Answer Book. Higher Chemistry. Enthalpy, Moles & RedOx. Topic 4: 4.1 A B C D 4.4 A B C D 4.8 A B C D 4.7 A B C D. CfE New Higher - Unit 3 - Topic4

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1 n Higher Chemistry C V Topic 4: Enthalpy, Moles & RedOx Answer Book 4. A B C D A B C D 4.4 A B C D A B C D A B C D A B C D A B C D KHS Aug 04 - Cheviot Learning Community - based on Challenge Chemistry R.I.S.E Page

2 Home Practice Answers 4. Q. The standard enthalpy of combustion of propane is -9 kj mol -. a) C H 8. b) C H 8(g) + 5 O (g) CO (g) + 4 H O (l) c) Molar mass of C H 8 = ( x ) + ( 8 x ) = 44g () -9 / 44 = kj g - ( d) kj x 000 = 50,400 kj kg - Q. a) any two from: (supports hypothesis) because when the fuel contains more carbon the temperature of the water went up more / faster (in minutes) (does not support hypothesis as) temperature change per gram decreases as the number of carbons increases (does not support hypothesis) because the more carbon in the fuel the more smoke or the dirtier / sootier it is b) The student only tested hydrocarbons / alkanes / fuels with between 5 and carbon atoms so - should have tested molecules from other familes Q. Large amount of energy released / very exothermic () High proportion (all) of products are gases - propulsion ( Total (0) KHS Aug 04 - Cheviot Learning Community - based on Challenge Chemistry R.I.S.E Page

3 Home Practice Answers 4./4. Q. ΔH = (Energy of Bond Breaking) + (Energy of Bond Forming) Bond Breaking = (4 x ΔH(C H) + ΔH(C=C) + ΔH(H Cl) ) (endothermic) = (4 x 467) + (6) + (4) = + 9 kj Bond Forming = (5 x ΔH(C H) + ΔH(C C) + ΔH(C Cl)) (exothermic) = (5 x 467) + (47) + (46) = - 08 overall ΔH = (+9) + (-08) = - 6 kj mol - Q. a) no effect b) E = m c ΔT = = 6.69 kj 000 cm of H O (aq) 0.88 mol 50 cm of H O (aq) 50/000 x 0.88 = mol mol 6.69 kj mol /044 x 6.69 = 5 kj include negative sign in final answer -5 kj (units not required; deduct ½ mark for incorrect units) Q. a) E = m c ΔT = = 67 kj Data Book, Butane 878 kj mol 67 kj x 67 / 878 = 0.8 mol b) 80% 0.8 mol 00% 00/80 x 0.8 = 0.75 mol Butane, C 4 H 0, mol 58g 0.75 mol 0.75/ x 58 = 5.8 g Total () KHS Aug 04 - Cheviot Learning Community - based on Challenge Chemistry R.I.S.E Page

4 Home Practice Answers 4.4 Q. ΔH c carbon x = -94 kj x = -788 kj ΔHc hydrogen as is = -86 kj reverse ΔHc ethyne = +00 kj addition = +6 kj ( sensible numbers required for ½ mark for addition based on following through; no units required; deduct ½ for incorrect units) Q. C 6 H 4 (OH(aq) C 6 H 4 O (aq) + H (g) kj mol - H O (aq) H (g) + O (g) reverse +9. kj mol- H (g) + O (g) H O (g) kj mol- H O (g) H O (l) kj mol- answer = -0.6 kj mol - (deduct ½ mark for incorrect addition based on numbers used; no units required; deduct ½ mark for incorrect units) Q. SiO (s) + H O (l) SiH 4(g) + O (g) +57 kj Si (s) + O (g) SiO (s) H (g) + O (g) H O (l) - 9 kj - 57 kj addition = 4 kj mol ( sensible numbers required for ½ mark for addition based on following through; no units required; deduct ½ mark for incorrect units) Q4. C + O CO 94 = 8 kj 4H + O 4H O 86 4 = 44 kj CO + 4H O C H 8 O + 7 / O = kj addition = 67 kj mol ( sensible numbers required for ½ mark for addition based on following through; no units required; accept kj; deduct ½ mark for incorrect units) Total (8) KHS Aug 04 - Cheviot Learning Community - based on Challenge Chemistry R.I.S.E Page 4

