QUESTIONSHEET 1 ENTHALPY OF ATOMISATION a) The standard enthalpy of atomisation of an element is the energy required (½) to produce 1 mole (½) of gaseous atoms (½) from the element in its standard state / most stable state under standard conditions (½) The standard enthalpy of atomisation of a compound is the energy required (½) to convert 1 mole (½) of the gaseous compound (½) into gaseous atoms under standard conditions (½) b) Enthalpy of atomisation of hydrogen = ½ H H bond dissociation enthalpy (1) c) (i) 920/2 = 460 kj mol-1 (1) (ii) Data book value includes values for the O H bond in other molecules (1) d) (i) C(s) + 2H 2 (g) CH 4 (g) (1 mark for formulae & balance; 1 for state symbols) (ii) H for bond breaking: 715 + (4 218) = +1587 kj mol-1 (1) H for bond making: -4 H be (C H) (1) Hê f (CH 4 ) = -74.9 = + 1587-4 H be (C H) (1) H be (C H) = (1587 + 74.9)/4 = 415 kj mol-1 (1) Or +715 kj mol C(g) -1 C(g) +2H 2 (g) 4(+218) kj mol -1 4H(g) -74.9 kj mol -1 4 H be (C H) CH 4 (g) (2) Deduct 1 mark for each error Or accept an enthalpy diagram (2) By Hess s law, -74.9 + 4 H be (C H) = +715 + 872 (1) H be (C H) = (715 + 872 + 74.9)/4 = 415 kj mol-1 (1)
QUESTIONSHEET 2 LATTICE ENTHALPY a) Enthalpy / heat required to separate the gaseous ions (1) in 1 mole of solid (1) to an infinite distance from one another (1) Or enthalpy / heat released when gaseous ions (1) infinitely far apart (1) join together to give 1 mole of solid (1) b) (i) Change Decreases (1) Explanation Ionic radii increase (1) Hence charge density / charge : radius ratio of the cations decreases (1) (ii) Melting point Decrease (1) Solubility in water Increase (1) c) (i) Ba 2+ (g) + 2F - (g) + aq (½) Hê hyd -1275 kj mol -1 (½) Hê lattice +2352 kj mol-1 (½) Ba 2+ (aq) + 2F - (g) + aq (½) 2 Hê hyd 2 (-530) = -1060 kj mol -1 (½) Ba 2+ (aq) + 2F - (aq) (½) Ba 2+ (F - ) 2 (s) + aq (½) Hê soln (½) By Hess s law, Hê soln = +2352 1275 1060 (1) = +17 kj mol-1 (1) (ii) Hê soln for BaF 2 (s) is endothermic while the value of Hê soln for BaCl 2 (s) is exothermic (1) dissolving of BaF 2 (s) is thermodynamically less favourable (1)
QUESTIONSHEET 3 BORN-HABER CYCLES I a) Born-Haber cycle Li + (g) + e + F(g) (1) +79.0 kj mol -1 (1) -334.0 kj mol -1 (1) Li + (g) + e + ½F 2 (g) (1) Li + (g) + F (g) (1) +520.0 kj mol -1 (1) Li(g) + ½F 2 (g) (1) +159.5 kj mol -1 (1) Li(s) + ½F 2 (g) (1) Hê lattice (1) -616.0 kj mol -1 (1) Li + F - (s) (1) Lattice enthalpy of lithium fluoride By Hess s law, 159.5 + 520.0 + 79.0 334.0 + Hê lattice = -616.0 (1) Hê lattice = -616.0 159.5 520.0 79.0 + 334.0 = -1040.5 kj mol-1 (1) b) Calculation is based on the assumption that bonding is purely ionic (1) In reality, bonding has some covalent character (1)
