4. State Hess s Law enthalpy/heat/heat energy change (1) independent of route (1) allow a clear diagram

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Edwin Greer
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1 1. Define the term standard enthalpy of combustion. one mole of (of a substance) (1) completely burnt (OR burned in excess 0 2 ) (1) under standard conditions (1) (3) 2. Calculate the enthalpy change for the reduction of propanoic acid to propanal: CH 3 CH 2 COOH + H 2 CH 3 CH 2 CHO + H 2 O given the following enthalpies of combustion / kj mol-1: propanoic acid ; hydrogen - 286; propanal ; H = H(propanoic) + H c +(H 2 ) - H(propanal) (1) = (-1527) + (-286) - (-1821) (1) = (+) 8 (kj mol-1) (1) correct answer with some working (3) correct answer with no working (1) incorrect sign in answer (max 1) 3. The thermal stability of group 2 carbonates increases from beyllium, the carbonate of which decomposes at room temperature, to barium, the carbonate of which is stable even at quite high temperatures. Explain this trend. Be2+ and CO 3 2 very different in size (1) whereas in the oxide produced are not so different (1) so H lat of oxide higher (1); differences with BaCO 3 not so marked (or mismatch in Ba2+ / O2 size) so BaCO 3 more stable relative to oxide (1) OR Be2+ smaller than Ba2+ Be2+ therefore very polarising (1) Co 3 2 more polarisable than O2 (1) BeCo 3 is more covalent or BaCO 3 more ionic (1) 4. State Hess s Law enthalpy/heat/heat energy change (1) independent of route (1) allow a clear diagram 5. (i) Write an equation the enthalpy change for which would be the enthalpy of formation of zinc sulphide, ZnS. Zn(s) + S(s) ZnS(s) (1) allow 1/ 8 S 8 must have correct state symbols (3) (Total 4 marks) (Total 2 marks) (1) NT Exampro 1

2 (ii) In the smelting of zinc ores, the following reaction occurs: ZnS(s) + 1½O 2 (g) ZnO(s) H = 441 kj mol 1 Use this, together with the data below, to calculate a value for the enthalpy of formation of ZnS. Data: ZnS(s) + 1½O 2 (g) ZnO(s) H = 441 kj mol 1 S(s) + O 2 (g) SO 2 (g) H = 297 kj mol 1 correct answer by any method with full sensible working (5) if correct answer but no working then (1) only incorrect answer means max 4 as follows: (2) for cycle of H statement (penalise ( 1) for each error) (2) for substitution of correct values and signs in the appropriate equation (penalise ( 1) for each error) Example Zn + S ZnS OR ZnO + SO ZnS + 1½O 2 ZnO + SO 2 H react = H f [ZnO] + H f [SO 2 ] H f [SnS] 441 = H f [ZnS] H f [ZnS] = 204 ignore units 6. (a) Enthalpy change / heat changes (1) is independent of route (1) 2 2 (5) (Total 6 marks) (b) Enthalpy or heat change for complete combustion / complete reaction with oxygen / burning in excess air (1) per mole of substance (1) 1 atm or standard states (1) 3 (c) (i) C C + 5C H + C O + O H + 3O=O = (1) 4C=O + 6O H = 5750 (1) H = = 1031 kj mol 1 (1) 3 NT Exampro 2

3 (ii) Diagram showing an exothermic reaction and labelled products and reactants / equations (1) Showing energy barrier (1) Consequential marking on (i) C H OH + 3O CO + 3H O (a) (280/2 = )140 (1) 5510/114 (1) = ( )48.3 (1) 3 (b) Eg. Hydrogen Less pollution (1) Greater energy per unit mass (1) renewable resource (1) Eg. Octane Liquid easily transported (1) More readily available (1) The two reasons must relate to the same fuel no mark for fuel itself 2 (c) The answer requires a disadvantage for the fuel chosen in part (b) Hydrogen needs to be made using electricity / idea of storage problems developed (1) Octane (air) pollution or non-renewable (1) 1 8. First table All weighings recorded in correct spaces to at least 2 dp (1) Correct subtraction (1) 2 Second table Both temperatures recorded in correct spaces to at least 1dp. (1) Correct subtraction including sign (1) Accuracy Multiply the candidate s - corrected if necessary - mass by 1.55_ and record this figure to 1 dp close to the candidates DT value (corrected if necessary) Record the difference between the two values on the script. Award accuracy marks as follows 2 [10] [6] Difference Mark (a) Correct arithmetic (1) NT Exampro 3

