ENTHALPY OF DISPLACEMENT

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Hollie Cooper
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1 QUESTIONSHEET 1 ENTHALPY OF DISPLACEMENT a) (i) = (1) (ii) 1.95/65 = (1) b) 0.02(1) Zn in excess/equivalent mol in reaction(1) c) (i) to ensure heat equilibrium/balance/steady temperature(1) (ii) regression to point of addition(1) (iii) to predict the value without heat loss(1) d) (i) (1) = 3318 J(1) (ii) 3318/0.02 (1) = J(1) = (1) kj mol -1 (1) e) more exothermic/greater (1) zinc more electropositive/reactive/higher in Reactivity Series (1) larger release of energy (1)

2 QUESTIONSHEET 2 THERMOMETRIC TITRATION a) (i) Axes: sensible choice of scales (1) labelling (1) Points correctly plotted (1) Two intersecting straight lines of best fit (1) Volume of acid = 22.5 ± 0.5 cm 3 (1) (ii) n (NaOH) = /1000 = 0.1 mol (1) n (HCl) = 0.1 mol (1) concentration of HCl = 0.1/ * = 4.44 M (1) * mark consequentially from graph (iii) q = m.θ. Τ = ( * ) 4.18 ( * ) (1) = J per 0.1 mol of H 2 O formed (1) H neut = -5576/0.1 = -55,760 J mol -1 or kj mol -1 (1) Maximum 2 marks if negative sign omitted * mark consequentially from graph. b) Quantity of energy produced by the coil = = 4350 J (1) Amount of water formed = 40 2/1000 = 0.08 mol (1) (Note: NaOH is in excess calculation based on acid) the quantity of energy produced by the formation of 1 mole of water = 4350/0.08 = 54,375 J mol -1 (1) H neut = -54,375 J mol -1 or 54.4 kj mol -1 (1)

3 QUESTIONSHEET 3 ENTHALPY OF FORMATION a) The heat change (not energy change (-1) but allow enthalpy change ) when 1 mole (1) of a compound is formed from its elements (1) under standard conditions / 1 atm (10 5 Pa) and 298 K (1) b) H θ = reaction Hθ (products) - f Hθ (reactants) (1) f = [(+90 4) + (-242 6)] - [-46 4] (1) = = -908 kj mol -1 (1) (Or accept a Hess s law solution) c) H θ = [(-394 4) + (-286 2)] - [(+228 2)] (1) reaction = = kj mol -1 (1) H θ (C H ) = -2604/2 = kj c 2 2 mol-1 (1) (Or accept a Hess s law solution) d) H θ f (MgO) = -602 kj mol-1 ; H θ f (CO 2 ) = -394 kj mol-1 (1) (Stated or implied) For MgCO 3 (s) MgO(s) + CO 2 H θ reaction = +100 = [(-602) + (-394)] - Hθ f (MgCO 3 ) (1) H θ f (MgCO 3 ) = = kj mol-1 (1) (Or accept a Hess s law solution)

4 QUESTIONSHEET 4 ENTHALPY OF COMBUSTION a) The heat evolved (accept enthalpy change, but not energy change or heat absorbed or evolved (-1)) when 1 mole (1) of a substance undergoes complete combustion / is burned in excess oxygen (1) under standard conditions / 1 atm (10 5 Pa) and 298 K (1) b) For 2C 2CO 2, H θ = (-566) = -788 kj mol -1 (1) for 1C 1CO 2, H θ = -788/2 = -394 kj mol -1 (1) c) H θ = reaction Hθ (reactants) - c Hθ (products) (1) c = [-6778] - [(-5470) + (-1411)] (1) = (-6881) = kj mol -1 (1) (Or accept a Hess s law solution) d) (i) H θ reaction = [(-2718) + (-286)] - [-2877] (1) = (-2877) = -127 kj mol -1 (1) (ii) Expected change Nil (1) Reason H depends only on the initial and final states of the system / is independent of the route taken (1)

5 a) The energy required (½) to break the covalent bonds in 1 mole (½) of gaseous hydrogen molecules (½) to give gaseous hydrogen atoms at 298 K (½) QUESTIONSHEET 5 BOND DISSOCIATION ENTHALPY b) Bond dissociation enthalpy is the energy required to break one O H bond (1) in each molecule in 1 mole of gaseous water molecules (1) whereas mean bond dissociation enthalpy is the average energy required to break all the O H bonds (1) c) (i) C C < C=C < C C (1) (ii) C I < C Br < C Cl (1) d) Bond dissociation enthalpy is directly related to bond strength (1) but inversely related to bond length (1) e) H for bond breaking Or 6 C H = = (½) 6 C H = (½) 2 C C = = (½) 2 C C = (½) 1 C=O = 804 = (½) 4 O=O = = (½) 4 O=O = (1) Total kj mol -1 (1) kj mol -1 (1) H for bond making 6 C=O = 6 (-804) = (½) 5 C=O = (½) 6 H O= 6 (-463) = (½) 6 H O = (½) Total kj mol -1 (1) kj mol -1 (1) H c = ( ) (1) H c = ( ) (1) = kj mol -1 (1) = kj mol -1 (1)