QUESTIONSHEET 4 BORN-HABER CYCLES II a) (i) Mg + (g) + e + Cl(g) (½) +121 kj mol -1 (½) -364 kj mol -1 (½) Mg + (g) + Cl (g) (½) Mg + (g) + e + ½Cl 2 (g) (½) +736 kj mol -1 (½) Mg(g) + ½Cl 2 (g) (½) -756 kj mol -1 (½) +150 kj mol -1 (½) Mg(s) + ½Cl 2 (g) (½) Hê f (MgCl) (½) MgCl(s) (½) (ii) Hê f (MgCl) = +150 + 736 + 121 + (-364) + (-756) (1) = -113 kj mol-1 (1) (iii) The formation of MgCl 2 is more exothermic than MgCl (1) Hence MgCl 2 is more stable / at a lower energy level than MgCl (1) b) Lattice enthalpy of MgCl (1) c) Second ionisation energy of magnesium (1) Lattice enthalpy of MgCl 2 (1) d) Greater or less than Hê lattice (MgCl 2 ) Less (1) Explanation Mg+ has a smaller charge than Mg 2+ (1) and is larger (1) Or Mg+ has a lower surface charge density than Mg 2+ (2)
QUESTIONSHEET 5 ENTROPY AND FREE ENERGY CHANGE a) Gê Standard free energy change / free energy change at 298 K and 1 atm (½) kj mol -1 (½) T Absolute temperature / temperature in kelvins (½) K (½) Hê Standard enthalpy change / enthalpy change at 298 K and 1 atm (½) kj mol -1 (½) Sê Standard entropy change / entropy change at 298 K and 1 atm (½) J K -1 mol -1 (½) b) (i) Decrease (1) O 2 gas (disordered) MgO solid (ordered) (1) (ii) Increase (1) Mg solid (ordered) H 2 gas (disordered) (1) (iii) Remain approximately the same (1) Reactants and products are all gaseous (½) No change in the number of gas molecules (½) (iv) Increase (1) Increased number of gaseous molecules (1) c) (i) Total entropy of products = (2 213.6) + (3 69.9) = 636.9 J K -1 mol -1 (1) Total entropy of reactants = (160.7) + (0) = 160.7 J K -1 mol -1 (1) Sê = 636.9 160.7 = 476.2 J K -1 mol -1 (1) (ii) Gê = -1367.3 (298 476.2 10-3) (1) = -1509.2 kj mol -1 (1)
a) (i) more ordered/less random structures (1) so fewer energy quanta available (1) (ii) sodium 2Na(s) + Cl 2 (g) 2NaCl(s) (1) 72-(51 + 223/2) (1) = -90.5 (1) hydrogen H 2 (g) + Cl 2 (g) 2HCl(g) (1) 187-(130+223) (1) 2 = 10.5 (1) (iii) NaCl : product more ordered structure (1) so loss of entropy (1) but entropy of surroundings increases (1) then as exothermic (1) or so overall loss of free energy (1) b) (i) HF 7.5 1000/293 (1) = 25.6 (1) HCl 16.2 1000/188 (1) = 86.2 (1) (ii) G = 0 (1) H = T/ S (1) (iii) HF retains order in gas state (1) giving lower change in randomness (1) due to strong H bonding (1) TOPIC 17 ANSWERS & MARK SCHEMES QUESTIONSHEET 6 ENTROPY AND STATE
QUESTIONSHEET 7 TEST QUESTION I a) Step 1 Enthalpy of atomisation / sublimation of sodium (1) Step 2 (First) ionisation energy of sodium (1) Step 3 Hydration enthalpy of the sodium ion (1) Step 4 Dissociation enthalpy of water (1) Step 5 Step 6 Step 7 b) (i) Negative(½) hydration enthalpy of the proton / hydrogen ion (½) Negative (½) ionisation energy of hydrogen (½) Negative (½) enthalpy of atomisation of hydrogen / half the bond dissociation enthalpy of the H H bond (½) Na + (g) + OH - (aq) + H + (aq) Na + (g) + OH - (aq) + H + (g) +1090 kj mol -1-406 kj mol -1 Na + (aq) + OH - (aq) + H(g) Hê x Na + (g) + H 2 O(l) -1316 kj mol -1 +500 kj mol -1 Na(g) + H 2 O(l) +109 kj mol -1 Na + (aq) + OH - (aq) + H + (g) Na(s) + H 2 O(l) Start -183.7 kj mol -1 Na + (aq) + OH - (aq) + ½H 2 (g) -218 kj mol -1 Finish 4 marks (Deduct ½ for each error or omission) (ii) By Hess s law, -183.7 = +109 + 500 + Hê x + 1090 406 1316 218 (1) Hê x = -183.7 109 500 1090 + 406 + 1316 + 218 = +57.3 kj mol -1 (1) c) Predicted value -57.3 kj mol -1 (1) Explanation Because neutralisation / H + (aq) + OH - (aq) H 2 O(l) is the reverse of the dissociation of water (1)
