1RAFFLES INSTITUTION Year 5 H2 CHEMISTRY 2011 Tutorial 6a – Chemical Energetics I Prepared by: Mrs Joan Siaw Self-Check Questions 1.a) By writing relevant thermochemical equations, define the following terms. (i) standard enthalpy change of reaction (ii) standard enthalpy change of formation of CH 3 OH(l) (iii) standard enthalpy change of combustion of C 3 H 8 (g) (iv) standard enthalpy change of neutralisation (v) standard enthalpy change of atomization of fluorine (vi) standard enthalpy change of solution of MgCl 2 (s) (vii) standard enthalpy change of hydration of Ca 2+ (g) (viii) bond energy of the C−O bond (ix) bond dissociation energy (x) first Ionisation energy (xi) second Ionisation energy (xii) first electron affinity (xiii) second electron affinity of oxygen (xiv) lattice energy of Na 2 O(s) b) The enthalpy change of combustion of butane, C 4 H 8 , is −3000 kJ mol −1 . Calculate the mass of water at 20 °C that could be brought to the boiling point by burning this 1.2 dm 3 of gaseous butane at room temperature and pressure. Assume that 80% of the heat from combustion of butane was absorbed by the water. c) Nitrogen and steam undergo a reaction to form ammonia and oxygen as given. 2 1 N 2 (g) + 2 3 H 2 O(g) → NH 3 (g) + 4 3 O 2 (g) Determine the enthalpy change for the above reaction using appropriate bond energy information from the Data Booklet. d) (i) (ii) (iii) 2 Practice Questions 2. a) Define Hess’ Law and explain why it plays such an important role in thermochemistry. b) Use the information given below to construct an energy cycle and calculate the standard enthalpy change of formation ∆H f θ of propane. [−104 kJ mol −1 ] c) 5.60 dm 3 of a mixture of propane and butane (measured at s.t.p) on complete combustion evolved 654 kJ of heat. Calculate the percentage composition of the mixture by volume. [% comp of propane = 39.7 %; % comp of butane = 60.3 % ] Given : ∆H c θ [C 3 H 8 (g)] = −2220 kJ mol −1 , ∆H f θ [H 2 O(l)] = −285.9 kJ mol −1 , ∆H f θ [CO 2 (g)] = −393.5 kJ mol −1 and ∆H c θ [C 4 H 10 (g)] = −2877 kJ mol −1 . 3. a) When 2.76 g of potassium carbonate was added to 30.0 cm 3 of approximately 2 mol dm −3 hydrochloric acid, the temperature rose by 5.20 o C. (i) Write an equation for this reaction. (ii) Calculate the enthalpy change for this reaction per mole of potassium carbonate. Assume that the specific heat capacity and density of all solutions are 4.18 J g −1 K −1 and 1.00 g cm −3 respectively. [− 32.7 kJ mol −1 ] (iii) Explain why the hydrochloric acid need only be approximately 2 mol dm −3 . b) When 2.00 g of potassium hydrogencarbonate was added to 30.0 cm 3 of the same hydrochloric acid, the temperature fell by 3.70 o C. (i) Write an equation for this reaction. (ii) Calculate the enthalpy change for this reaction per mole of potassium hydrogencarbonate. [+23.2 kJ mol −1 ] c) When potassium hydrogencarbonate is heated, it decomposes into potassium carbonate, carbon dioxide and water. By applying Hess’ Law and your results in (a) and (b), calculate the enthalpy change for the decomposition of potassium hydrogencarbonate. [+39.6 kJ mol −1 ] 4. To determine the enthalpy change of neutralisation of sodium hydroxide with sulfuric acid, 50.0 cm 3 of 0.400 mol dm −3 sodium hydroxide was