Edexcel Jan 2017 (IAL) Paper 1 Q23 (with explained solutions) |

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Edexcel Jan 2017 (IAL) Paper 1 Q23

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23 This question is about the gases propane, C3H8, and butane, C4H10. (a) (i) Propane and butane are both alkanes. Alkanes are said to belong to the same homologous series. Give two characteristics associated with homologous series. (2)D O N O T W R I T E I N T H I S A R E A(ii) Butane has a structural isomer but propane does not. State what is meant by a structural isomer and explain why butane has a structural isomer but propane does not. (2) Structural isomerExplanation(b) Bottled propane is used as the fuel for the burners in hot air balloons. A hot air balloon carries 80 kg of liquefied propane. (i) Write the equation for the complete combustion of propane in air under standard conditions. State symbols are not required. (2) 14 *P48367A01424* D O N O T W R I T E I N T H I S A R E A D O N O T W R I T E I N T H I S A R E A<br />
(ii) Calculate the number of moles of propane in 80 kg. (2) (iii) The standard enthalpy change of combustion of propane, (cid:507)H (cid:57) c,298 = 2220 kJ mol1. Calculate the heat energy, in joules, given out when 80 kg of propane burns completely. (1) (iv) The burners have a maximum power rating of 4800 W. (1 W = 1 J s1) Calculate the maximum time, in hours, that the balloons fuel would last if the burners are used continuously on full power with 80 kg of fuel. (1) A E R A S I H T N I E T I R W T O N O D A E R A S I H T N I E T I R W T O N O D A E R A S I H T N I E T I R W T O N O D *P48367A01524* 15 Turn over<br />
(v) A student suggests that butane would be a better fuel for hot air balloons than propane because it has a more negative enthalpy change of combustion, (cid:507)H (cid:57) c,298 = 2880 kJ mol1. Suggest two reasons why butane is not a better fuel than propane for hot air balloons. (2) Reason oneD O N O T W R I T E I N T H I S A R E A Reason twoD O N O T W R I T E I N T H I S A R E A D O N O T W R I T E I N T H I S A R E A 16 *P48367A01624*<br />
(c) The standard enthalpy changes of atomisation of propane and butane can be calculated. The calculation requires their standard enthalpy changes of formation and the standard enthalpy changes of atomisation of carbon and hydrogen. (i) Complete the Hess cycle for the calculation of the standard enthalpy change of atomisation of propane. 3C(s, graphite) + 4H2(g) (cid:111)(cid:3)(cid:3)(cid:3)C3H8(g)() +() (ii) Calculate the standard enthalpy change of atomisation of propane, (cid:507)H (cid:57) at,298[C3H8(g)] Use the data below. (cid:507)H (cid:507)H (cid:507)H f,298[C3H8(g)] = 104.5 kJ mol1 (cid:57) at,298[H2 (g)] = +218 kJ mol1 (cid:57) at,298[C(s, graphite)] = +716.7 kJ mol1 (cid:57) (1) (3) A E R A S I H T N I E T I R W T O N O D A E R A S I H T N I E T I R W T O N O D A E R A S I H T N I E T I R W T O N O D *P48367A01724* 17 Turn over<br />
(iii) The standard enthalpy change of atomisation of butane can be calculated using the same method as for propane. This value, together with the carbon-hydrogen bond energy, can be used to calculate the carbon-carbon bond energy (cid:507)H at,298[C4H10(g)] = +5173.3 kJ mol1. E(C H) = +412.3 kJ mol1 Calculate the carbon-carbon bond energy. (iv) Suggest why your answer differs from the mean bond energy for the carbon-carbon bond given in data books. (2) (1)(Total for Question 23 = 19 marks) 18 *P48367A01824* D O N O T W R I T E I N T H I S A R E A D O N O T W R I T E I N T H I S A R E A D O N O T W R I T E I N T H I S A R E A<br />