5 Home Practice Answers 4.5/4.6 Q. mol Cu(NO mol NO 87.5 g 48 l 0 g 0/ = 0.5 litres or mol Cu(NO mol NO no. of moles of Cu(NO = 0/87.5 = 0.07 mol 0.07 mol 0.07 = 0.4 mol 0.4 x 4 l = 0.5 litres Q. mol CaF mol HF 78 g 48 l 000 g 000/78 48 l = 65 litres or mol CaF mol HF no. of moles of CaF = 000/78 =.8 mol.8 mol.8 x =5.6 mol 5.6 mol 4 = 65 litres (deduct ½ mark for no or incorrect units) Q. O O mol / mol g / x 6 x 0 = 4 x 0 molecules 6 g 6/ x 4 x 0 = 7.5 x 0 molecules Q4. moles of LiOH = = 0.04 mol moles of CO = 0.4/4 = 0.0 mol 0.0 mol of LiOH needed to react with 0.0 mol of CO excess LiOH = 0.0 Q5. mole Ca(OCl moles Cl 4 g 48 litres litres 4 x 0.096/4 = 0 86 g or 0 9 g (deduct half mark for missing or incorrect unit Total (0) KHS Aug 04 - Cheviot Learning Community - based on Challenge Chemistry R.I.S.E Page 5

6 Home Practice Answers 4.7 Q. a) NH, 7g mol CO, 5 l mol 7.5g 7.5/7 = 0.44 mol 4.5 l 4.5/5 = 0.8 mol NH CO 0.44/ and 0.8/ = 0. and 0.8 so CO limiting reagent b) CO (NH CO 0.8 mol 0.8 mol 0.8 x 60 = 0.8 g Q. a) Na PO 4, 0. x mol Ba(NO, 0.5 x mol Na PO 4 Ba(NO 0.0/ 0.05 / = and so Ba(NO is limiting reagent b) Ba(NO Ba (PO 4 (s) 0.05 mol 0.05/ = mol x 60 =.5 g Q. a) TiO, 80 g mol 4.5 g 4.5/80 = 0.05 mol 0.05/ = 0.07 C, g mol 5.67 g 5.67/ = 0.47 mol 0.47/4 = 0.8 Cl, 7 g mol 6.78 g 6.78/7 = mol 0.096/6 = 0.06 so Cl is limiting reagent. b) 6Cl TiCl mol mol x 90 =.5 g Total (0) KHS Aug 04 - Cheviot Learning Community - based on Challenge Chemistry R.I.S.E Page 6

7 Home Practice Answers 4.8 Q. a) i) Mg (s) Mg + + (aqe- ii) Sn 4+ + (aqe- Sn + (aq) iii) ClO - + (aqh+ + (aqe- Cl - (aq) + H O (l) iv) FeO 4 -(aq) + 8H+ + (aq) e- Fe + (aq) + 4H O (l) b) i) Oxidation - ii) Reduction iii) Reduction iv) Reduction Q. a) i) Zn (s) Zn + (aq) + e- iial (s) Al + (aq) + 6e- iiii - (aq) I (s) + e- iv) Zn (s) Zn + (aq) + e- b) i) Pb + + (aqe- Pb (s) ii) 6H + + 6e - H (aq(g) iii) Cl (g) + e - Cl - (aq) iv) S (s) + e - S - (s) Q. Cd (s) + OH - (aq) Cd(OH(s) + e- Ni(OH) (s) + e - Ni(OHs) + OH - (aq) multiply by a) Cd (s) + Ni(OH) (s) Cd(OH(s) + Ni(OHs) b) Ni(OH) (s) Total (0) KHS Aug 04 - Cheviot Learning Community - based on Challenge Chemistry R.I.S.E Page 7

8 Home Practice Answers 4.9/4.0 Q. a) H O (aq) + H + (aq) + I- (aq) H O (l) + I (aq) (State symbols not required; accept I (s) on right hand side of equation; deduct ½ if e - shown on each side). b) no. of moles of thiosulphate = x = 7.45 x 0-5 mol thiosulphate : mol I () no. of moles of I = ½ x 7.45 x 0-5 =.75 x 0-5 mol I = x 6.9 = 5.8 g mass of I =.75 x 0-5 x 5.8 = g (9.45 x 0 - g) (Deduct ½ for no or incorrect units) Q. a) MnO 4 -(aq) + 8H + (aq) + 5e- Mn + (aq) + 4H O (l) (state symbols not required) b) there is a colour change from colourless to purple (or purple to colourless) or the reaction is self-indicating (or a colour change shows the end of the reaction) c) i) first titre is a rough (or approximate) result or not accurate or an estimate or too far away from the others ii) no. of moles of MnO 4 = = (aq) ratio of (COOH(aq) : MnO 4 -(aq) = 5 : no. of moles of (COOH(aq) in 5 cm = 5/ 0.00 = no. of moles of (COOH(aq) in 500 cm = /5 = 0.05 (no units required; deduct ½ mark for incorrect units) Total (9) KHS Aug 04 - Cheviot Learning Community - based on Challenge Chemistry R.I.S.E Page 8

