Unit 5. Answers Marks Examiner s tips. 1 a) ΔG = ΔH TΔS 1. b) Boiling is a spontaneous change at boiling point. 1 Accept: ΔH = TΔS

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Accept more energy is needed to overcome the intermolecular forces. a) 1 N (g) + H (g) NH (g) 1 Since you are making 1 mole of ammonia this is the only equation. You cannot have multiples here.

1 1 a) ΔG = ΔH TΔS 1 b) Boiling is a spontaneous change at boiling point. 1 Accept: ΔH = TΔS ΔH c) When ΔG = 0, ΔS = T 1000 =.4 40 = 97.5 J K 1 mol 1 d) H bonding in both. H bonding is stronger in HF, because H F is more polar than H N, since F is more electronegative than N. You were told in part b) that ΔG = 0. Accept more energy is needed to overcome the intermolecular forces. a) 1 N (g) + H (g) NH (g) 1 Since you are making 1 mole of ammonia this is the only equation. You cannot have multiples here. b) ΔS = ΣS(products) ΣS(reactants) 19 = = 99.5 J K 1 mol 1 Accept but only award one mark. c) i) ΔG = ΔH TΔS = = +.45 kj mol 1 It would decrease. You could use the entropy change value suggested in the question ( 15 J K 1 mol 1 ) and this would give +41. kj mol 1 as the answer. d) to speed up the reaction 1 This also supplies the activation energy. a) i) W: Pt X: KCl, KNO Y: Mg Z: MgCl (aq), MgSO 4 (aq), and Mg(NO ) (aq) 6 X and W are examples of metal salts, other possible Mg and K salts can be given. Pt H (g) H + (aq) Mg + (aq) Mg In this cell diagram, one mark is for all the species being correct and the other mark is for the correct order. AQA Chemistry A Nelson Thornes Ltd 009 1

2 b) i) V 6 Remember: E = E right E left + i MnO + H O + Zn MnO(OH) + OH + Zn + iv) oxidising agent: MnO reducing agent: Zn v) Zn is used up. c) i) 4H + + SO 4 + e SO + H O Br Br + e 4H + + SO 4 + Br SO + H O + Br 5 Accept: H SO 4 + KBr K SO 4 + SO + H O + Br if you include the spectator ions and join the sulfuric acid ions together. Accept: H SO 4 + KCl K SO 4 + HCl H SO 4 cannot oxidise Cl. Accept: Cl is a stronger oxidising agent than H SO 4. H SO 4 + KCl KHSO 4 + HCl 4 a) The order is 1. gradient = rate When the concentration of iodine is double the gradient doubles. b) Curve X starts at the origin and is below curve B. Curve X finishes at the same level as curve B. c) Curve Y starts at the origin and is steeper than curve A. A lower temperature means a slower reaction, but still makes the same amount of product. A catalyst speeds up the reaction but still makes the same amount of product. Curve Y finishes at the same level as curve A. AQA Chemistry A Nelson Thornes Ltd 009

3 d) i) S O 8 + Fe + SO 4 + Fe + Fe + + I Fe + + I 4 These equations would allow the electrons to cancel out if the overall equation was written. You do not need to have the s in the second equation. alternative route not used up Accept: regenerated instead of not used up. Another alternative would be that it speeds up the reaction. e) i) a different phase from the reactants 4 Reactants adsorb weakly onto the surface of silver i The reaction may be too fast, leading to an explosion 5 a) ΔH o sol = ΔH o lattice + ΔH o hyd so ΔH o lattice = ΔH o sol ΔH o hyd ΔH o lattice(mgcl ) = = +49 kj mol 1 ΔH o lattice(nacl) = = kj mol 1 b) Attraction between ions is weaker. The charge on Na + is only half that on Mg +. c) Al + ions have higher charge/size ratio so can attract water molecules more strongly. [H ][A ] d) K a = [HA] + + [H ] = [HA] = = ph =.0 There are stronger ion dipole forces between the aluminium ion and the water molecules. 4 [H + ]=[A ] or [H + ] = K a [HA] The correct answer of.0 would gain the four marks. ph value must be given to two decimal places. AQA Chemistry A Nelson Thornes Ltd 009