6 QUESTIONSHEET 6 HESS S LAW AND ENTHALPY DIAGRAMS a) The enthalpy change (not energy change ) (1) when one chemical system is converted into another (1) is independent of the route taken (1) but depends only on the initial and final states of the system (1) b) (i) Energy can be neither created nor destroyed (1) but can be converted from one form to another (1) (ii) Consider a conversion and reconversion by two routes (1) If Hess s law were obeyed, conversion by route 1 and reconversion by route 2 would give overall H of 0, consistent with the first law (1) If Hess s law were not obeyed, conversion by route 1 and reconversion by route 2 would give overall H > 0 or < 0, which would violate the first law (1) c) (i) C 3 H 8 + 5O 2 3CO 2 + 4H 2 O(l) (2) Award (1) for balance and (1) for state symbols. (ii) Enthalpy 0 3C(s) + 4H 2 + 5O 2 (1) 3(-394) θ H f (C 3H 8) = kj mol -1 3C 3 H 8 + 5O 2 3CO 2 + 4H 2 + 2O 2 Route 2 (1) Route 1 (1) 4(-286) = kj mol kj mol -1 3CO 2 + 4H 2 O(l) (1) (2 for diagram, 2 for start & finish labels, 2 for route labels) (iii) By Hess s law, H(Route 1) = H(Route 2) (1) (-1144) = H θ (C H ) + (-2222) (1) f 3 8 H θ (C H ) = (-1144) f 3 8 = -104 kj mol -1 (1)

7 QUESTIONSHEET 7 HESS S LAW CALCULATIONS a) (i) S(monoclinic) + O 2 S(rhombic) + O kj mol -1 SO 2 H conversion kj mol -1 (2) Or H conversion S(rhombic) + O 2 + O kj mol -1 SO kj mol -1 (2) By Hess s law, H conversion + (-297.2) = (1) H conversion = = +0.3 kj mol -1 (1) (ii) More stable form Rhombic (1) Reason The conversion of rhombic to monoclinic sulfur is endothermic / absorbs energy (1) b) + 3S(monoclinic) + O 2 2(-20.2) = kj mol -1 S + S(monoclinic) + O 2 Or SO 2 + S Hê reaction O + 3S(mono) kj mol -1 S + SO 2 2(-286) = -575 kj mol -1 2(-20.2) = kj mol -1 2(-286) = kj mol kj mol -1 2(+41.1) = kj mol -1 + O 2 + 3S(monoclinic) Hê reaction O + 3S(monoclinic) O(l) + 3S(monoclinic) 2(+41.1) = kj mol -1 (3) For either scheme. Deduct 1 mark for each error By Hess s law, (-297.2) + Hê reaction = (1) Hê reaction = = kj mol -1 (1) c) S(monoclinic) + 1½O kj mol -1 SO 2 + 1½O 2 (-196.2)/2 = kj mol -1 Hê f (SO 3 ) Or S(monoclinic) + 1½ O kj mol -1 Hê f (SO 3 ) SO3 (-196.2)/2 = kj mol -1 SO 3 SO 2 + ½O 2 (2) By Hess s law, Hê f (SO 3 ) = (-98.1) (1) = kj mol -1 (1)

8 QUESTIONSHEET 8 HESS S LAW WITH CALORIMETRY a) Graph: axes appropriately labelled (½ each) suitable scales to match size of graph paper - at least 2 / 3 of each side used (½ each) points correctly plotted (1 but -½ for each error) two extended straight lines drawn (½ each) Maximum temp. change (taken as the vertical distance at 3½ min) between the two straight lines = 25.0 ± C (1) b) q = (-25)* *Or value from graph = J (1) n (Mg) = 0.24/24 = 0.01 mol (1) H reaction = -5275/0.01 = -527,500 J mol -1 / kj mol -1 (1) Maximum of 2 marks if ve sign omitted c) q = (-5.2) = J (1) H reaction = -1110/0.01 = -111,000 J mol -1 / -111 kj mol -1 (1) Maximum (1) if ve sign omitted d) Mg(s) + C(s) + 3 / 2 O 2 MgCO 3 (s) (1 for formulae and balance; 1 for state symbols) e) Mg(s) + C(s) + O 2 + 2HCl(aq) H 1 (1) 3 2 H f (MgCO 3 ) MgCO 3 (s) + 2HCl(aq) (1) H 2 (1) MgCl 2 (aq) + H 2 + C(s) O 2 H 3 (1) MgCl 2 (aq) + H 2 O(l) + CO 2 Where H 1 = answer from b), H 2 = answer from c), and H 3 = (-286 +(-394)) = kj mol -1 for H 2 O(l) + CO 2 f) By Hess s law, H f (MgCO 3 ) + H 2 = H 1 + H 3 (1) H f (MgCO 3 ) = = kj mol -1 (1)* *marked consequentially on answers in b) and c)

