14 Entrpy and Gibbs energy Answers t wrked examples WE 14.1 Predicting the sign f an entrpy change (n p. 658 in Chemistry 3 ) What will be the sign f the value f S fr: (a) crystallizatin f salt frm a slutin; (b) cndensatin f a vapur t a liquid; (c) disslving sugar in water? Cnsider whether the individual prcesses lead t increase in disrder, which wuld crrespnd t a psitive value fr S, r a decrease in disrder, in which case S will be negative. (a) Crystallisatin f salt frm a slutin: In slutin, the ins can mve arund freely whereas in a crystal they are fixed in the slid. There is thus an increase in rder s that S is negative. (b) Cndensatin f a vapur t a liquid: Althugh mlecules can mve arund in a liquid, they are less free t d s than in a gas which is highly disrdered. There is therefre a decrease in rder, and S is negative. (c) Disslving sugar in water: The psitin f sugar mlecules are fixed relative t each ther in a slid but the mlecules can mve arund in slutin: There is an increase in disrder, s that S is psitive.
WE 14.2 Entrpy change f vaprizatin (n p. 662 in Chemistry 3 ) Calculate the entrpy change when 1.00 ml f water at 0 C freezes t frm ice. Fr water, fush = +6.02 kj ml 1. Use Equatin 14.6, which relates the entrpy change fr a system t the enthalpy change and the temperature. Using Equatin 14.6, and remembering that the temperature shuld be expressed in units f kelvin, where 0 C = 273 K Δ fus S 298 = Δ fush 298 = +6.02 103 J ml 1 = +22.1 J K 1 ml 1 T 273 K This is fr fusin, i.e. melting. Freezing is the ppsite prcess, fr which ΔS 298 = Δ fus S 298 = 22.1 J K 1 ml 1 WE 14.3 Calculating entrpy changes n heating (n p. 663 in Chemistry 3 ) Calculate the entrpy change when 3.00 ml f nitrgen gas are heated frm 25 C t 50 C at cnstant pressure. (Cp fr N2 (g) is 29.13 J K 1 ml 1 ) Rearrange Equatin 14.7, remembering that temperatures must be in kelvin. Frm Equatin 14.7 then the mlar entrpy change S Tf = S Ti + C p ln(t f /T i ) ΔS m = S Tf S Ti = C p,m ln(t f /T i ) Remembering that the temperatures must be in units f kelvin, s that T i = (273 + 25) K = 298 K T f = (273 + 50) K = 323 K then the entrpy change fr an amunt n ml is ΔS = nδs m = nc p,m ln(t f /T i )
= 3.0 ml 29.13 J K 1 ml 1 ln(323 K/298 K) = +7.04 J K 1 WE 14.4 Finding a standard entrpy (n p. 667 in Chemistry 3 ) Draw a diagram, similar t Figure 1 in this example, t identify the stages invlved in finding S 298 (H2O). Starting with Figure 1 in the wrked example, decide what phase changes ccur when H2O(s) at 0 K is cnverted t H2O(l) at 298 K, and what temperatures they ccur at.
WE 14.5 Entrpy changes fr reactins at 298 K (n p. 669 in Chemistry 3 ) Calculate the standard entrpy changes at 298 K fr the fllwing reactins: (a) (b) CaCO3 (s) CO2 (g) + CaO (s) N2O4 (g) 2 NO2 (g) Use Equatin 14.11 t determine the entrpy change frm the entrpies f the individual prducts and reactants. Using Equatin 14.11 Δ r S 298 = ν i S 298 (prducts) ν i S 298 (reactants) where the terms in i represent the stichimetric cefficients f the prducts and