titrated thermometrically with 0.500 mol dm −3 sulfuric acid. The results of the titration were given below: Vol of H 2 SO 4 used / cm 3 6.00 12.00 18.00 24.00 30.00 Temperature rise / o C 1.40 2.60 3.50 3.60 3.30 a) Define the term enthalpy change of neutralisation. b) Plot a graph of temperature rise ( o C) against volume of acid used (cm 3 ). Account for the shape of the graph. c) Calculate the enthalpy change of neutralisation of NaOH with sulfuric acid. State the assumptions made in your calculations. [−56.3 kJ mol −1 ] d) A similar experiment carried out with hydrogen cyanide, HCN(aq), and aqueous ammonia gave a value of −5.4 kJ mol −1 for the enthalpy change of neutralisation. The corresponding value for sodium hydroxide with hydrochloric acid was −57.0 kJ mol −1 . Suggest a possible explanation for the difference in these two values. 3 5. a) By referring to CH 4 , explain what you understand by the term ‘bond energy’. b) The standard heat change of formation of the following four hydrocarbons are given below: Hydrocarbon C 2 H 6 (g) C 2 H 4 (g) C 2 H 2 (g) C 6 H 6 (l) ∆H f θ / kJ mol −1 −84.7 +52.3 +227 +82.9 Using bond energy data from the data booklet and given that the standard enthalpy change of atomisation of graphite is +715 kJ mol −1 and the standard enthalpy change of vaporization of benzene is +34.7 kJ mol −1 , calculate the C−C bond energies of the four hydrocarbons and comment on their values. [BE(C−C) for C 2 H 6 = 363 kJ mol −1 ; C 2 H 4 = 610 kJ mol −1 ; C 2 H 2 = 819 kJ mol −1 ; C 6 H 6 = 503 kJ mol −1 ] 6. One of the most important uses of alkanes is for fuels. In some countries, where crude oil is either scarce or expensive, biofuels such as ethanol are increasingly being used for fuels instead of hydrocarbons. a) Use the bond energies given in Data Booklet to calculate a value for the enthalpy change of combustion of octane, C 8 H 18 . [−4090 kJ mol −1 ] b) The accurate experimental enthalpy changes of combustion of three hydrocarbons are given in the table below. Alkane Formula ∆ ∆∆ ∆H c /kJ mol −1 Heptane C 7 H 16 −4817 Octane C 8 H 18 −5470 Nonane C 9 H 20 −6125 (i) Suggest a reason for the discrepancy between ∆H c for octane you calculated in (a) and that given in the table. (ii) Suggest what the regular increase in the values of ∆H c given in the table represents. c) The enthalpy change of combustion of enthanol is −1367 kJmol −1 and the densities of enthanol and octane are 0.79 g cm −3 and 0.70 g cm −3 respectively. Calculate the heat produced by the complete combustion of 1.0 dm 3 of each fuel. [2.35 × 10 4 kJ and 3.36 × 10 4 kJ] N2005/III/3 or 7. a) Using the data provided, construct a Born−Haber cycle for magnesium chloride, MgCl 2 , and from it determine the first electron affinity of chlorine. [1 st EA of Cl = −348 kJmol −1 ] ∆H / kJ mol −1 Enthalpy change of atomisation of chlorine +122 Enthalpy change of atomisation of magnesium +148 First ionisation energy of magnesium +738 Second ionisation energy of magnesium +1451 Lattice energy of magnesium chloride −2526 Enthalpy change of formation of magnesium chloride −641 b) The theoretically calculated value for the lattice energy of magnesium chloride is –2326 kJ mol −1 . Explain the difference between the theoretically calculated value and the experimental value given in the data in (a), in terms of bonding of magnesium chloride. 