CHEM1101 Answers to Problem Sheet 8. where c is the specific heat capacity a property of the substance involved.

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1 CEM1101 Answers to Problem Sheet 8 1. The energy q, required to heat a substance of mass m by a temperature ΔT is given by the equation: q = c m ΔT where c is the specific heat capacity a property of the substance involved. If q = 78.2 J, m = 45.6 g and ΔT = 13.3 K, then c = q 78.2J = = 0.129J K g mδ T (45.6 g) (13.3 K) The atomic mass of lead is g mol -1. The molar heat capacity, C, is therefore: C = (0.129 J K -1 g -1 ) (207.2 g mol -1 ) = 26.7 J K -1 mol The neutralization reaction is a 1:1 reaction with chemical equation: Cl(aq) + Na(aq) à NaCl(aq) + 2 (l) After the reactants mix the solution has a volume of 200 ml. Assuming the solution the same density as pure water, this corresponds to a mass of: mass (g) = density (g ml -1 ) volume (ml) = (0.997 g ml -1 ) (200 ml) = 199 g The temperature change, ΔT = (31.1 C 24.6 C) = 6.7 K. The heat capacity of water is J K -1 g -1. The heat change is therefore: q = c m ΔT = (4.184 J K -1 g -1 ) (199 g) (6.7 K) = 5600 J or 5.6 kj (heat released) The number of moles of + and - present are both the same: number of moles = concentration (M or mol L -1 ) volume (L) = (1.0 mol L -1 ) (0.100 ml) = 0.1 mol The heat change for a mole would therefore be 5.6 kj / 0.1 mol = 56 kj mol -1. The reaction gives out heat (as the temperature rises) so it is an exothermic reaction with a negative enthalpy change: Δ = 56 kj mol Reaction Δ r (kj mol 1 ) A 2B(s) + 3 /2 2 (g) B 2 3 (s) B B 2 6 (g) (g) B 2 3 (s) (g) C 2 (g) + ½ 2 (g) 2 (l) -286 D 2 (l) 2 (g) +44

2 The enthalpy change for a reaction such as B can be written as the difference between the enthalpy of formation of reactants and products: Δ rxn = ΣmΔ f (products) - ΣnΔ f (reactants) For reaction B this becomes: Δ rxn = [Δ f (B 2 3 (s) + 3Δ f ( 2 (g))] [Δ f (B 2 6 (g) + 3Δ f ( 2 (g))] Reaction A corresponds to the heat of formation of B 2 3 (s) so Δ f (B 2 3 (s)) = kj mol -1. Reaction C corresponds to formation of 2 (l) and reaction D corresponds to the energy that must then be supplied to vapourize the liquid into 2 (g): Δ f ( 2 (g)) = ( ) kj mol -1 = -242 kj mol -1 The enthalpy change for reaction B is Δ rxn = kj mol -1. As 2 exists as a gas under standard conditions, its enthalpy of formation is zero. Putting all this together gives: ence, kj mol -1 = ([ ] [Δ f (B 2 6 (g) + 3 0]) kj mol -1 Δ f (B 2 6 (g)) = +36 kj mol As in Q1, the enthalpy change for a reaction can be written as the difference between the enthalpy of formation of reactants and products: Δ rxn = ΣmΔ f (products) - ΣnΔ f (reactants) (a) 2N 3 (g) (g) + 2C 4 (g) 2CN(g) (g) As 2 exists as a gas under standard conditions, its enthalpy of formation is zero. The remaining enthalpies of formation are given. Δ rxn =[2Δ f (CN(g)) + 6Δ f ( 2 (g))] [2Δ f (N 3 (g)) + 2Δ f (C 4 (g))] = ([ ] [ ]) kj mol -1 = -940 kj mol -1 (b) N 3 (g) + Cl(g) N 4 Cl(s) Δ rxn =[Δ f (N 4 Cl(s))] [Δ f (N 3 (g)) + Δ f (Cl(g))] = ([-314] [ ]) kj mol -1 = -176 kj mol (a) Δ comb = ΣΔ atom (reactants) ΣΔ atom (products) The actual values obtained will vary somewhat with the bond enthalpies used. The calculations below use the values in the table below:

3 Average Bond Enthalpies at 298 K. Bond Δ / kj mol 1 Bond Δ / kj mol 1 C C 346 C 358 C C= 804 = 498 Ethane - the balanced equation for combustion is: 2C à 4C C C + 7 à 4 C + 6 For the reaction as written: Δ = {2 [Δ atom (C 2 6 )] + 7 [Δ atom ( 2 )] } - {4 [Δ atom (C 2 )] + 6 Δ atom ( 2 )] } = {2 [346 (C-C) (C-)] + 7 [498 (=)] } - {4 [2 804 (C=)] + 6 [2 463)] } kj mol -1 = kj mol -1 This is for the combustion of two moles of C 2 6 so the heat of combustion for one mole of ethane is: Δ comb = kj mol -1 Butane - the balanced equation for combustion is: 2C à 8C For the reaction as written: Δ = { [2 Δ atom (C 4 10 )] + 13 [Δ atom ( 2 )] } - {8 [Δ atom (C 2 )] + 10 Δ atom ( 2 )] } = {2 [3 346 (C-C) (C-)] + 13 [498 (=)] } - {8 [2 804 (C=)] + 10 [2 463)] } kj mol -1 = kj mol -1 This is for the combustion of two moles of C 4 10 so the heat of combustion for one mole of ethane is: Δ comb = kj mol -1 exane - the balanced equation for combustion is: 2C à 12C