reactants, then (a) (b) Δ r S 298 = S 298 (CO 2 (g)) + S 298 (CaO(s)) S 298 (CaCO 3 (s)) = 213.7 J K 1 ml 1 + 39.8 J K 1 ml 1 92.9 J K 1 ml 1 = +160.6 J K 1 ml 1 Δ r S 298 = 2 S 298 (NO 2 (g)) S 298 (N 2 O 4 (g)) = 2 240.1 J K 1 ml 1 304.3 J K 1 ml 1 = +175.9 J K 1 ml 1 Nte that it is gd practice, as with enthalpy changes, t qute the sign f the change, even if it is psitive. WE 14.6 Entrpy changes fr reactins at ther temperatures (n p. 671 in Chemistry 3 ) Calculate the standard entrpy changes f reactin at 100 C (373 K) fr the fllwing reactins: (a) CaCO3 (s) CO2 (g) + CaO (s) ΔrS 298 = +160.6 J K 1 ml 1 (b) N2O4 (g) 2 NO2 (g) ΔrS 298 = 175.9 J K 1 ml 1
Calculate the change in the heat capacity fr the reactin using Equatin 13.11 and the data in Appendix 7. Use this value, alng with Equatin 14.12 t determine the entrpy change n reactin at the new temperature. (a) Frm Equatin 13.11 Δ r C p = ν i C p (prducts) ν i C p (reactants) = {C p (CO 2 (g)) + C p (CaCO 3 (s))} C p (CaO(s)) = {42.8 J K 1 ml 1 + 37.1 J K 1 ml 1 } 81.9 J K 1 ml 1 = 2.0 J K 1 ml 1 Using Equatin 14.12 ΔS T2 = ΔS T1 + ΔC p ln(t 2 /T 1 ) then ΔS 373 = ΔS 298 + ΔC p ln(373 K/298 K) (b) Frm Equatin 13.11 Using Equatin 14.12 = +160.6 J K 1 ml 1 2.0 J K 1 ml 1 ln(373 K/298 K) = +160.2 J K 1 ml 1 Δ r C p = ν i C p (prducts) ν i C p (reactants) = 2 C p (NO 2 (g)) C p (N 2 O 4 (g)) = 2 37.2 J K 1 ml 1 77.3 J K 1 ml 1 = 2.9 J K 1 ml 1 ΔS T2 = ΔS T1 + ΔC p ln(t 2 /T 1 ) then ΔS 373 = ΔS 298 + ΔC p ln(373 K/298 K) = +175.9 J K 1 ml 1 2.9 J K 1 ml 1 ln(373 K/273 K) = +175.0 J K 1 ml 1
WE 14.7 Gibbs energy change and spntaneity (n p. 676 in Chemistry 3 ) Fr the melting f sdium chlride, NaCl, fush = +30.2 kj ml 1 and fuss = +28.1 J K 1 ml 1. Estimate the melting pint f NaCl. Use Equatin 14.16, remembering that at the melting pint, slid and liquid are in equilibrium s that fus G = 0. Using Equatin 14.16, since at the melting pint we may rearrange Δ fus G = Δ fus H T m Δ fus S = 0 T m = Δ fush Δ fus S = +30.2 103 J ml 1 +28.1 J K 1 = 1075 K ml 1 WE 14.8 Calculating the Gibbs energy change f reactin at 298 K (n p. 678 in Chemistry 3 ) Calculate the standard Gibbs energy change at 298 K fr the reactin 4 HCl (g) + O2 (g) 2 Cl2 (g) + 2 H2O (l) (Use Appendix 7 (p.1350) t find values f fh 298 and standard entrpies fr reactants and prducts.) Use the data in Appendix 7, alng with Equatins 13.6 and 14.11 t determine the enthalpy and entrpy f reactin. Then use Equatin 14.16 t determine the Gibbs energy f reactin frm these values. Using Equatin 13.6 Δ r H 298 = ν i Δ f H 298 (prducts) ν i Δ f H 298 (reactants)