4 8. a) Describe, in terms of the energetics involved, the process of dissolution of an ionic solid in water. b) Given that the lattice energy of lithium chloride is −843 kJ mol −1 and its hydration energy is −883 kJ mol −1 , draw an energy level diagram and use it to calculate the enthalpy change of solution of lithium chloride. [−40 kJ mol −1 ] Specific heat capacity of water = 4.20 J g −1 K −1 c) 50.0 cm 3 of water was placed in a thin, well-lagged calorimeter of negligible heat capacity. The temperature of the water was 18.2 o C. 3.00 g of potassium chloride, KCl, was added and the mixture well-stirred until all the crystals had dissolved. The minimum temperature reached was 14.9 o C. Calculate the enthalpy change of solution of potassium chloride. [+17.2 kJ mol −1 ] d) Comment on the results obtained in (b) and (c). 9. Hydrazine is used as rocket fuel and to prepare gas precursors used in air bags. Approximately 260 thousand tonnes of hydrazine are manufactured annually. Liquid hydrazine undergoes combustion according to the following equation: N 2 H 4 (l) + O 2 (g) → N 2 (g) + 2H 2 O(l) A chemist conducted an experiment to determine the standard enthalpy change of combustion of hydrazine. In the experiment, 0.210 g of hydrazine was burnt as fuel to heat up a beaker containing 200 cm 3 of water. The temperature of water rose by 4 °C. You may assume the process has 80 % efficiency. a) Calculate the standard enthalpy change of combustion of hydrazine.[−6.37 x 10 2 kJ mol -1 ] b) Given the following data: Enthalpy change of formation of steam = −242 kJ mol -1 Enthalpy change of vapourisation of water = +44 kJ mol -1 and using the value you have calculated in (a), draw an appropriate energy cycle to determine the standard enthalpy of formation of hydrazine. [+ 65 kJmol -1 ] c) The standard enthalpy change of formation of hydrazine gas is +235 kJ mol -1 . i) Using appropriate data from the Data Booklet, draw an energy level diagram to calculate the average bond energy of N−H bond in hydrazine. ii) Suggest a reason for the difference in the N-H bond energy value obtained from (c)(i) with the value given in the Data Booklet. [368 kJ mol -1 ] MJC08 Prelim/III Challenging Questions (Optional) 10. a) Using the information below and the Data Booklet, construct a Born−Haber diagram and calculate the lattice energy of Cu(I) oxide (Cu 2 O). [−3232 kJ mol −1 ] Energy change / kJ mol −1 Atomisation energy of Cu 339 First electron affinity of O − 141 Second electron affinity of O 791 Heat of formation of Cu(I) oxide − 166 b) The lattice energy of Cu(II) oxide is − 4143 kJ mol −1 . Explain why this value is different from the one calculated in (a). 