4 Positive sign (1) Value to 2sf or 3sf and units (1) 3 (b) The principle is general change (1) reason for change (1) Note: If a candidate changes the apparatus give 1 mark for reasoning Examples might include: e.g Take a series of temperature readings / repeat experiment (and average results ) (1) e.g To allow for odd / erratic temperatures / heat gain from surroundings (1) e.g Use more solid (1) larger T value (1) e.g Put a lid on apparatus (0) to prevent heat losses (1) e.g Use the same thermometer / different thermometer (0) reading to an accuracy of 0.1 C (1) 2 9. (a) Three possible reasons Reason 1 Any fluctuations in temperature smoothed out / minimizes reading error / allows line of best fit to be drawn (1) Reason 2 Able to allow for cooling effect / able to calculate more accurate temperature change / need to find highest temperature (1) Reason 3 Reference to problem of how the experiment is carried out practically e.g. reaction slow / too much to do all at once (1) Any 2 2 (b) Extrapolation of lower and upper temperatures at 3.5 min (1) T correct to C according to candidate s method (1) Consequential on sensible method of finding temp change between 3 and 4 minutes 2 (c) H = candidate s T H correctly calculated (1) with consequentially correct sign and to 3 or 4 SF (1) This is consequential on part (b) (a) (i) enthalpy change /heat energy released on complete combustion of 1 mole of substance under standard conditions (3) (ii) activation energy needed to be present before reaction can proceed at reasonable rate (2) (iii) methane plus oxygen at higher level than products activation energy peak shown. H shown on diagram (consequential on above) (3) 8 (b) (i) 2C 2 H O2 16CO H 2 O 1 mark for correct species 1mark for correct balance (2) (ii) 1 mole of octane needs 12.5 moles of oxygen 1 mole of octane needs moles of air (1) 1 mole of octane needs dm³ of air = 1500 dm³(1) (2) [13] [6] NT Exampro 4

5 (c) (iii) 890/16 (1) = 55.6 kj (1) (2) (iv) 890/24 = 37.1 kj (2) (v) More energy per unit of volume Methane difficult to liquefy/containers for compressed gases heavy Plus 1 mark for clarity of argument and communication of ideas (3) 11 Source of hydrogen Source of heat (energy)to run process (2) (a) Mass ethanol + burner : initial (1) Mass ethanol + burner : final (1) Initial water temperature (1) Final water temperature (1) Mass water (1) 5 (b) Criticism 1: shown as standard value (1) Reason: Experiment not conducted under standard conditions (1) Criticism 2: Too many d.ps/significant figures (1) Reason: accuracy of apparatus/expt does not warrant (1) Criticism 3: Not shown as negative (1) Reason: Exothermic reaction (1) 6 (c) Loss of heat to surroundings (1) Loss of ethanol by evaporation (1) Incomplete combustion (1) Max (a) (i) Enthalpy/heat change for formation of 1 mole of a compound (1) from its elements (1) in their standard states / or stated temperature of 298K (25 C) and 1 atm (or suitable unit) (1) 3 (ii) = ( ) (1) = (Kj mol 1) (1) 2 (iii) negative sign means reaction exothermic/gives out heat (1) if no answer given to part (ii) must give general explanation that negative means exothermic and positive means endothermic reaction 1 (b) Energy in = ( ) = 978 (1) Energy out = = 1036 (1) Energy change = = -58 (1) consequential If candidates choose to include the four C-H bonds the above figures are 2626, 2684 and [21] [13] NT Exampro 5