= {2 Δ f H 298 (Cl 2 (g)) + 2 Δ f H 298 (H 2 O(l))} {4 Δ f H 298 (HCl(g)) + Δ f H 298 (O 2 (g))} = {2 0 + 2 285.8 kj ml 1 } {4 92.3 kj ml 1 + 0} = 202.4 kj ml 1 Using Equatin 14.11 Δ r S 298 = ν i S 298 (prducts) ν i S 298 (reactants) = {2 S 298 (Cl 2 (g)) + 2 S 298 (H 2 O(l))} {4 S 298 (HCl(g)) + S 298 (O 2 (g))} = {2 223.1 J K 1 ml 1 + 2 69.9 J K 1 ml 1 } = 366.7 J K 1 ml 1 {4 186.9 J K 1 ml 1 + 205.1 J K 1 ml 1 } Using Equatin 14.16 t calculate rg 298, Δ r G = Δ r H TΔ r S = 202.4 10 3 J ml 1 (298 K 366.7 J K 1 ml 1 ) = 93.1 10 3 J ml 1 = 93.1 kj ml 1 WE 14.9 Using Gibbs energy changes f frmatin t calculate Gibbs energy change f reactin (n p. 681 in Chemistry 3 ) Calculate the standard Gibbs energy change fr the xidatin f ammnia at 298 K. 5 O2 (g) + 4 NH3 (g) 6 H2O (l) + 4 NO (g) Use Equatin 14.18, substituting the data fr the Gibbs energy change f frmatin frm Appendix 7. Using Equatin 14.18 Δ r G 298 = ν i Δ f G 298 (prducts) ν i Δ f G 298 = {6 Δ f G 298 (H 2 O(l)) + 4 Δ f G 298 (NO(g))} (reactants) {5 Δ f G 298 (O 2 (g)) + 4 Δ f G 298 (NH 3 (g))}
= {6 237.1 kj ml 1 + 4 86.6 kj ml 1 } {5 0 + 4 16.5 kj ml 1 } = 1010 kj ml 1 WE 14.10 Calculating the Gibbs energy change f reactin at ther temperatures (n p. 681 in Chemistry 3 ) Calculate the standard Gibbs energy change fr the xidatin f ammnia at 650 K. Cmpare yur answer with the value calculated fr 298 K in the questin at the end f Wrked Example 14.9. 5 O2 (g) + 4 NH3 (g) 6 H2O (l) + 4 NO (g) Calculate the standard enthalpy and entrpy f reactin at 298 K using Equatins 13.6 and 14.11. Als determine the change in the heat capacity n reactin using Equatin 13.11. Hence use Equatins 13.10 and 14.12 t determine the standard enthalpy and entrpy f reactin at 650 K. Finally, use Equatin 14.16 t determine the change in the Gibbs energy at 650 K. Using Equatin 13.6, Δ r H 298 = ν i Δ f H 298 (prducts) ν i Δ f H 298 (reactants) Then, using Equatin 14.11, = {6 Δ f H 298 (H 2 O(l)) + 4 Δ f H 298 (NO(g)) +} {5 Δ f H 298 (O 2 (g)) + 4 Δ f H 298 (NH 3 (g))} = {6 285.8 kj ml 1 + 4 90.3 kj ml 1 } {5 0 + 4 46.1 kj ml 1 } = 1169.2 kj ml 1 Δ r S 298 = ν i S 298 (prducts) ν i S 298 (reactants)
Using Equatin 13.11, = {6 S 298 (H 2 O(l)) + 4 S 298 (NO(g)) +} {5 S 298 (O 2 (g)) + 4 S 298 (NH 3 (g))} = {6 69.9 J K 1 ml 1 + 4 210.8 J K 1 ml 1 } {5 205.1 J K 1 ml 1 + 4 192.5 J K 1 ml 1 } = 532.9 J K 1 ml 1 Δ r C p = ν i C p (prducts) ν i C p (reactants) = {6 C p (H 2 O(l)) + 4 C p (NO(g)) +} {5 C p (O 2 (g)) + 4 C p (NH 3 (g))} = {6 75.3 J K 1 ml 1 + 4 29.8 J K 1 ml 1 } {5 29.4 J K 1 ml 1 + 4 35.1 J K 1 ml 1 } = +283.6 J K 1 ml 1 Then, frm Equatin 13.10 Δ r H T2 = Δ r H T1 + ΔC p (T 2 T 1 ) s that Δ r H 650 = Δ r H 298 + ΔC p (650 K 298 K) Using Equatin 14.12 = 1169.2 10 3 J ml 1 + 283.6 J K 1 ml 1 352 K = 1069 10 3 J ml 1 = 1069 kj ml 1 ΔS T2 = ΔS T1 + ΔC p ln(t 2 /T 1 ) then ΔS 650 = ΔS 298 + ΔC p ln(650 K/298 K) = 532.9 J K 1 ml 1 + 283.6 J K 1 ml 1 ln(650 K/298 K) = 311.7 J K 1 ml 1 Cmbining the values fr the enthalpy and entrpy changes at 650 K using Equatin 14.16, Δ r G 650 = Δ r H 650 TΔ r S 650 = 1069.4 10 3 J ml 1 650 K 311.7 J K 1 ml 1
= 866.8 10 3 J ml 1 = 866.8 kj ml 1 Cmparing the value with that calculated in the previus Wrked Example at a temperature f 298 K, Δ r G 298 = 1010.2 kj ml 1, it can be seen that the standard Gibbs energy f reactin increases, that is, it becmes less negative, as the temperature increases. The reactin therefre must becme less spntaneus at higher temperatures.