9 a) 2C(s) QUESTIONSHEET 9 HESS S LAW WITH BOND DISSOCIATION ENTHALPY 2(+715) = kj mol -1 (1) 3(+436) = kj mol -1 (1) 3H 2 (+496)/2 = +248 kj mol -1 (1) ½O 2 2C 6H O (-413) = kj mol -1 (1) 2(-464) = -928 kj mol -1 (1) 2CH 2 =CH 2 + H 2 O H θ f (C 2 H 2 OH) = -278 kj mol -1 H reaction C 2 H 5 OH(l) (1) for cycle By Hess s law, -278 = (-2264) + (-928) + H reaction (1) H reaction = = -72 kj mol -1 (1) Or 2C(s) + 2(+715) = kj mol -1 2(+436) = +872 kj mol -1 2C + 4H C 2 H 4 Hê f (C 2 H 4 ) (-413) = kj mol -1 By Hess s law, Hê f (C 2 H 4 ) = (-2264) (1) = + 38 kj mol -1 (1) (1) H 2 + ½O kj mol -1 (+496)/2 = kj mol -1 2H + O H 2 O Hê f (H 2 O) 2(-464) = -928 kj mol -1 By Hess s law, H θ (H O) = (-928) (1) f 2 = -244 kj mol -1 (1) For the hydration of ethene, H reaction = -278 [+38 + (-244)] (1) = -72 kj mol -1 (1) b) H θ = reaction Hθ (reactants) - c Hθ (products) (1) c = (-1368) (1) = - 43 kj mol -1 (1) (Or accept a Hess s law solution) (1) c) Mean/average bond dissociation enthalpies are used and not the dissociation enthalpies of the bonds in the molecules in the equation (1) d) More or less negative H is less negative (1) Reason For the change C 2 H 5 OH (l) C 2 H 5 OH, enthalpy of vaporisation / latent heat of vaporisation (1) is required / absorbed / positive (1)

10 QUESTIONSHEET 10 TEST QUESTION I a) (i) Heat change / enthalpy change (½) when 1 mole of a compound (1) is dissolved in water (½) to form an infinitely dilute solution (1) (ii) n (NH 4 Cl) = 5.35/53.5 = 0.1 mol (1) Heat change = ( ) ( ) = kj (1) H solution = +1.52/0.1 = kj mol -1 (1) (iii) endothermic (1) b) Heat change = ( ) ( ) = kj (1) per /1000 = mol (1) H neut = /0.025 = kj mol -1 (1) c) (i) ½N ½Cl 2 NH 4 Cl(s) (1) (ii) Enthalpy cycle Or accept an enthalpy diagram Application of Hess s law H f (NH 3 ) + H f (HCl) + H solution (NH 3 ) + H solution (HCl) + H neut = H f (NH 4 Cl) + H solution (NH 4 Cl) (1) Or (-92.3) + H solution (NH 3 ) + (-187) + (-52.2*) = (1) * mark consequentially from b) mark consequentially form a) Molar enthalpy of solution of ammonia H solution (NH 3 ) = (1) = kj mol -1 (1)

11 QUESTIONSHEET 11 TEST QUESTION II a) (i) = 0.48g (1) (ii) = 16.7 o C (1) b) (i) 80x4.2x16.7 (1) = 5611(.2)J (1) (ii) 5611 /0.48(1) = J (1) (iii) Mr = 46(1) mol ethanol = 0.48/46 = (1) = 1124 kj(1) enthalpy of combustion = kj mol -1 (1) (sign must be correct) If different stages of figures, allow use of J not kj Mark consequentially (iv) standard conditions not stated (1) 298K(25 o C)/1 atm P(most stable form)(1) (v) Density = 1g cm -3 (1) or no account of container s heat absorption (1) c) Reason 1 Heat losses to atmosphere(1) Explanation Temp rise less than correct value(1) Calculated J less than correct value(1) Reason 2 Incomplete combustion Explanation Mol used less than from mass loss (1) so energy/mol smaller than correct value (1)

QUESTIONSHEETS ENERGETICS I ENTHALPY OF DISPLACEMENT BOND DISSOCIATION ENTHALPY HESS S LAW AND ENTHALPY DIAGRAMS HESS S LAW WITH CALORIMETRY

QUESTIONSHEETS ENERGETICS I ENTHALPY OF DISPLACEMENT BOND DISSOCIATION ENTHALPY HESS S LAW AND ENTHALPY DIAGRAMS HESS S LAW WITH CALORIMETRY CHEMISTRY QUESTIONSHEETS AS Level AS TOPIC 5 ENERGETICS I Questionsheet 1 Questionsheet 2 Questionsheet 3 Questionsheet 4 Questionsheet 5 Questionsheet 6 Questionsheet 7 Questionsheet 8 Questionsheet 9