QUESTIONSHEET 8 TEST QUESTION II a) Mg(s) + ½O 2 (g) MgO(s) (2) Award (1) for correctly balanced equation and (1) for correct state symbols. b) (i) Mg 2+ (g) + O 2 (g) Mg 2+ (g) + 2e + O(g) +1451 kj mol -1-142 kj mol -1 +844 kj mol-1 Mg 2+ (g) + e + O (g) Mg + (g) + e + O(g) +738 kj mol -1 Mg(g) + O(g) +249 kj mol -1-3884 kj mol -1 Mg(g) + ½O 2 (g) +147 kj mol -1 Mg(s) + ½O 2 (g) Hê f (MgO) MgO(s) (4) Deduct 1 mark for each error. (ii) Hê f (MgO) = [+147] + [+249] + [+738] + [+1451] + [-142] + [+844] + [-3884] (1) = -597 kj mol -1 (1) c) Sê = [27.0] [(32.7) + ½(205)] (1) = -108.2 J K -1 mol-1 (1) d) (i) Gê = Hê - T Sê (1) = [-597 103] [(298) (-108.2)] (1) = -564 800 J mol -1 = -564.8 kj mol -1 (1) (ii) Reaction has high activation energy (1) Energy is required to form Mg 2+ and O 2- ions before lattice enthalpy can be released (1) reaction is infinitely slow (1) Maximum 2 marks
QUESTIONSHEET 9 TEST QUESTION III a) (i) The enthalpy change when 1 mole of gaseous ions (1) becomes hydrated by a large / infinitely large quantity of water (1) (ii) The enthalpy change when 1 mole of the solid (1) is dissolved in a large quantity of water / to give an infinitely dilute solution (1) b) (i) Mg 2+ (g) + 2Cl (g) Hê hyd (Mg 2+ (g)) +2493 kj mol -1 Mg 2+ (aq) + 2Cl (g) MgCl 2 (s) -728 kj mol -1-155 kj mol -1 (3) Deduct 1 mark for each error. Mg 2+ (aq) + 2Cl (aq) (ii) Hê solution = Hê lattice + Hê hyd(cation) + Hê hyd(anion) (1) (iii) -155 = +2493 + Hê hyd (Mg 2+ (g)) + [-728] (1) Hê hyd (Mg 2+ (g)) = [-155] + [-2493] + [+728] = -1920 kj mol -1 (1) c) (i) Value becomes more negative (1) (ii) Size of the metal cation increases (1) charge density of the metal cation decreases (1) (iii) Decrease in lattice enthalpy is much greater than decrease in hydration enthalpy (1) Hê solution becomes more negative and solubility increases (1)
QUESTIONSHEET 10 PRINCIPLES OF ELECTROCHEMICAL CELLS a) (i) Oxidation Zn(s) Zn 2+ (aq) + 2e - (1) Reduction Cu 2+ (aq) + 2e - Cu(s) (1) (ii) Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) (1) (iii) Solution changes colour from blue to colourless (1) Pink / red solid is precipitated (1) b) (i) Zinc rod becomes smaller (1) Copper rod becomes larger (1) Blue colour of CuSO 4 (aq) fades (1) (ii) With reaction in a cell, energy is released as electricity (1) With reaction in a beaker, energy is released as heat (1) c) (i) For a redox reaction to occur, there must be oxidation at one half-cell (1) and reduction at the other (1) Or electrons are released at one (1) and have to be accepted at the other (1) (ii) To complete the circuit (1) Contains saturated KNO 3 (aq) / KCl(aq) (1) as a gel or saturated on filter paper or in sintered glass salt bridge(1) (iii) Electrons (1) which flow from Zn to Cu (1) (iv) K + ions to the Cu half-cell (1) NO 3 - / Cl - ions to the Zn half-cell (1) SO 4 2- ions from the Cu half-cell to the Zn (1)