9 Consolidation Answers A Q. a) 0 kj b) kj c) Exothermic Because ΔH is negative. or Products have less energy than reactants Q. a) Loss of heat to beaker/ thermometer/ air/ surroundings b) Any one from: Use plastic / better insulated beaker. Use a vacuum flask for reaction. Have a lid on top (to reduce heat loss). Use a more accurate thermometer. Q. NH, mole = 7g.4g =.4/7 = 0. mol of NH mol NH mol of N atoms 0. mol of NH 0.6 mol of N atoms C mol 6.0 x 0 N atoms 0.6 mol 0.6 x 6.0 x 0 =.6 x 0 N atoms Q4. a) C H 8(g) CO (g) mol mol 0. mol 0.6 mol b) mol l 0.6 mol 0.6 x =.8 l Q5. a) N (g) + O (g) NO (g) mol mol vol vol So 80 l of N would require 60 l of O - only 0 l available so O is limiting reagent 0 l of O would react with 0 l of N so 80-0 = 70 l of N in excess. b) N (g) + O (g) NO (g) mol mol vol vol 0 l 0 l plus 70 l of unreacted N, so total volume = = 90 l of gas. KHS Aug 04 - Cheviot Learning Community - based on Challenge Chemistry R.I.S.E Page 9

10 Consolidation Answers A Q6. a) I (aq) + e - I - (aq) b) C 6 H 8 O 6(aq) + I (aq) H + I - (aq) + C 6 H 6 O 6(aq) or C 6 H 8 O 6(aq) + I (aq) H + (aq) + I- (aq) + C 6 H 6 O 6(aq) c) Reducing agent Q7. a) None or no effect b) Any purple colour would disappear. KHS Aug 04 - Cheviot Learning Community - based on Challenge Chemistry R.I.S.E Page 0

11 Consolidation Answers B Q. a) The overall enthalpy change in a reaction (or sequence of reactions) depends only on the reactants and the products and not on the route taken. b) Combustion of magnesium sulphide or MgS + / O MgO + SO or MgS + O MgO + SO b) (ΔH for the) combustion of magnesium sulphide. Q. Ammonium sulphate fertiliser can be manufactured by neutralising sulphuric acid with ammonia. NH (g) + H SO 4(l) (NH 4 SO 4(s) (GFM = 7 g) (GFM = 98 g) (GFM = g) a) g mol (NH 4 SO 4(s) mol NH 4 + ions mol 6.0 x 0 ions mol.04 x 0 4 ions b) though 0 tonnes (0,000,000 g) and 49 tonnes (49,000,000 g) could be converted into grammes and actual numbers of moles calculated, it is the ratio that counts and both calculations can be done with original numbers retained. NH, 7g mol 0g 0/7 =.8 mol mole ratio =.8/ = 0.59 H SO 4, 98g mol 49g 49/98 = 0.5 mol mole ratio = 0.5/ = 0.5 Since H SO 4 has the lower mole ratio it is the limiting reagent and will run out first. An alternative method is to work out the amount of product produced by each chemical and whichever produces the least is the limiting reagent. NH (g) + H SO 4(l) (NH 4 SO 4(s) mol mol 4 g g 4 tonnes tonnes 0 tonnes x 0/4 = tonnes NH (g) + H SO 4(l) (NH 4 SO 4(s) mol mol 98 g g 98 tonnes tonnes 49 tonnes x 49/98 = 66 tonnes KHS Aug 04 - Cheviot Learning Community - based on Challenge Chemistry R.I.S.E Page

12 Consolidation Answers B Q. H (g) + / O (g) H O (l) (state symbols are required) Not H (g) + O (g) H O (l) Q4. a) Initial temperature of water Final (or maximum) temperature of water Initial mass of ethanol ( + container) Final mass of ethanol ( + container) b) i) -67 kj mol - (-½ if unit incorrect or missing) ii) Heat loss to air or Heat loss to beaker [not inaccuracy in thermometer or balance] Q5. Fe O + H Fe + H O mol mol mol of H =. l mol of H = x. = 69.6 l Q6. a) The purple colour just remains / appears. b) ( ) / = 5.7 cm Q7. a) Dissolve the tablet in a little deionised water and make up exactly to the mark in a graduated flask. b) Add a little starch solution to the flask. The end point is when the mixture just turns a permanent blue/black colour. Q8. a) ClO 4 -(aq) + 8H + (aq) + 8e- Cl - (aq) + 4H O (l) b) H + ions are involved/ used up. (aq) KHS Aug 04 - Cheviot Learning Community - based on Challenge Chemistry R.I.S.E Page