Answers t bxes Bx 14.4 What happens when prteins dn t fld crrectly? (n p. 673 in Chemistry 3 ) The thermdynamics f prtein flding can be cnsidered using the relatinship in Equatin 14.3: ΔS(ttal) = ΔS(system) + ΔS(surrundings) (a) The flding f a prtein leads t a highly rdered structure. What is the sign f the value f ΔS(system) fr the prcess? (b) Flding is a spntaneus prcess under physilgical cnditins. What des this tell yu abut the value f ΔS(ttal)? What implicatins des this have fr the value f ΔS(surrundings)? Explain hw this happens. (c) Suggest why the frmatin f the misflded frm f PrP cannt take place spntaneusly frm the nrmal frm f the prtein. (a) Cnsider whether the entrpy f a flded prtein is greater than that f an unflded prtein. (b) Use the Secnd Law f Thermdynamics t predict the sign f ΔS(ttal), and therefre ΔS(surrundings). (c) Predict the sign f the ttal entrpy change and use the Secnd Law f Thermdynamics. (a) The flded prtein is mre highly rdered than the unflded prtein. The entrpy f the prtein must therefre decrease n flding s that ΔS(system) < 0. (b) The value f ΔS(ttal) must be psitive fr the prcess t bey the Secnd Law. Thus, if ΔS(ttal) = ΔS(system) + ΔS(surrundings) > 0 yet ΔS(system) < 0 then ΔS(surrundings) > 0 and the magnitudes f the entrpy changes fr the surrundings and system, ΔS(surrundings) > ΔS(system)
Thus, ΔS(system) must be less negative than ΔS(surrundings) is psitive. Heat is released when hydrgen bnds r ther nn-cvalent interactins frm. Prteins are large mlecules s lts f nn-cvalent interactins frm, making the heat change large. (c) The flding f PrP is endthermic, s that Thus, The entrpy change fr the system and s the ttal entrpy change Δ r H > 0 ΔS(surrundings) = Δ rh T < 0 ΔS(system) = Δ r S < 0 ΔS ttal = ΔS(system) + ΔS(surrundings) < 0 Fr a reactin t be spntaneus, the ttal entrpy change must be psitive. Bx 14.6 Thermdynamics f additin plymerizatin (n p. 679 in Chemistry 3 ) At 298 K, plh (the enthalpy change f plymerizatin) fr styrene is - 68.5 kj ml -1. The value f plh fr -methyl styrene is 35.1 kj ml -1. pls298 = - 105 J K -1 ml -1 fr bth reactins. (a) Calculate the Gibbs energy change f plymerizatin at 298 K fr bth mnmers. Estimate the ceiling temperature at which plg = 0 in each case. (b) What are the implicatins f the ceiling temperatures fr these mnmers? Use Equatin 14.16 t determine the Gibbs energy f reactin frm the enthalpy and entrpy f reactin. Assume that the enthalpy and entrpy f reactin d nt vary with temperature and hence determine the temperature at which the Gibbs energy change is zer.
(a) Fr styrene, When rearranging, Δ pl G 298 = Δ pl H 298 TΔ pl S 298 = 68.5 10 3 J ml 1 (298 105 J K 1 ml 1 ) = 37.2 10 3 J ml 1 = 37.2 kj ml 1 Δ pl G 298 = Δ pl H 298 TΔ pl S 298 = 0 T = Δ pl H 298 /Δ pl S 298 = 68.5 10 3 J ml 1 / 105 J K 1 ml 1 = 652 K Fr -methyl styrene: When rearranging, Δ pl G 298 = Δ pl H 298 TΔ pl S 298 = 35.1 10 3 J ml 1 (298 K 105 J K 1 ml 1 ) = 3.8 10 3 J ml 1 = 3.8 kj ml 1 Δ pl G 298 = Δ pl H 298 TΔ pl S 298 = 0 T = Δ pl H 298 /Δ pl S 298 = 35.1 10 3 J ml 1 / 105 J K 1 ml 1 = 334 K (b) The calculatins shw that that plystyrene is thermdynamically stable relative t its mnmer up t 652 K (379 C). With -methyl styrene, the mnmer is thermdynamically stable abve 334 K (61 C) s that the plymer will nt frm frm mnmers abve this temperature. Since plymers are ften made by heating, this means that care must be taken during the synthesis. It als has implicatins fr the highest temperature at which the plymer can be used. Bx 14.7 Obtaining silicn fr use in silicn chips (n p. 682 in Chemistry 3 ) Use the thermdynamic data in the fllwing table t answer the fllwing questins.