QUESTIONSHEET 11 ELECTRODE POTENTIALS AND CELL EMF a) (i) Metal tends to ionise / form hydrated ions in solution (½) releasing electrons to the rod / giving the rod a negative charge (½) Reverse change also tends to occur (½) accepting electrons from rod / giving rod a positive charge (½) Equilibrium is established between the opposing reactions (1) (ii) Enthalpy of atomisation (1) Ionisation energy / energies (1) Enthalpy of hydration (of ions) (1) (iii) 298 K (1) 1 M / 1 mol dm -3 (1) (iv) Different reactivity (1) Highly reactive metals have a high tendency to ionise (½) giving a high electron density on the metal rod / rod acquires a high ve potential (½) Metals of low reactivity converse argument (1) Maximum 2 marks b) (i) Platinium (foil) (1) (ii) 1M / 1 mol dm -3 (1) c) (i) The voltage developed when the cell delivers zero current / the maximum voltage the cell can develop (1) (ii) High resistance (1) voltmeter (1) in the external circuit (1) (iii) E cell = E + E - (1) Or E RHS E LHS (applied to cell notation) (1) Or E reduction electrode E oxidation electrode (1)
QUESTIONSHEET 12 MEASUREMENT OF ELECTRODE POTENTIALS a) (i) It is impossible to find the potential of a single electrode (1) because we can only measure a potential difference (1) Or because a voltmeter has two terminals and must be connected to two electrodes (1) (ii) H 2 (g) (½) 1 atm (½) HCl(aq) or H 2 SO 4 (aq) (½) [H] + = 1 mol dm -3 (½) at 298 K (1) Pt foil (1) b) (i) Reduction Fe 3+ (aq) + e - Fe 2+ (aq) (1) Oxidation Cu(s) Cu 2+ (aq) + 2e - (1) (ii) 2Fe 3+ (aq) + Cu(s) 2Fe 2+ (aq) + Cu 2+ (aq) (1) (iii) E cell = E + E +0.43 = E + 0.34 (1) E + = + 0.77 V (1) (iv) E cell = E + E E cell = 0.43 ( 0.37) (1) E cell = 0.71 V (1)
QUESTIONSHEET 13 FACTORS INFLUENCING ELECTRODE POTENTIALS AND CELL EMF a) (i) Pt(s) H 2 (g), H+ (aq) Ag + (aq) Ag(s) (2) (1for species in correct order, 1 for state symbols) (ii) E cell = + 0.80 0.00 = + 0.80 V (1) (iii) From hydrogen to silver (1) (iv) H 2 (g) + 2Ag + (aq) 2H + (aq) + 2Ag(s) (2) (1 for species, 1 for balance & state symbols) (v) ph decreases (1) because H + ions are formed in the cell reaction (1) Or because [H + ] increases (1) b) (i) Effect on electrode potential Nil (1) Effect on e.m.f. Nil (1) (ii) Effect on electrode potential More +ve / higher (1) Effect on e.m.f. Decreases (1) (iii) Effect on electrode potential More +ve / higher (1) Effect on e.m.f. Decreases (1) (iv) Effect on electrode potential More ve / becomes ve (1) Effect on e.m.f. Increases (1) (v) Effect on electrode potential More +ve / higher (1) Effect on e.m.f. Increases (1)
QUESTIONSHEET 14 PREDICTION OF REDOX CHANGES FROM STANDARD ELECTRODE POTENTIALS a) (i) I (½) & S 4 O 6 2 (½) (ii) Cl (½) & SO 4 2 (½) (iii) None (1) (iv) Fe 3+ and I 2 (½) & Cr 3+ (½) b) (i) Eê cell =+ 0.77 (+ 1.23) = 0.46 V (1) Outcome No reaction (because Eê cell is ve) (1) (ii) Eê cell = + 1.33 (+ 0.54) = + 0.79 V (1) Outcome Reaction goes to completion (because Eê cell > 0.3 V) (1) (iii) Eê cell =+ 1.36 (+ 1.23) = + 0.13 V (1) Outcome Dynamic equilibrium (because Eê cell < 0.3 V) (1) (iv) Eê cell =+ 1.36 (+ 