6 (c) Average values from many compounds used in bond enthalpies (1) Actual values for these compounds probably slightly different / or, calculation in (a) (ii) uses real / actual / experimental /standard/ values (1) n.b. do not accept arguments based on error 2 [11] 13. Thermochemistry Table 2 All weighings recorded in correct spaces to 2 d.p. (1) correct subtraction (1) 2 Table 3 Both temperatures recorded in correct spaces to at least one d.p. (1) Correct subtraction including sign (1) Accuracy Compare candidate s value of T with that of the examiner or supervisor, and record the difference between the two as d =... on the script. Mass (rounded to 1 d.p.) gives T record on script T = 8.4 C T = 8.5 C Award marks for accuracy as follows: difference d= ±0.50 ±0.80 ±1.00 ±1.50 mark Calculation and Question (a) Calculation correct arithmetic (1) with +ve sign (1) (units given) Value to 2 or 3 sig. figs. (1) This calculation is based on the candidate s own value 3 (b) One mark for suggestion, second mark for linked reason: Take a series of temperature readings and plot graph (1) to allow for heat gain from surroundings (1) Repeat experiment and average results (1) to allow for odd /erratic temperatures (1) Use more solid (1) larger T value (1) Put a lid on apparatus / increase insulation (1) to prevent heat gains (1) Use a different thermometer (1) reading to an accuracy of 0.1 C (1) (a) Enthalpy / heat (energy) change on the neutralisation / reaction of one mole of a monobasic acid / hydrogen ions (by an alkali) or Enthalpy / heat (energy) change on the formation of one mole of water when an acid is neutralised Or Enthalpy change per mole for reaction H+ + OH, H 2 O (1) 1 (b) q = mc T (1) other unambiguous symbols/names = (1) 6 [15] NT Exampro 6

7 = 2884 J including units (1) 3 Consequential on sensible chemistry in line 2 i.e. use of 50 for mass or temp in K or data for temperature, transposed(max2). Ignore sign of answer Allow 3 or 4 significant figures (c) 2884/0.05 (1) answer from (b) 0.05/allow answer from (b) 20 = 57.7 kj mol 1 (1) accept If wrong sign (max 1) If wrong units (max 1) (d) Ensures all acid reacts / neutralisation (of acid) 1 completed / reaction (of acid) completed / all H+ reacted (1) 15. (a) (i) Enthalpy or heat change / released when 1 mol of substance (1) is burned in excess oxygen / completely (1) all substances in standard states (at a specified temp)/ at a pressure of 1 atm. (1) 3 (ii) Suitable cycle (need not be labelled but if labelled, these must be correct) (1) working (1) answer (1) e.g. C + 1 / O CO ( H CO) ( H Carbon) comb 2 2 CO ( H Carbon monoxide) 2 comb = 394 ( 283) (1) = 111 (kj mol 1) (1) Penalise 1 mark if units incorrect 3 (iii) (some) CO 2 is always produced in the reaction (1) 1 (iv) (Energy) C (+ O ) 1 /2 2 CO form Consequential on (a) (ii) n.b. if no answer in(a) (ii), correct diagram can still score (1) (1) 1 [7] NT Exampro 7

8 (b) Methane (and oxygen) / reactants thermodynamically unstable w.r.t. products (1) Must be a comparison Since reactants are at a higher energy level (than products) (1) or reverse argument Reactants / methane, oxygen kinetically stable (1) Due to high activation energy (1) 4 If no reference to methane in the answer (max 3) (c) Comparison of energy per unit volume (1) Positive comment on storage / transport of octane e.g. octane is a liquid therefore easy to store/ transport/ transfer (1) Negative comment on storage/ transport of methane e.g. methane is gas and will need to be stored under pressure in a special container as a liquid / in a large container as a gas (1) (a) Note 1 mark for improvement 1 mark for related reason in each case to max 4 marks. Reason must relate to improvement. Max 2 for improvement. Max 2 for reason. Improvement insulate beaker / polystyrene cup / plastic cup / use lid (1) Reason Prevents / reduces heat loss or absorbs less heat (1) Improvement Use pipette / burette (1) Reason More accurate (than measuring cylinder) (1) Improvement Measure temperature for several minutes before the addition (1) Reason Allows more accurate value for the initial temperature (1) Improvement Measure temperature more often (1) Reason Allows for better extrapolation / more accurate temperature change from graph (1) Improvement Read thermometer to 1 dp / use more precise thermometer/ digital thermometer (1) Reason Gives more accurate temperature change (1) Improvement Stir mixture (1) Reason Ensure even temperature / reaction faster less heat loss with time (1) Improvement Use finely divided iron / smaller pieces (1) Reason Reaction faster less heat loss with time (1) Not speeds up alone 4 (b) (i) Heat change = J = / 1000kJ = 3.18kJ or 3180J (1) Ignore sig. fig. Allow mark if units omitted 1 If units quoted but wrong eg 3.18 J score 0. (ii) No of mols of copper sulphate = / 1000 = (1) 1 [15] NT Exampro 8