5 c) Using the information in (a) and (b), draw the appropriate energy cycle and calculate the enthalpy change of reaction for: Cu 2 O(s) CuO(s) + Cu(s) Based on the value obtained, predict whether the reaction will proceed under standard conditions. [−35 kJ mol −1 ] d) Consider the following reaction: Cu 2 O(s) + H 2 SO 4 (aq) CuSO 4 (aq) + Cu(s) + H 2 O(l) When 2.86 g of solid Cu 2 O is added to 60.0 cm 3 of dilute sulfuric acid of approximately 2 mol dm -3 , the temperature of the solution was raised by 8.9 K. Calculate the enthalpy change for the above reaction, stating the assumptions made. [−112 kJ mol −1 ] e) Hence, calculate the enthalpy change of neutralisation for the reaction between H 2 SO 4 (aq) and CuO(s). [−77 kJ mol −1 ] Suggested Answers to Self− −− −Check Questions 1a) Refer to lecture notes. Please give specific definition when asked for the enthalpy change for a specific substance. Example : (ii) Standard enthalpy change of formation of CH 3 OH(l) is the energy change when 1 mole of CH 3 OH(l) is formed from its constituents elements, C(s), O 2 (g) and H 2 (g) at 298 K and 1 atm. C(s) + ½ O 2 (g) + 2H 2 (g) CH 3 OH(l) ∆ ∆∆ ∆H f θ θθ θ [CH 3 OH(l)] b) No. of moles of butane = 1.2/24 = 0.0500 Heat released by reaction = ∆H c (butane) x n butane = 3000x10 3 x0.0500 = 150000 J Heat absorbed by water = (80/100) x heat released by reaction (80% efficiency) = 80/100 x 150000 = 120000 J 120000 = m x 4.18 x (100 – 20) Mass of water heated, m = 359 g c) 2 1 N 2 (g) + 2 3 H 2 O(g) → NH 3 (g) + 4 3 O 2 (g) ∆H r = [ 2 1 BE(N≡N) + 3BE(O-H)] – [3BE(N-H) + 4 3 BE(O=O)] = [ 2 1 (994) + 3(460)] – [3(390) + 4 3 (496)] = +335 kJ mol − −− −1 d) (i) C, (ii) B, (iii) D 6 Suggested Solutions to Tutorial 6a: Chemical Energetics I 1a) Refer to lecture notes. Please give specific definition when asked for the enthalpy change for a specific substance. Example : (ii) Standard enthalpy change of formation of CH 3 OH(l) is the enthalpy change when 1 mole of CH 3 OH(l) is formed from its constituents elements, C(s), O 2 (g) and H 2 (g) at 298 K and 1 atm. C(s) + ½ O 2 (g) + 2H 2 (g) CH 3 OH(l) ∆ ∆∆ ∆H f θ θθ θ [CH 3 OH(l)] b) No. of moles of butane = 1.2/24 = 0.05 Heat released by reaction = ∆H c (butane) x n butane = 3000x10 3 x0.05 = 150000 J Heat absorbed by water = (80/100) x heat released by reaction (80% efficiency) = 80/100 x 150000 = 120000 J 120000 = m x 4.18 x (100 – 20) Mass of water heated, m = 359 g c) 2 1 N 2 (g) + 2 3 H 2 O(g) → NH 3 (g) + 4 3 O 2 (g) ∆H rxn = [ 2 1 BE(N≡N) + 3BE(O-H)] – [3BE(N-H) + 4 3 BE(O=O)] = [ 2 1 (994) + 3(460)] – [3(390) + 4 3 (496)] = 335 kJ mol − −− −1 d) (i) C, (ii) B, (iii) D 2a) Hess’ Law states that the enthalpy change of a reaction is determined by the initial and final states of the system and is independent of the pathways taken. Beauty of it is allowing one to calculate ∆H r of a reaction that is difficult or impossible to perform. b) ∆H f o 3C(s) + 4H 2 (g) C 3 H 8 (g) 3∆H f o [CO 2 (g)] 4∆H f o [H 2 O(l)] ∆H c o [C 3 H 8 (g)] 3 CO 2 (g) + 4 H 2 O (l) ∆H f o [C 3 H 8 (g)] = 3 ∆H f o [CO 2 (g)] + 4 ∆H f o [H 2 O(l)] – ∆H c o [C 3 H 8 (g)] = − −− −104 kJ mol − −− −1 (to 3 sf) c) Enthalpy, ∆H = heat change per mol Actual heat change = ∆H x n, where n is the no of mol of compound burnt (reactant) Total amt = 5.6 / 22.4 = 0.25 mol Let the amt of propane be n mol, then amt of butane is (0.25 – n) mol. Total heat change = {n.∆Hc o [C 3 H 8 (g)] + (0.25-n).