1.20) = 0.16 V (1) Outcome Dynamic equilibrium (because Eê celll < 0.3 V) (1) c) (i) Brown colour / iodine is expected (1) from a comparison of Eê values / because Eê cell is positive (1) Reason: high activation energy (1) Both reacting ions are negatively charged / repel each other (1) Maximum 3 marks (ii) Comparison of Eê values / Eê cell negative suggests no reaction (1) but hydrochloric acid is concentrated / [Cl ] is > 1M (1) hence conditions are non-standard / Eê values do not give a reliable prediction (1)
QUESTIONSHEET 15 FUEL CELLS a) Reactants / oxidising and reducing agents are fed in continuously / from outside the cell (1) Reaction products are removed continuously (1) b) (i) H 2 (g) + 2OH (aq) 2H 2 O(l) + 2e (1) (ii) O 2 (g) + 2H 2 O(l) + 4e 4OH (aq) (1) (iii) 2H 2 (g) + O 2 (g) 2H 2 O(l) (1) (iv) To increase surface area (1) to assist adsorption of gas (1) c) (i) 1 kwh 4.3 10 6 / 96 500 = 44.56 mol electrons (1) 2 mol e 1 mol H 2 44.56 mol e 22.28 mol H 2 (1) V = 22.28 24 = 534.7 / 535 dm 3 H 2 (1) (ii) n (H 2 ) required in practice = (22.28 100) / 72 = 30.94 mol (1) m (H 2 ) = 30.94 2 = 61.9 / 62 g H 2 (1) d) (i) The reaction product is water which is non-polluting (1) Because H 2 (g) and O 2 (g) can be obtained from water, there is no need to use non-renewable fossil fuels (1) (ii) Storage as a hydride (1) or adsorbed on to a solid adsorbant (1) (Not stored as compressed / liquefied gas )
QUESTIONSHEET 16 TEST QUESTION IV a) The e.m.f. of a cell (1) comprising a half-cell consisting of Pt foil (1) immersed in a solution of an aqueous halogen and halide ions (1) both at concentrations of 1.00 mol dm -3 (½) at 298 K (½) and a second half-cell which is a standard hydrogen electrode / SHE (1) b) Oxidising power decreases as Group 7 is descended (1) c) (i) Equation 2Fe 2+ (aq) + Cl 2 (aq) 2Fe 3+ (aq) + 2Cl - (aq) (1) e.m.f. = Eê reduction half-cell - Eê oxidation half-cell = +1.36 (+0.77) = +0.59 V (1) Reaction is energetically feasible (½) because the e.m.f. is positive (½) (ii) Equation 2Fe 2+ (aq) + I 2 (aq) 2Fe 3+ (aq) + 2I - (aq) (1) e.m.f. = +0.54 (+0.77) = -0.23 V (1) Reaction is not energetically feasible (½) because the e.m.f. is negative (½) (iii) Equation 2Fe 3+ (aq) + 2I - (aq) 2Fe 2+ (aq) + I 2 (aq) (1) e.m.f. = +0.77 (+0.54) = +0.23 V (1) Reaction is energetically feasible (½) because the e.m.f. is positive (½) In the whole of c) state symbols are necessary, but deduct 1 mark only if they are not shown. d) FeI 3 is an unstable compound / it would decompose into FeI 2 and I 2 (1)
QUESTIONSHEET 17 TEST QUESTION V a) A = electrochemical B = negative C = positive D = oxidant E = electrolyte F = catalysis G = carrier (7) b) advantages renewable (1) non-polluting/produces only water (1) disadvantages low energy yield per unit volume (1) not liquefiable under normal pressures (1) c) (i) H 2 (g) 2H + (aq) + 2e (1) equation (1) state subscripts (ii) Increased cell e.m.f (1) Increased [Cu 2+ ] moves (reversible) half-cell reaction to Cu (1) Increases e acceptance/+ half-cell potential (1) (iii) 0.34V (1)