(a) Explain, in terms f entrpy, why it is s difficult t prepare ultra-high purity silicn crystals. Cnsider the degree f disrder and hence the magnitude and sign f the entrpy change. Ultra-high purity silicn crystals have a very high degree f rder. The entrpy change fr turning impure silicn t ultra-high purity silicn is thus large and negative. (b) Silicn culd be extracted using the reactin: SiO2 (s) Si (s) + O2 (g) (Reactin 1) What are the standard enthalpy change and the standard entrpy change at 298 K fr this reactin? Cmment n the value f rs 298 that yu btain. Use Equatin 13.6 t determine the enthalpy change and Equatin 14.12 t determine the entrpy change fr the Reactin 1 using the enthalpies f frmatin and entrpies in the table. Using Equatin 13.6 and Equatin 14.11 Δ r H 298 = ν i Δ f H 298 (prducts) ν i Δ f H 298 (reactants) = {Δ f H 298 (Si(s)) + Δ f H 298 (O 2 (g))} Δ f H 298 (SiO 2 (s)) = 0 + 0 ( 910.9 kj ml 1 ) = +910.9 kj ml 1 Δ r S 298 = ν i S 298 (prducts) ν i S 298 Δ r S 298 = {S 298 (Si(s)) + S 298 (O 2 (g))} S 298 (SiO 2 (s)) (reactants) = 18.8 J K 1 ml 1 + 205.1 J K 1 ml 1 41.8 J K 1 ml 1 = +182.1 J K 1 ml 1
(c) Calculate the value f rg 298 fr the reactin. Why des elemental silicn nt exist in the Earth s crust? Use the values fr the standard enthalpy and entrpy change frm part (b) and Equatin 14.16. Using Equatin 14.16 Δ r G 298 = Δ r H 298 TΔ r S 298 = +910.9 10 3 J ml 1 298 K +182.1 J K 1 ml 1 = +856.6 10 3 J ml 1 = +856.6 kj ml 1 Elemental silicn des nt exist in the Earth s crust because G 298 fr the cnversin f silicn(iv) xide t silicn thrugh reactin 1 has a large psitive value, s the prcess des nt ccur spntaneusly. The xide is therefre thermdynamically much mre stable. (d) Estimate the temperature at which rg = 0. What is the significance f this fr preparing elemental silicn using this reactin? Rearrange Equatin 14.16 and slve fr the temperature at which rg = 0. Frm Equatin 14.16 then, rearranging, Δ r G 298 = Δ r H 298 TΔ r S 298 = 0 T = Δ r H 298 /Δ r S 298 = +910.9 10 3 J ml 1 /+182.1 J K 1 ml 1 = 5002 K Achieving such a high temperature wuld need very large energy input and it wuld be unrealistic and unecnmic t perate under these cnditins. (e) Calculate the value f rg 298 fr this reactin and cmment n the value yu btain. SiO2 (s) + 2 C (s) Si (s) + 2 CO (g) (Reactin 2)
Use Equatin 14.6 t determine the enthalpy change and Equatin 14.12 t determine the entrpy change fr Reactin 2 using the enthalpies f frmatin and entrpies in the table. Use the values and Equatin 14.16 t determine the standard Gibbs energy change. Using Equatin 13.6 Δ r H 298 = ν i Δ f H 298 (prducts) ν i Δ f H 298 (reactants) and Equatin 14.11 = {Δ f H 298 (Si(s)) + 2 Δ f H 298 (CO(g))} {Δ f H 298 (SiO 2 (s)) + 2 Δ f H 298 (C(s))} = {0 + 2 110.5 kj ml 1 } {( 910.9 kj ml 1 ) + 2 0} = +689.9 kj ml 1 Δ r S 298 = ν i S 298 (prducts) ν i S 298 (reactants) Δ r S 298 = {S 298 (Si(s)) + 2 S 298 (CO(g))} {S 298 (SiO 2 (s)) + 2 S 298 (C(s))} = {18.8 J K 1 ml 1 + 2 197.7 J K 1 ml 1 } {41.8 J K 1 ml 1 + 2 5.7 J K 1 ml 1 } = +361 J K 1 ml 1 Using Equatin 14.16 Δ r G 298 = Δ r H 298 TΔ r S 298 = +689.9 10 3 J ml 1 298 K +361 J K 1 ml 1 = +582.3 10 3 J ml 1 = +582.3 kj ml 1 rg 298 is thus still psitive, s the reactin is nt spntaneus, but it is significantly less psitive than fr Reactin 1. (f) Estimate the temperature at which rg = 0 fr this reactin. Cmment n yur answer. Rearrange Equatin 14.16 and slve fr the temperature at which rg = 0.
Frm Equatin 14.16 then, rearranging, Δ r G 298 = Δ r H 298 TΔ r S 298 = 0 T = Δ r H 298 /Δ r S 298 = +689.9 10 3 J ml 1 /+361 J K 1 ml 1 = 1911 K The temperature at which the reactin becmes spntaneus is thus lwer than fr reactin 1, but still very high. (g) Repeat the calculatins f rg 298 and the temperature at which rg = 0 fr this reactin SiO2 (s) + C (s) Si (s) + CO2 (g) (Reactin 3) Use Equatin 13.6 t determine the enthalpy change and Equatin 14.12 t determine the entrpy change fr Reactin 3 using the enthalpies f frmatin and entrpies in the table. Use the values and Equatin 14.16 t determine the standard Gibbs energy change. Using Equatin 13.6 Δ r H 298 = ν i Δ f H 298 (prducts) ν i Δ f H 298 (reactants) = {Δ f H 298 (Si(s)) + Δ f H 298 (CO 2 (g))} {Δ f H 298 (SiO 2 (s)) + Δ f H 298 (C(s))} = {0 + ( 393.5 kj ml 1 )} {( 910.9 kj ml 1 ) + 0} = +517.4 kj ml 1 and Equatin 14.11 Δ r S 298 = ν i S 298 (prducts) ν i S 298 (reactants) Δ r S 298 = {S 298 (Si(s)) + S 298 (CO 2 (g))} {S 298 (SiO 2 (s)) + S 298 (C(s))} = {18.8 J K 1 ml 1 + 213.7 J K 1 ml 1 } {41.8 J K 1 ml 1 + 5.7 J K 1 ml 1 } = +185 J K 1 ml 1