∆H c o [C 4 H 8 (g)]} = −654 J % composition of propane = 39.7 % (to 3 sf) % composition of butane = 60.3 % (to 3 sf) + 5 O 2 (g) + 3 O 2 (g) + 2 O 2 (g) 7 3a) (i) K 2 CO 3 (s) + 2HCl(aq) → 2KCl(aq) + CO 2 (g) + H 2 O(l) (ii) Heat released = mc∆T = 30.0 x 4.18 x 5.20 = 652.1 J M r of K 2 CO 3 = 2(39.1) + 12.0 + 3(16.0) = 138.2 Amt of K 2 CO 3 = 2.76 / M r =2.76 / 138.2 = 0.01997 mol Amt of HCl = 30/1000 x 2 = 0.06 mol; Hence, HCl is in excess. Enthalpy change of reaction, ∆H 1 = − −− −652.1/0.01997 =−32650 J mol −1 (to 4 sf) = − −− −32.7 kJ mol − −− −1 (to 3 sf) (iii) The only requirement is that the H + be in excess. A metal carbonate reacts effectively with a dilute acid. b) (i) KHCO 3 (s) + HCl(aq) → KCl(aq) + CO 2 (g) + H 2 O(l) (ii) Heat absorbed = 30.0 x 4.18 x 3.70 = 464.0 J M r of KHCO 3 = 39.1 + 1.0 + 12.0 + 3(16.0) = 100.1 Amt of KHCO 3 = 2.00 / M r = 0.01998 mol Amt of HCl = 30 / 1000 x 2 = 0.06 mol; Hence, HCl is in excess. Enthalpy change of reaction, ∆H 2 = +464.0 /0.01998 = +23220 J mol −1 (to 4 sf) = +23.2 kJ mol − −− −1 (to 3 sf) c) 2KHCO 3 (s) K 2 CO 3 (s) + H 2 O (l) + CO 2 (g) heat +2HCl (aq) +2HCl (aq) 2∆H 2 ∆H 1 2KCl (aq) + 2H 2 O(l) + 2CO 2 (g) By Hess’ Law, 2 ∆H r = 2∆H 2 − ∆H 1 = +79.1 kJ mol −1 ∴Heat change for the decomposition of 1 mole KHCO 3 = +39.6 kJ mol − −− −1 (to 3 sf) 4a) Enthalpy change of neutralisation is the enthalpy change when an amount of acid or alkali is neutralised to form 1 mole of water, the reaction being carried out using dilute aqueous solutions. b) Temp (20.0,3.85) rise/ o C Volume of acid /cm 3 • Initially, temperature rose as heat of neutralisation is released. • Maximum temperature rise signified end of the neutralisation reaction. • Further addition of acid resulted in drop in temperature as addition of room temperature acid to the salt solution cooled it down. c) Heat released = (50.0 + 20.0) x 4.18 x 3.85 = 1126.5 J ∆H neut = −1126.5 / (0.0500 x 0.400) = −56.3 kJ mol −1 (to 3 sf) Assumptions: 1. No heat loss to surroundings 2. Density of solution ≅ density of water 3. Specific heat capacity of soln. approximates that of water. 2∆H r Points plotted traces out a curve. But we extrapolate straight lines to find the maximum temperature that can be reached (heat is lost to the surroundings through the heating and cooling portions of the heat change) 8 d) HCN is a weak acid and energy is required to ionise it to produce one mole of H + ions before neutralisation. NH 3 is a weak base and energy is required to ionise it to produce one mole of OH - ions before neutralisation can take place. Heat of ionisation of the weak acid is endothermic and that resulted in a less exothermic heat of neutralisation of the weak acid−weak base as compared to that of a strong acid−strong base. 5a) The bond energy of a C-H bond is the average energy absorbed when 1 mole of C-H bonds of methane are broken. The successive breaking of each of the 4 C-H bonds has a different bond dissociation energy. Thus CH 4 (g) → C(g) + 4H(g) ; ∆H /4 = bond energy BE(C-C) + 6BE(C-H) b) C 2 H 6 (g) 2C(g) + 6H(g) ∆H f o = −84.7 kJ mol −1 2 ∆H atm o [C (g)] + 3 BE(H-H) 2C(s) + 3H 2 (g) ∆H f θ = energy consumed in bond breaking + energy released in bond formation −84.7 = 2 ∆ atm o [C(s)] + 3 BE(H−H) − {BE(C−C) + 6 BE(C−H)} BE(C−C) = + 363 kJ mol − −− −1 (to 3 sf) Hydrocarbon C 2 H 6 C 2 H 4 C 2 H 2 C 6 H 6 BE(C-C) / kJmol -1 363 610 819 503 Note: Regardless of the multiplicity of the bond, the BE(C-C) should not be divided by the bond order as the bond energy of the σ and π-bonds are