9 (iii) Enthalpy change per mol = 3.18/.025 = 127kJ (1) negative sign (1) stand alone consequential on (i) and (ii) max 4 sig fig and answer must be in kj mol 1 even if units omitted. 2 [8] 17. (a) Enthalpy or heat change or heat energy / released when 1 mol of substance / element or compound (need to say both) (1) is burned in excess oxygen / completely / reacts completely (1) at 1 atm pressure and specified temperature (1) 3 (b) H = 2 H c (C) + 2 H c (H 2 ) H c (CH 3 COOH) (1) for this or equivalent cycle drawn; H = ( 394 2) + ( 286 2) ( 874) (1) = 486 kj mol 1 (1) 3 (c) (Enthalpy of) formation / H f (1) (d) correct orientation of energy levels / labelled (at least one) (1) H shown - number allowed (1) reaction profile showing E a (1) [if based on (b) max 2] Enthalpy C (+ O ) 2 H CO activation energy (e) Reactants are at a higher energy level (than products) (1) so the CO 2 / products are thermodynamically stable with respect to reactants / C (+ O 2 ) (1), (or the inverse argument) activation energy high enough to prevent appreciable reaction (at room temperature) (1) so mixture / C + O 2 is kinetically stable (1) (max 3 if no ref to graphite / C) Part 3 Table 2 Full set of temperatures recorded (1) Each temperature recorded to one dp (1) allow one slip 2 (a) Appropriate scales (1) Plotting points (1) allow one slip 2 (b) Drawing smooth / continuous lines which join at maximum (1) Allow curves or straight lines of best fit. (c) Reading off temperature (1) and volume correctly (1) 2 2 [14] NT Exampro 9

10 (d) (i) T value correct (1) Accuracy Compare candidate s T with that of the examiner/supervisor and record the difference between the two as d = on the script. Expected value = 6.5 C Award marks for accuracy as follows. Difference d = ±0.5 ±0.7 ±0.9 ±1.0 Mark (ii) V T correctly calculated (1) 1 (e) (i) 10 answer to part 2 (a) (1) 1 (ii) H correctly calculated to no more than 3 sig fig (1) 1 answer must include - sign but ignore units unless wrong (f) Accuracy would be increased improved (1) Temperature rise would be greater so less error in temperature readings. (1) 2 First mark consequential on an attempt at an explanation 19. (a) (i) Reaction is complete (1) addition of cooler NaOH causes temp to fall (1) 2 (ii) 20.0 cm3 (1) 1 (iii) / 1000 (1)= mol 1 (iv) / 1000 (1) = mol 1 (v) 1 : 2 (1) MUST be consequential on working in (iii) to (iv) 1 (vi) Cu(OH) 2 (1) Consequential provided that the ratio of Cu to OH is a whole number 1 (b) (i) 7.2 C (or K) (1) 1 (ii) q = 1210 J / 1.21 kj (1) Consequential on (b)(i) 1 (iii) H = 1210 J / (1) ie. method Mark consequentially on (a)(iv) and (b)(ii). sign (1) Correct units (1) (*) 2 max if numerical error (*) In final answer 3 (c) No stirring / poor mixing (1) Specified method of stirring or mixing e.g. magnetic stirrer / swirl cup between additions (1) or Solutions at different initial temperatures (1) Allow them to stabilise at room temperature (1) Do not allow anything to do with heat loss. Do not allow more accurate thermometer` since the one specified is good enough. 2 [17] [14] 20. (a) (i) It is the enthalpy / heat (energy) change / evolved for the formation of 1 mol of urea (1) NT Exampro 10

11 (b) from its elements (1) in their standard states / at 1 atm and stated temperature {298K} (1) 3 AMENDED (ignore units) {(-333.0) + ( 285-8)} {(2 46.2) + ( 393.5)} = = kj (3) Correct answer with some correct working (3) Correct answer alone (1) kj (2) Omitting the 2 gives kj (2) kj (1) Incorrect application of Hess s Law gives kj (2) kj (1) Incorrect Hess s Law and omit 2 gives kj (1) kj (0) NOT AMENDED (ignore units) {(632.2) + ( 285.8)) ((2 46.2) + ( 393.5)} (1) = = kj (3) Correct answer with some correct working (3) Correct answer alone (1) kj (2) Omitting the 2 gives kj (2) kj (1) Incorrect application of Hess s Law gives kj (2) kj (1) Incorrect Hess s Law and omit x2 gives kj (1) kj (0) 21. (a) A species with a lone pair / pair of electrons (1) NOT negative ion alone or as an alternative which it uses / donates to form a (dative) covalent bond (1) 2 (b) (i) Ammonia / NH 3 (in ethanol) (1) heat (1) NOT heat under reflux UNLESS in a sealed tube If a temperature is quoted it must be greater than 100 C [6] in sealed tube / under pressure / concentrated (1) If a pressure is quoted it must be greater than 1 atm (ii) Conditions are dependent on correct reagent. If ammonia and an additional reagent max (1) for two correct conditions. 3 Carbon-bromine bond stronger / higher bond enthalpy than carbon iodine / Ea for C-Br is higher than C-I IGNORE any extra explanations involving the alkyl groups 1 NT Exampro 11