Using Equatin 14.16 Frm Equatin 14.16 then, rearranging, Δ r G 298 = Δ r H 298 TΔ r S 298 = +517.4 10 3 J ml 1 298 K +185 J K 1 ml 1 = +462.3 10 3 J ml 1 = +462.3 kj ml 1 Δ r G 298 = Δ r H 298 TΔ r S 298 = 0 T = Δ r H 298 /Δ r S 298 = +517.4 10 3 J ml 1 /+185 J K 1 ml 1 = 2797 K (h) Hw des the change in entrpy influence which reactin is actually used? Cnsider the magnitude f rs 298 fr Reactins 2 and 3. A large psitive entrpy change lwers the temperature at which rg 298 = 0. Reactin 2, with its larger entrpy change, ccurs at a lwer temperature than Reactin 3 and s at 2500 K, Reactin 2 will be the spntaneus prcess. Bx 14.8 Energetics f bichemical reactins (n p. 683 in Chemistry 3 ) (a) The enthalpy change fr the hydrlysis f ATP under bichemical standard cnditins at 37 C (310 K) is rh = 20 kj ml 1 (Reactin 2). Calculate the entrpy change under these cnditins and cmment n the value. Use Equatin 14.16, and substitute the knw values fr the Gibbs energy and enthalpy changes f reactin. Using Equatin 14.16 Δ r G = Δ r H TΔ r S and rearranging
Δ r S = Δ rh Δ r G T = 20 103 J ml 1 ( 30.5 10 3 J ml 1 ) (273 + 37) K = +33.9 J K 1 ml 1 The reactin has a psitive entrpy change. Since the enthalpy change is als negative, the reactin is spntaneus under all cnditins. (b) Calculate the verall Gibbs energy change fr the fllwing reactin at 37 C glucse (aq) + ATP (aq) glucse-6-phsphate (aq) + ADP (aq) + H + (aq) Express the reactin as a cmbinatin f Reactins 1 and 2. Hence determine the Gibbs energy change as the equivalent cmbinatin f the Gibbs energy changes fr Reactins 1 and 2 The reactin C6H12O6 (aq) + ATP (aq) C6H11O6Pi (aq) + ADP (aq) + H + (aq) is the sum f Reactins 1 and 2 C6H12O6 (aq) + Pi (aq) C6H11O6Pi (aq) + H2O (l) ATP (aq) + H2O (l) ADP (aq) + H + (aq) + Pi (aq) Δ r G 1 = +13.4 kj ml 1 Δ r G 2 = 30.5 kj ml 1 The Gibbs energy change is thus Δ r G = Δ r G 1 + Δr G 2 = +13.4 kj ml 1 30.5 kj ml 1 = 17.1kJ ml 1 Bx 14.9 Hw much wrk can yu get frm glucse? (n p. 686 in Chemistry 3 ) The Shard in Lndn has an bservatin platfrm at a height f 244 m. Rather than taking the lift, a persn weighing 65 kg climbs the stairs t the platfrm. Hw much glucse wuld the persn need t cnsume t replace the energy used in climbing the stairs?
Determine the wrk dne, which is equivalent t the increase in ptential energy, in climbing the stairs frm Equatin 1.16. Calculate the mass f glucse required as a multiple f the energy release by 1 g. The ptential energy f an bject at a height h abve the surface f the Earth is, frm Equatin 1.16, E PE = mgh The wrk dne in climbing the stairs is thus w = 65 kg 9.81 m s 2 244 m = 156 10 3 J = 156 kj A mass f 1 g glucse prvides 16 kj s 156 kj/16 kj g 1 = 9.8 g glucse wuld be needed. In practice, mre than this wuld be needed because f inefficiencies in the transfer f energy; the bdy des nt use sugars with 100% efficiency. Bx 14.10 Extractin f metals frm res (n p. 689 in Chemistry 3 ) (a) Predict the sign f the entrpy changes fr Reactins 1 3. Cnsider in turn, whether each reactin results in an increase, r a decrease in the amunt f gas. An increase in the amunt f gas wuld crrespnd t an increase in disrder, and hence a psitive value fr the change in entrpy. Cnversely, a decrease wuld crrespnd t a decrease in disrder and hence a negative value fr the change in entrpy. Reactin 1: M(s) + ½ O2 (g) MO (s)
Gas is cnverted int slid s ngas = ½. There is thus an increase in rder, s that the value f S is negative. Reactin 2: C(s) + ½ O2 (g) CO (g) The reactin results in an increase in the amunt f gas, ngas = +½, s there is an increase in the disrder and the value f S is psitive. Reactin 3: C(s) + O2 (g) CO2 (g) There is n change in the amunt f gas, ngas = 0, s any change in the entrpy will be small. (b) Explain why the line fr CO slpes upwards, that fr CO2 is almst hrizntal while thse fr the metal xides slpe dwnward. Differentiate Equatin 14.16 and s btain an expressin that shws hw the gradients f the lines n the Ellingham diagram depend n the standard entrpy change f reactin. Differentiating Equatin 14.16, Δ r G = Δ r H TΔ r S with respect t temperature results in an expressin fr the rate f change f the Gibbs energy with temperature, which is just the gradient f the lines n the Ellingham diagram. Thus ( Δ rg T ) p = ( Δ rh T ) p ( (TΔ rs ) ) Δ T r S p because the enthalpy and entrpy f reactin vary little with temperature, s that ( Δ rh T ) 0 p