different. Comments: • Data booklet: BE[H-H] = +436 kJ mol −1 and BE[C-H] = +410 kJ mol −1 • For calculation involving benzene, C 6 H 6 (l) needs to be vaporized. Enthalpy change of vaporization is +34.7 kJ mol − −− −1 • Bond strength: C≡C > C=C > C-C • C-C bond in benzene is intermediate in bond length and strength between C=C and C-C bonds due to resonance. • Bond energy of C=C is not twice that of C-C ⇒ π-bond is weaker than a σ-bond. 6a) C 8 H 18 + 25/2 O 2 → 8CO 2 + 9H 2 O ∆H c [C 8 H 18 ] = Σ[bond energy broken in reactants] − Σ[bond energy formed in products] = [7BE(C−C) + 18BE(C−H) + 25/2BE(O=O)] − [8BE(C=O) + 18BE(H−O)] = (7 x 350 + 18 x 410 + 25/2 x 496) − (16 x 740 + 18 x 460) = −4090 kJ mol −1 b)(i) ∆H c assumes that all substances are in their standard states eg H 2 O formed is in the liquid state at 298 K. However, bond energy involves breaking bonds (including O−H bonds in H 2 O) in the gaseous phase. Thus ∆H c calculated from bond energy data is different from that obtained experimentally. Alternative Answers: 1. Bond energy data are average bond energies and may differ from bond energies of actual bonds broken and formed in the combustion reaction. OR 2. Can also state that octane is liquid at standard conditions. Hence actual ∆H c is different from experimental value as bond energy involves breaking bonds in gaseous phase. (ii) From heptane to nonane, there is an additional −CH 2 − group. Hence ∆H c increases from heptane to nonane regularly as the increase represents the ∆H c value for the combustion of the –CH 2 – group. 9 c) Mass of 1 dm 3 of ethanol = 0.79 x 1000 g = 790 g. Amount of ethanol in 1 dm 3 = 790/46.0 = 17.17 mol. Heat produced by 1 dm 3 of ethanol = 17.17 x 1367 kJ = 2.35 x 10 4 kJ Mass of 1 dm 3 of octane = 0.70 x 1000 g = 700 g. Amount of ethanol in 1 dm 3 = 700/114.0 = 6.140 mol. Heat produced by 1 dm 3 of octane = 6.140 x 5470 kJ = 3.36 x 10 4 kJ 7a) Mg 2+ (g) + 2e - + 2Cl(g) Mg(g) + 2Cl(g) 2∆H atom [Cl 2 (g)] Mg 2+ (g) + 2Cl - (g) Mg(g) + Cl 2 (g) ∆H atom [Mg(s)] Mg(s) + Cl 2 (g) MgCl 2 (s) By Hess’ Law, ∆H f [MgCl 2 (s)] = ∆H atom [Mg(s)] + 2∆H atom [Cl 2 (g)] + 1 st IE + 2 nd IE + 2(1 st EA) + LE ∴ 1 st EA = ½ {∆H f [MgCl 2 (s)]- ∆H atom [Mg(s)] - 2∆H atom [Cl 2 (g)] – 1 st IE – 2 nd IE – LE } = −348 kJmol -1 b) The calculated theoretical value for the LE is less exothermic than the experimental LE. This is because theoretically, the ionic bond is assumed to be purely ionic. In reality, due to polarisation of the anion by the cation, the bond between the two oppositely charged ions is strengthened. 8a) When an ionic solid dissolves in water, two enthalpy terms are involved. 1. The ions must be separated from the ionic lattice. The energy required is the lattice dissociation enthalpy (-LE). 