12 (c) Identify bonds broken and made (1) (d) e.g. Energy in or AND Energy out (-) 656 or (-) 3532 (1) Energy needed to break bonds energy released to make bonds = 36 (1) e.g. C-I = + 36 or C-I = + 36 (1) Correct evaluation dependent on use of 36 (1) i.e. C-I = 228 kj mol 1 (1) Correct answer with some correct working (3) If final answer is negative max (2) If 36 is on the wrong side, then 156 max 2 (-156 (1)) If miss out 36, then ±192 max 1 3 H H C C O O H ALLOW OH H 22. (a) for energy level of product lower than reactant both labelled with species (1) for activation hump H (1) for E a being correctly marked (1) Arrow must not point downwards for H being correctly marked (1) e.g. Enthalpy N + (3)H 2 2 E a H (2)NH 3 1 [10] Extent of reaction 4 NT Exampro 12

13 (b) (i) starts at or near origin (not on x or y axis) skewed distribution that is reasonably asymptotic to the x-axis (1) for E uncat (must not be to the left of the peak) (1) 3 for E cat to the left of E uncat (1) and still not to the /eft of the peak e.g. E cat a No of Molecules E uncat a Energy (ii) With the catalyst a greater number/proportion of the molecules (1) This point must be linked to the diagram in some way, e.g. labelled shading have E E cat / sufficient energy to react (1) and so more collisions result in reaction / more successful collisions (1) 3 (c) At 25 C no / few molecules / collisions have E E a./ sufficient energy to react (1) At higher temperature the (average) energy of the molecules increases (1) At 400 C a greater proportion of the molecules/collisions have energy greater than/equal to the activation energy (1) Table 2 All weighing in correct spaces (1) All masses to at least 2 dp (1) Correct subtraction (1) 3 Table 3 both temperatures recorded in correct spaces to 1 dp (1) Correct subtraction including sign to 1 dp also (1) 2 dp in table (0) but T to 2 dp (1) ie transferred error Accuracy multiply the candidates mass corrected if necessary by and record this figure to 1 dp close to the T value Record the difference between the two values on the script Award accuracy mark as follows [13] Difference ±0.2 ±0.3 ±0.4 ±0.5 Mark NT Exampro 13

14 (a) correct arithmetic using candidate s value of T (1) if T > 0, then H < 0 this is OK as it is consistent correct sign (1) 3 ACCEPT 2 or 3 significant figures (1) (b) No (1) not stand alone, dependent on some attempt at a reason value of T controlling factor (1) 2 [14] 24. Table 2 All weighing in correct spaces (1) All masses to at least 2 dp (1) Correct subtraction (1) 3 Table 3 both temperatures recorded in correct spaces to 1 dp (1) Correct subtraction including sign to 1 dp also (1) 2 dp in table (0) but T to 2 dp (1) ie transferred error Accuracy multiply the candidates mass corrected if necessary by and record this figure to 1 dp close to the T value Record the difference between the two values on the script Award accuracy mark as follows Difference ±0.2 ±0.3 ±0.4 ±0.5 Mark (a) correct arithmetic using candidate s value of T (1) if T > 0, then H < 0 this is OK as it is consistent correct sign (1) 3 ACCEPT 2 or 3 significant figures (1) (b) Yes (1) not stand alone, must be supported by an attempt at a reason Temperature would then be known to a comparable precision to the other factors in the equation (1) 2 6 [14] NT Exampro 14

1.4 Energetics. N Goalby chemrevise.org 1. Standard Enthalpy Change of Formation. Standard Enthalpy Change of Combustion

1.4 Energetics. N Goalby chemrevise.org 1. Standard Enthalpy Change of Formation. Standard Enthalpy Change of Combustion 1.4 Energetics Definition: Enthalpy change is the amount of heat energy taken in or given out during any change in a system provided the pressure is constant. In an exothermic change energy is transferred