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Question Number 23(a)(i) Acceptable Answers Any two from:The same general formula ALLOW example of any general formula aloneEach member differs by CH2 / by a mass of 14Contains the same functional group or have similar / same chemical propertiesSmooth gradation of / gradual change / trend in physical properties IGNORE Same physical properties / same properties Mark Reject Same empirical formula / same molecular formula<br />
Acceptable Answers Question Number 23(a)(ii) Structural isomers have the same molecular formulae but different structural formulae / structures / displayed formulae / skeletal formula ALLOW Just different arrangement of same atoms IGNORE General formula (1) When the last carbon atom is added to the chain, in butane there are two possible points of attachment but in propane there is only one. ALLOW Butane can form a branched chain / but propane cannot form a branched chain OR Two isomers of C4H10 correctly drawn (alone) There is only one way of arranging the atoms in propane but two ways in butane (1) IGNORE More (carbon) atoms in butane Any other comments providing it is non-contradictory Reject Mark<br />
Acceptable Answers Question Number 23(b)(i) C3H8 + 5O23CO2 + 4H2O Correct reactants and products (1) Balancing (1) 2nd mark dependent on first BUT ALLOW one mark for correctly balanced equation for C3H6 IGNORE state symbols even if incorrect Reject Mark<br />
Acceptable Answers Question Number 23(b)(ii) Molar mass of propane = (3 x 12 + 8 x 1 =) 44 (1) 80 x 1000 = 1818 (mol) (1) 44 ALLOW TE from first to second mark Correct answer no working 2 marks IGNORE SF except 1 Acceptable Answers Question Number 23(b)(iii) 1818 x 2220 = 4.04 x 106(kJ) = 4036363000 / 4.04 x109 (J) TE from (ii) e.g. 1.8 give 3996(kJ) IGNORE sign even if incorrect / IGNORE SF Acceptable Answers Question Number 23(b)(iv) 4036363000 = 8.41 x 105 s 4800 =1 4000 min = 234 hours Ignore SF ALLOW 230234 (early rounding) TE from (iii) Reject Mark Reject Mark Reject Mark<br />
Reject Mark Acceptable Answers Question Number 23(b)(v) Both marks can be awarded in either space Any two from MP1 For butane 80 kg produces 3.97 X 106 kJ whereas propane produces 4.04 X106 kJ OR Butane has a higher molar mass so energy per g/kg is about the same (slightly less) (values propane 50 J g-1, Butane 48 J g-1) (1) MP2 Incomplete combustion more likely for butane ALLOW Butane needs more oxygen for complete combustion OR Butane does not burn as easily / quickly / efficiently ALLOW Butane / longer hydrocarbon is less efficient Butane takes longer to burn (1) MP3 Butane may not vaporise at high altitudes / if it is cold / butane is less volatile OR Butane has higher boiling temperature (1) IGNORE More carbon dioxide formed / butane has greater density More energy required for burning<br />
Acceptable Answers Question Number 23(c)(i) 3C(g) + 8H(g) Acceptable Answers Question Number 23(c)(ii) 3 x 716.7 + 8 x 218 (-104.5) (2) (2 150.1 + 1 744) + 104.5 = (+)3 998.6 (kJ mol-1) (1) Correct answer no working (3) Ignore SF and units even if incorrect (+)1039.2 scores 1 mark (missing both multiples) (+)2472.6 scores 2 marks (missing x 8) (+)2565.2 scores 2 marks (missing x 3) -3998.6 scores 2 marks (Hess wrong way round) (+)3789.6 scores 2 marks (-104.5) (+)2690.6 scores 2 marks (x 2 instead of x 8) -2690.6 scores 1 mark (x 2 instead of x 8, Hess wrong way round) (+)3126.6 scores 2 marks (x 4 instead of x 8) -3126.6 scores 1 mark (x 4 instead of x 8, Hess wrong way round ) Reject Reject Mark Mark<br />
Reject Mark Acceptable Answers Question Number 23(c)(iii) 3 X E(CC) = 5 173.310 X E(CH) = 1050.3 (1) E(CC) = (+) 350.1(kJmol-1) (1) IGNORE SF except 1 IGNORE units even if incorrect Acceptable Answers Question Number 23(c)(iv) Bond energies vary with Reject Mark environment (whereas bond energies in data book values are mean bond energies) ALLOW Bond energies averaged over (many / different) compounds (TOTAL FOR QUESTION 23 = 19 MARKS)<br />

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