g) ∆H = -242kJ/mole_______________________________________+H2O(l) → H2O(g) ∆H = -44kJ/mole4,5 grams of H2O = 0,25 mole0,25 x 44 = 11From liquid to gas needs heat so it absorbs 11 kJ of heat (B)
  • Number 34C + O2 → CO2 ∆H= -395,22kJ2H2 + O2 → 2H2O ∆H= -573,72kJIf the enthalpy change for the formation of C2H4 is +27,30kJ, the enthalpy of combustion of C2H4 is...kJa. -654,78b. -663,18c. -709,38d. -1.336,86e. -1.391,46
  • C + O2 → CO2 ∆H= -395,22kJ2H2 + O2 → 2H2O ∆H= -573,72kJ2C + 2H2 → C2H4 ∆H= 27,30kJC2H4 + 3 O2 → 2CO2 + 2H2O ∆H=...?2C + 2O2 → 2CO2 ∆H= -395,22 x 2 kJ2H2 + O2 → 2H2O ∆H= -573,72kJC2H4 → 2C + 2H2 ∆H= -27,30kJ______________________________________+C2H4 + 3 O2 → 2CO2 + 2H2O ∆H=-1.391,46kJ (E)
  • Number 35The enthalpy change for the formation of CO2 is - 395kJ. The enthalpy change for the combustion of glucose is -28,20kJ. The enthalpy change for the combustion of ethanol is -1.368 kJ. What is ∆H calue for the following equation?C6H12O6 → 2C2H5OH + 2CO2a. -874kJb. +706kJc. -1.057 kJd. +1.057 kJe. -4.583 kJ
  • C + O2 → CO2 ∆H= -395kJC6H12O6 + 6O2 → 6CO2 + 6H2O ∆H= -28,20kJC2H5OH+ 3O2 → 2CO2 + 3H2O ∆H= -1.368kJC6H12O6 → 2C2H5OH + 2CO2 ∆H= ...?C6H12O6 + 6O2 → 6CO2 + 6H2O ∆H= -28,20kJ4CO2 + 6H2O → 2C2H5OH+ 6O2 ∆H= 1.368kJ____________________________________________+C6H12O6 → 2C2H5OH + 2CO2 ∆H= 2707,8kJ
  • 36. H2 + ½O2 H2O ΔH=-242kJ The bond energy of H-H and O=O is 436 kj/mole and 500 kj/mole respectively. The mean bond entalpy of H-O is…… kj (a)121 (d) 464 (b)222 (e) 589 (c)363
  • No 36H2 + O₂ H20 ΔH=-242kJΔH = energy separation – energy formation-242=(H-H + (O=O) ) - 2 (O-H)-242=(436 +250) – 2(O-H)O-H =(786+ 242)/2 = (1028)/2 = 464
  • 37. Bond energy data C=C = 611 kj/mole C-Cl = 339 kj/mole C-H = 414 kj/mole C-C = 347 kJ/mole H-Cl = 431 kj /moleUsing the data, the entalpy change for the following reaction is ..kJ C2H4 + HCl C2H5Cl (a) +46 (d) -92 (b) -46 (e) -138 (c) -58
  • No 37 C=C = 611 kj/mole C-Cl = 339 kj/mole C-H = 414 kj/mole C-C = 347 kJ/mole H-Cl = 431 kj /mole• C2H4 + HCl C2H5ClΔH = energy separation – energy formation =(2(C=C) + 4(C-H) +H-Cl ) -((C-C) + 5(C-H) +C-Cl) =(611) + 4(414) +431) – ((347) + 5(414) + 339) =(611+1656+431)-(347+2070+339) =(2698)-(2756) = -58 kJ
  • 38 .the enthalpy of formation of NO is +90kJ/mole. If the bond energy of N=N is 418 kJ/mole and O=O is 498 kJ/mole. Energy required to breakdown 2 mole NO bond is….. (a)413 (d) 826 (b)765 (e) 911 (c)720
  • solution NO 38• ΔHf NO = +90kJ/mole ;N=N 418 kJ/mole ; O=O 498kJ/moleAnswer: NO ½ N2 + ½ O2 ΔHd = -90 kJ/mole ½N2 N ΔH = 209 kJ/mole ½O2 O ΔH = 249 kJ/mole ₊ NO N +O 368 kJ/mole ΔH atomisasi = 368 kJ/mole N-O = 368 kJ/mole2 N-O = 2 x 368 = 736 kJ
  • 39. Bond energy data H-Cl = 431 kj/mole H-H = 436 kJ/mole Cl-Cl = 243 kj /mole According to the data, the heat required to decompose 73 grams of HCl (Mr = 36,5 kJ/mole) to its element is (a) 336 kJ (d) 139,5 kJ (b) 69,75 kJ(e) -183 kJ (c) 100 kJ
  • solution NO 39H-Cl = 431 kj/moleH-H = 436 kJ/moleCl-Cl = 243 kj /mole HCl H2 + Cl2 ΔH = energy separation – energy formation = HCl –(1/2 H2+1/2 Cl2) = 431 -(218+121,5) = 431 – 339.5 = -91,5 kJ/mole mole : 73 / 36.5 = 2 mole ΔH : 2 (-91,5)= -183 kJ
  • 40. Bond energy data H-F = 565 kj/mole H-H = 436 kJ/mole F-F = 158 kj /mole energy released for the formation of 5 grams HF (Mr=20kJ/mole) from its elements is…. (a) -268 kJ (d) -67 kJ (b) -201 kJ (e) -33,5 kJ (c) -124
  • PEMBAHASAN NO 40H-F = 565 kj/moleH-H = 436 kJ/moleF-F = 158 kj /mole H2 + F2 HF ΔH = P – R = (1/2 H2 + ½ F2) – H-F = (218+79) - 565 = 297 -565 =- 268 kJ (A)
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