2. The specific ions interact with water molecules. The energy released (Σ∆H hyd ) as these attractive forces come into play is compensation for the energy required to dissociate the lattice. ∆H soln θ = – LE + Σ∆H hyd θ (a positive quantity) (a negative quantity) b) ∆H soln [LiCl(s)] = ∆H hyd [Li + (g)] + ∆H hyd [Cl − (g)] − lattice energy = − −− −40 kJ mol − −− −1 LE[MgCl 2 ] ∆H f [MgCl 2 (s)] 1 st IE (Mg) + 2 nd IE (Mg) 2{1 st EA (Cl)} LiCl(s) Li + (aq) + Cl − (aq) Li + (g) + Cl − (g) LE [LiCl] ∆H hyd [Li + (g)] + ∆H hyd [Cl − (g)] ∆H soln [LiCl(s)] E / kJ mol −1 0 10 c) Heat absorbed = mc∆T = 50.0 x 4.20 x 3.3 = 693 J Enthalpy change of solution, ∆H soln [KCl(s)] = +693 / (3.00/74.6) = +17.2 kJ mol − −− −1 (to 3 sf) d) Since LE is inversely proportional to interionic radius, the lattice energy of LiCl is slightly more exothermic than that of KCl. However, since hydration energy is proportional to charge density, Li + will have a much more exothermic hydration enthalpy compared to K + . Therefore, the heat change of solution of LiCl is more exothermic that that of potassium chloride. 9a) Amount of heat absorbed by water, Q = 200 x 4 x 4.18 = 3344 J Amount of heat released by reaction, Q’ = Q / 0.8 = 4180 J No of moles of N 2 H 4 = 6.56 x 10 -3 Standard enthalpy change of combustion of hydrazine =−6.37 x 10 2 kJ mol -1 b) N 2 (g) + 2H 2 (g) N 2 H 4 (l) N 2 (g) + 2H 2 O (g) N 2 (g) + 2H 2 O (l) By Hess’ law, ∆H f (N 2 H 4 ) = 2(-242) – (-637) – 2(+44) = + 65 kJmol -1 c) i) By Hess Law, 160 + 4 x B.E(N-H) + 235 = 994 + 2(436) B.E (N-H) = 368 kJ mol -1 ii) The bond energy values obtained from the Data Booklet are average values and would differ from the experimental values. +O 2 ∆H c (N 2 H 4 ) +O 2 2 x ∆H f (H 2 O(g)) 2 x ∆H v (H 2 O) ∆H f (N 2 H 4 ) N 2 H 4 (g) N 2 (g) + 2H 2 (g) ∆H f (N 2 H 4 ) = + 235 kJmol -1 B.E(N-N) + 4 x B.E(N-H) = + 65 kJmol -1 B.E(N≡N) + 2x B.E(H-H) = + 994 + 2(436) kJmol -1 2N (g) + 4H (g) 11 10a) Born-Haber diagram 2Cu + (g) + O 2− (g) 2Cu + (g) + O(g) + 2e − 1 st EA [O] = − 141 kJ mol -1 2Cu + (g) + O − (g) + e − 2 nd EA [O] = +791 kJ mol -1 ½ BE [O 2 ] = ½(+496) kJ mol -1 2Cu + (g) + ½ O 2 (g) + 2e − 1 st IE [Cu] x 2 = 2(+745) kJ mol -1 2Cu(g) + ½ O 2 (g) LE [Cu 2 O] ∆H atom [Cu(g)] x 2 = 2(+339) kJ mol -1 2Cu(s) + ½ O 2 (g) ∆H f [Cu 2 O] = − 166 kJ mol -1 Cu 2 O(s) Lattice energy of Cu(I) oxide, Cu 2 O = −3232 kJ mol -1 (b) LE for CuO is more exothermic (− 4143 kJ mol -1 ) ⇒ stronger ionic bonds formed! for CuO, • higher charge [Cu 2+ vs Cu + ] ⇒ q + larger • smaller cationic radius (r + ) as e − clouds are attracted more strongly to the nucleus ⇒ smaller interionic distance (r + + r - ) (c) Cu 2 O(s) CuO(s) + Cu(s) LE [Cu 2 O] = −3232 kJ mol -1 LE [CuO] = − 4143 kJ mol -1 2Cu + (g) + O 2− −− − (g) Cu 2+ (g) + O 2− −− − (g) + Cu(s) 2 nd IE (Cu) 1 st IE (Cu) ∆H atom = +339 kJ mol -1 = +1960 kJ mol -1 = +745 kJ mol -1 Cu + (g) + Cu 2+ (g) + O 2− −− − (g) +e − −− − Cu 2+ (g) + O 2− −− − (g) + Cu(g) ∆H = −35 kJ mol -1 ⇒ exothermic reaction ⇒ thermodynamically favourable ⇒ reaction probably spontaneous (d) Heat evolved = mc∆T = 60 x 4.18 x 8.9 = 2243 J Amt of Cu 2 O (s) = 0.02 mol Cu 2 O (s) + H 2 SO 4 (aq) CuSO 4 (s) + Cu (s) + H 2 O (l) Enthalpy change = Q/n = − −− −112 kJ mol -1 (to 3sf) Assuming specific heat capacity of solution = 4.18 J g -1 K -1 Density of solution = density of water = 1.00 g cm -3 (e) Enthalpy change of neutralisation Heat change when 1 mole of water is formed when an acid neutralises a base (in dilute aqueous solution) at 1 atm and 298K CuO (s) + H 2 SO 4 (aq) CuSO 4 (aq) + H 2 O (l) Cu 2 O (s) + H 2 SO 4 (aq) ∆H = − −− −77 kJ mol -1 − Cu (s) ∆H = −112 kJ mol -1 − Cu (s) ∆H = −35 kJ mol -1 ∆H − + − + + = r r q q LE ∆H E / kJ mol −1 0