and ( Δ rs T ) 0 p Ellingham diagrams are cnventinally drawn s that the Gibbs energy is pltted as running frm negative t psitive, s that the gradient f the plt is m = ( Δ rg T ) p = +Δ r S The gradient f the line fr a reactin fr which the entrpy change is negative, Δ r S < 0, thus has a negative slpe. Fr a reactins such as Reactin 1, fr which Δ r S > 0, then we shuld expect the line n the diagram t slpe upwards, which it des. Cnversely, fr Reactin 2, because Δ r S < 0, the line slpes dwnwards. Because fr Reactin 3, Δ r S 0, then the line is almst hrizntal. (c) Use the Ellingham diagram t estimate the lwest temperature at which zinc xide can be reduced t the metal by carbn. Find the temperature at which the tw lines crss. This will be the pint at which the Gibbs energy f the tw reactins are equal. The ZnO and CO lines crss ver at arund 1300 K, s this is the lwest temperature at which reductin will ccur spntaneusly. (d) Suggest why aluminium is prduced using an electrchemical methd rather than by reductin f the xide. Using the Ellingham diagram, decide which metals culd be used t reduce aluminium xide and under what cnditins. It appears frm the Ellingham diagram that aluminium xide may nly be reduced by carbn and magnesium, and even then nly at high temperatures. The line fr aluminium xide crsses that fr carbn at a temperature f
apprximately 2300 K. Achieving this temperature wuld make the prcess t expensive t be ecnmically feasible.
Answers t end f chapter questins 1. (a) Explain the meaning f the term spntaneus as applied t a chemical r physical change. (b) Explain why sme reactins which appear t be spntaneus frm thermdynamic data d nt in fact take place. Illustrate yur answer by referring t the fllwing changes: (i) H2O(s) H2O(l) (iii) Oxidatin f aluminium in air. Whether a reactin is thermdynamically favurable depends n the cnditins, particularly temperature and pressure accrding t the Secnd Law. Even if a reactin is thermdynamically favurable it may be kinetically unfavurable and s prceed s slwly as t be unbservable. (a) Spntaneus changes are thse that, nce started, will cntinue withut any utside interventin. (b) A reactin needs t be bth thermdynamically and kinetically favurable in rder t take place under a given set f cnditins. In example (i), H2O(s) nly melts when the temperature is abve 273 K, because this is the temperature at which the entrpy changes in the system and surrundings are equal. Abve this temperature the ttal entrpy change fr melting is psitive and the H2O(s) melts spntaneusly. Kinetic factrs can play a part in the reverse prcess freezing f water t frm ice. If very pure water is cled slwly it can exist as a supercled liquid at temperatures lwer than 0 C. In example (ii), the reactin is thermdynamically favurable at rm temperature, but it is kinetically unfavurable because the aluminium
becmes cvered in a thin layer f aluminium xide which prevents xygen reaching the metal surface. Further reactin ccurs extremely slwly, and the aluminium effectively remains unxidized. 2. Predict the sign f the change in entrpy in the system in the fllwing prcesses. Use the general rule that gases, and slids in aqueus slutin, have a higher entrpy than liquids, which have a higher entrpy than slids. a Steam cndensing n a cld windw. Negative because entrpy decreases as a less rdered gas frms a liquid. b c A clud frming in the atmsphere. As fr (a). Inflating a bicycle tyre with air Negative because the entrpy f a gas decreases with pressure. d Disslving sugar in ht cffee. Psitive (a slid changes t an aqueus slutin with an increase in disrder.) e PCl3(g) + Cl2(g) PCl5(g) Negative because the number f mlecules f gas decreases, increasing the rder in the system. f H2O(g) + CaSO4(s) CaSO4 H2O(s) Negative because the number f mlecules f gas decreases and s des the disrder. g SO3(g) + H2O(l) H2SO4(aq) As the equatin is written this will be negative (ne mlecule each f a gas and a liquid are cnverted t a mlecule in