13 Consolidation Answers C Q. a) ΔH = - cm ΔT () = -4.8 x 0. x (-) () = +.7 kj () [- if no, or incorrect unit] b) Any one from: no lid/cover to flask (to prevent convection) not waiting to ensure minimum temperature reached not ensuring that the vacuum flask was also at 4 C Q. a) N (g) + H (g) mol mol NH (g) vol vol () 5 cm x 5 = 75 cm () b) N (g) + H (g) NH (g) mol mol vol vol () x 50 = 75 cm 50 cm () Q. Twice (as many methane molecules) Q4. Magnesium reacts with mol l - hydrochloric acid to form hydrogen. Mg (s) + HCl (aq) MgC (aq) + H (g) mol mol () (4. g 4 l () 4.86 g 4 x 4.86/4. = 4.8 l ( Q5. a) Endothermic. b) ΔH = - cm ΔT () = -4.8 x 0.5 x (-8) () = +6.7 kj () [- if no, or incorrect unit] KHS Aug 04 - Cheviot Learning Community - based on Challenge Chemistry R.I.S.E Page

14 Consolidation Answers C Q6. Zn (s) + H SO 4(aq) ZnSO 4(aq) + H (g) Zn, 0.65 mol mole ratio = 0.65/ = 0.65 () H SO 4, x 0.50 = 0.5 mol () mole ratio = 0.5/ = 0.5 () So H SO 4 is the limiting reagent and Zn is the chemical in excess. An alternative method is to work out the amount of product produced by each chemical and whichever produces the least is the limiting reagent. Zn (s) + H SO 4(aq) ZnSO 4(aq) + H (g) mol mol 0.65 mol 0.65 mol = 0.65 x =.0 g Zn (s) + H SO 4(aq) ZnSO 4(aq) + H (g) mol mol 0.5 mol 0.5 mol = 0.5 x =.0 g Q7. C H 5 OH (l) + O (g) CO (g) + H O (l) (state symbols are required) Q8. An electrochemical cell in which the following changes took place was set up. Cr O 7 - (aq) + 4 H+ (aq) + 6 e- Cr + (aq) + 7 H O (l) Br - (aq) Br (aq) + e- a) The second equation must be tripled / multiplied by. or Both of the equations must have an equal number of electrons. b) Cr O 7 -(aq) + 4 H+ (aq) + 6 Br- (aq) Cr+ (aq) + 7 H O (l) + Br (aq) c) Br - / Br - / bromide (ion) / bromide solution (aq) Q9. a) 5 Fe (aq) H+ + MnO - 5 (aq) 4 (aq) Fe+ + (aq) Mn+ + 4 H O (l) () 5 x C x = x 0.05 () so 5 x C x =. so C x =./5 = 0.46 mol l - ( 5 [- if no, or incorrect unit] b) None / no effect or Reduce the margin of error increase the accuracy. KHS Aug 04 - Cheviot Learning Community - based on Challenge Chemistry R.I.S.E Page 4

15 Consolidation Answers D Q. a) b) CH 4(g) + O (g) CO (g) + H O (g) ΔH = - 88 kj x / rev CO (g) CO (g) + / O (g) ΔH = + 8 kj x rev H O (g) H (g) + / O (g) ΔH = kj Combined: CH 4(g) + H O (g) CO (g) + H (g) + 59 kj [Correct equations (), [Correct ΔH values (), () if only one error] () if only one error] Q. CO (g) + O (g) CO (g) mol mol vol vol So 80 cm of CO would require 40 cm of O - 50 cm available so O is in excess and CO is the limiting reagent. CO (g) + O (g) CO (g) mol mol vol vol 80 cm of CO would produce 80 cm of CO Q. a) l of water = kg () ΔH = - cm ΔT () = -4.8 x x (-.5) () = +0.5 kj () [- if no, or incorrect unit] b) KBr, mol = 9 g () 8 g +0.5 kj () 9 g +0.5 x 9/8 () = +5.5 kj mol - ( Q4. a) L (Data Booklet) b) Al(NO ) so mol of nitrate ions per mol of aluminium nitrate c) MgSO 4 MgBr () 4 mol of SO 4 means 4 mol of MgSO 4 4 mol Mg () Therefore, Mg in MgBr = 7-4 = mol () So Br in MgBr = x = 6 mol ( KHS Aug 04 - Cheviot Learning Community - based on Challenge Chemistry R.I.S.E Page 5

16 Consolidation Answers D Q5. a) It is reduced / receives electrons / gains electrons. b Al (s) + 6 H + (aq) Al+ (aq) + H (g) Q6. a) Either of the following: The error in reading the larger volume is a smaller portion of the total volume. The error in reading the longer time is a smaller portion of the total time. b) The volume occupied by a gas changes with temperature. Q7. a) The addition of ions from the potassium iodide b) All of the vitamin C has been oxidised / reacted so no more H + or I - ions are being produced. KHS Aug 04 - Cheviot Learning Community - based on Challenge Chemistry R.I.S.E Page 6

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