4 For the reaction as written: Δ = {2 [Δ atom (C 6 14 )] + 19 [Δ atom ( 2 )] } - {12 [Δ atom (C 2 )] + 14 Δ atom ( 2 )] } = {2 [5 346 (C-C) (C-)] + 19 [498 (=)] } - {12 [2 804 (C=)] + 14 [2 463)] } kj mol -1 = kj mol -1 This is for the combustion of two moles of C 6 14 so the heat of combustion for one mole of hexane is: Δ comb = kj mol -1 ctane - the balanced equation for combustion is: 2C à 16C For the reaction as written: Δ = {2 [Δ atom (C 8 18 )] + 25 [Δ atom ( 2 )] } - {16 [Δ atom (C 2 )] + 6 Δ atom ( 2 )] } = {2 [7 346 (C-C) (C-)] + 25 [498 (=)] } - {16 [2 804 (C=)] + 18 [2 463)] } kj mol -1 = kj mol -1 This is for the combustion of two moles of C 8 18 so the heat of combustion for one mole of hexane is: Δ comb = kj mol -1 (b) Ethane: Butane: exane: molar mass = ( (C) ()) g mol -1 = g mol -1 Δ comb = kj mol -1 or 421kJ mol g mol = 47.26kJ g -1 molar mass = ( (C) ()) g mol -1 = g mol -1 Δ comb = kj mol -1 or 2647kJ mol 58.12g mol = 45.54kJ g -1 molar mass = ( (C) ()) g mol -1

5 = g mol -1 Δ comb = kj mol -1 or 3873kJ mol g mol = kj g -1. (c) ctane: molar mass = ( (C) ()) g mol 1 = Δ comb = kj mol kJ mol or = kj g g mol n the basis of these values, ethane appears to be the most efficient fuel on a mass basis, although the values are very similar. ther considerations, such as the phase and transport costs are also relevant ethane, being a gas, is more difficult to transport. Ethanol - the balanced equation for combustion is: C à 2C C C + 3 à 2 C + 3 Δ comb = { [Δ atom (C 2 5 )] + 3 [Δ atom ( 2 )] } - {2 [Δ atom (C 2 )] + 3 Δ atom ( 2 )] } } = { [346 (C-C) (C-) (C-) (-)] + 3 [498 (=)] - {2 [2 804 (C=)] + 3 [2 463)] } kj mol -1 = kj mol -1 Ethanol: molar mass = ( (C) () ()) g mol 1 = g mol -1 Δ comb = kj mol -1 or 263kJ mol g mol = kj g -1 The heat content is somewhat less than that of the hydrocarbons. The main advantage of ethanol is that it can be obtained, in principle, from renewable sources. The present practice of using agricultural land to grow plants to make ethanol from has serious consequences for food supplies.

6 (d) Assuming that the densities of octane and ethanol are the similar, there will be ~0.1 g of ethanol and ~0.9 g of octane in 1.0 g of E10. (The actual densities differ slightly but are close enough for this comparison). 1 g of E10 will therefore provide (octane) = 42.9 kj g -1. As pure octane provides kj g -1, the percentage efficiency is ~96 % 6. (a) C 3 (C 2 ) 14 C à 16C (b) Using Δ comb = ΣmΔ f (products) - ΣnΔ f (reactants) o o o o comb f 2 f 2 f Δ = [16 Δ (C ) + 16 Δ ( )] [ Δ (palmitic acid)] as Δ f o ( 2) = 0 for the formation of an element in its standard state. As the combustion is an exothermic process, Therefore: or Δ comb o = kj mol -1. [ ] [ Δ (palmitic acid)] = 9980kJmol Δ f o (palmitic acid) = 889kJmol f o 1 (c) C 3 (C 2 ) 14 C C The molar mass of palmitic acid is: molar mass = ( (C) () ()) g mol -1 = g mol -1 So, Δ comb = kj mol -1 or 9980kJ mol g mol = 38.9kJ g Entropy describes the dispersal of energy. An entropy increase occurs when heat energy becomes more dispersed. By definition, ΔS = q / T so ΔS > 0 for the process in which q > 0. (a) 2 (g) + ½ 2 (g) 2 (l) = -286 kj mol -1 Δ surr S > 0. The surrounding receive heat from the exothermic reaction. This heat energy is dispersed in the surroundings and its entropy increases. (b) the evaporation of water evap = +44 kj mol -1 Δ sys S > 0. The evaporation is endothermic: heat is removed from the surroundings and transferred to the system. The system receives heat. This heat energy is dispersed in the system and its entropy increases. (c) heat flowing from a hot body to a colder one = 0

7 ΔS (cold body) > 0. The cold body receives heat energy from the hot body. This heat energy is dispersed in the cold body and its entropy increases. (d) Ba() (s) + 2N 4 N 3 (s) Ba 2+ (aq) + 2N 3 (aq) + 2N 3 - (aq) (l) = +62 kj/mol Δ sys S > 0. The reaction is endothermic: heat is removed from the surroundings and transferred to the system. The system receives heat. This heat energy is dispersed in the system and its entropy increases. (e) condensation of water from the vapour cond = - 44 kj mol -1 Δ surr S > 0. The surrounding receive heat from the exothermic condensation process. This heat energy is dispersed in the surroundings and its entropy increases.