4 e) SiCl 4 + 4H O Si(OH) 4 + 4HCl ph = 0 Since the products are so acidic, the ph range allowed is 1 to 1. 6 a) Mg + (g) + e + Cl(g) 4 Mg + (g) + e + Cl (g) Mg + (g) + e + Cl (g) Mg(g) + Cl (g) b) nd IE ( 11) = ( 64) + 49 One mark is for the factors of. IE = kj mol 1 c) ΔH = ΔH o lattice formation + ΣΔH o hydration = ( 64) = 155 kj mol 1 Accept: a cycle with state symbols or labels instead. If you get the sign of the answer wrong, you can score a maximum of one mark. d) i) increase in disorder on dissolving moles of NH 4 Cl = = heat absorbed = = Q = mcδt Q ΔT = mc = ( )/(50 4.) =.6 C Final temperature = 0.6 = 17.4 C 5 Accept: ΔS positive, ΔG negative, TΔS > ΔH Answers in a range of.5 to.7 are allowed, since you may have rounded your calculations on the top and bottom line. You must put temperature change to at least 1 dp. 7 a) the ability of an atom or element to attract electrons or electron density from a covalent bond Do not put electron, i.e. singular, in the first line. b) It increases. more protons similar or same shielding If the trend is wrong you get no marks. Accept: greater nuclear charge and electrons are in the same shell or they have a similar radius or a smaller radius. AQA Chemistry A Nelson Thornes Ltd 009 4

5 c) i) MgO is ionic P 4 O 10 is covalent If you mention molecules you score 0. The electronegativity difference is small. Accept: the electronegativities are similar, but you cannot say they are the same. d) Na O + H O Na + + OH SO + H O H SO e) MgO + HCl MgCl + H O 1 Accept: the ionic equation MgO + H + Mg + + H O f) P 4 O NaOH 4Na PO 4 +6H O 1 Accept: the ionic equation P 4 O OH 4PO 4 + 6H O 8 a) i) SO + V O 5 SO + V O 4 V O O V O 5 V(IV) or +4 and V(V) or +5 Must have both oxidation states. MnO 4 + 8H + + 4Mn + 5Mn + + 4H O Mn + + C O 4 Mn + + CO Mn(III) or + and Mn(II) or + Must have both oxidation states. b) [Co(NH ) 6 ] + (formed) complex easier to oxidise H O Accept: air or oxygen 9. c) moles of dichromate = 1000 = You may have rounded to , which is OK since there are three significant figures. AQA Chemistry A Nelson Thornes Ltd 009 5

6 moles of Q + = = Each mole of dichromate needs 6 electrons or half equation with 6e. moles of electrons = = moles of electrons per mole of Q = =.00 = Q has an oxidation state of 4. moles Q +: moles dichromate = : 1 Accept: Q 4+ 9 a) CH CH Cl two peaks integration ratio : split into triplet and quartet CH CHCl two peaks integration ratio : 1 split into doublet and quartet b) i) Cl + Br Cl + Br KBr orange-brown solution KI brown solution or black solid 6 If two peaks not stated then no marks for CH CH Cl. If two peaks not stated then no marks for CH CHCl. 9 Accept: Cl + I Cl + I since only one correct equation is required. You must state that it is a solution. Just the colour is insufficient. Any mention of purple loses the mark since iodine is only purple as a vapour. Ba + + SO 4 BaSO 4 BaCl : white precipitate MgCl : no precipitate or no change i CoCl solution goes blue. CuCl solution goes green. [Co(H O) 6 ] + + 4Cl CoCl 4 + 6H O Accept equation, i.e. BaCl + H SO 4 BaSO 4 + HCl Do not accept: nothing or no observation. Accept: yellow or yellow green Accept: [Cu(H O) 6 ] + + 4Cl CuCl 4 + 6H O AQA Chemistry A Nelson Thornes Ltd 009 6

abc Mark Scheme Chemistry 6421 General Certificate of Education Thermodynamics and Further Inorganic Chemistry 2008 examination - June series

abc Mark Scheme Chemistry 6421 General Certificate of Education Thermodynamics and Further Inorganic Chemistry 2008 examination - June series Version 1.0 abc General Certificate of Education Chemistry 6421 CHM5 Thermodynamics and Further Inorganic Chemistry Mark Scheme 2008 examination - June series Mark schemes are prepared by the Principal