aqueus slutin). Hwever, if the slutin is dilute the H2SO4 will dissciate int 2H + (aq) and SO4 2- (aq), s the entrpy change is likely t be psitive. h 2 KCl(s) + H2SO4(l) K2SO4(s) + 2 HCl(g) Psitive because tw mlecules f gas are frmed increasing the disrder. i C2H4(g) + H2O(g) C2H5OH(l). Negative because tw mlecules f gas are cnverted t a mlecule f liquid with a decrease in disrder. 3. An apparatus cnsists f tw bulbs f the same vlume cnnected by a tap. Initially, the tap is clsed with ne bulb cntaining nitrgen gas and the ther xygen gas. Bth bulbs are at the same temperature and pressure. (a) What happens when the tap is pened? What will be the equilibrium state f the system? (b) What are the signs f H, S and G fr the prcess in (a)? (c) Is this cnsistent with the Secnd Law f Thermdynamics? (a) Decide what will happen t the gases when the tap is pened and whether this is likely t be a spntaneus prcess. (b) Cnsider hw the enthalpy changes fr a system cntaining ideal gases in which the temperature and pressure remain cnstant. Determine the sign f the entrpy change by deciding whether the prcess results in an increase r decrease in disrder. Apply Equatin 14.16 t determine the change in the Gibbs energy. (c) Cnsider the cnditins n the ttal entrpy r Gibbs energy f the system fr a prcess t be spntaneus. (a) The tw gases gradually diffuse int each ther and mix until the cncentratins f each gas are the same in bth bulbs. (b) There is n temperature change and the mixing ccurs at cnstant pressure, s the enthalpy change fr ideal gases where n interactins ccur is Δ r H = 0
The mixed gases have greater disrder (randmness) than when in separate bulbs s the entrpy change will be psitive Δ r S > 0 Frm Equatin 14.16, Δ r G = Δ r H TΔ r S = TΔ r S s Δ r G < 0 (c) The diffusin and mixing f gases is a spntaneus prcess. The Secnd Law states that fr a change t be spntaneus ΔS ttal > 0 which is equivalent t the cnditin (dg) p,t < 0 Thus, this is all cnsistent with the Secnd Law. 4. Estimate the change in entrpy when 1.00 ml f argn is heated frm 300 K t 1200 K. What assumptins have yu made and hw culd yu make yur estimate mre accurate? (The heat capacity, Cp f argn gas is 20.8 J K 1 ml 1.) Use Equatin 14.7 t calculate hw the mlar entrpy changes n heating. Frm Equatin 14.7 S Tf = S Ti + C p ln(t f /T i ) then the mlar entrpy change ΔS m = S Tf S Ti = C p,m ln(t f /T i ) Thus, the entrpy change fr 1 ml is ΔS = nδs m = nc p,m ln(t f /T i ) = 1.0 ml 20.8 J K 1 ml 1 ln(1200 K/300 K)
= +28.8 J K 1 ml 1 The calculatin assumes that the heat capacity remains cnstant acrss the whle temperature range. The estimate culd be imprved by accunting fr the variatin using a series functin t describe hw Cp varies with temperature. 5. Fr each f the fllwing reactins, suggest whether the entrpy change in the system wuld be (i) near zer; (ii) psitive r (iii) negative. Explain yur answers. (a) (b) (c) (d) (e) N 2 (g) + 3H 2 (g) 2 NH 3 (g) Zn (s) + Cu 2+ (aq) Zn 2+ (aq) + Cu (s) 3 Mg (s) + 2 Fe 3+ (aq) 3 Mg 2+ (aq) + 2 Fe (s) CH 4 (g) + 2O 2 (g) CO 2 (g) + 2 H 2 O (g) C (diamnd) C (graphite) Cnsider whether there is an increase r decrease in the disrder f the system. An increase in disrder, such as that expected when there is an increase in the amunt f gas r ins in slutin, will lead t a psitive value fr the entrpy change. Cnversely, a decrease in disrder will lead t a negative entrpy change. (a) The amunt f gas decreases as a result f the reactin, Δn gas = 2 4 = 2 s the entrpy is reduced. Thus, S is negative. (b) There is little change in disrder and s nly a small change in entrpy. S 0 (c) The reactin increases the number f ins in slutin and s results in greater disrder. S is psitive. (d) There is n change in the amunt f gas Δn gas = 3 3 = 0
s nly a small change in entrpy, S 0 (e) Diamnd has a mre rganised structure than graphite s the entrpy increases. S is psitive. 6. Disslving slid ptassium idide in water results in a lwering f the temperature. Explain why this endthermic prcess can be spntaneus. Cnsider the entrpy changes in the system and the surrundings fr this prcess. The entrpy change in the system is psitive because a slid is being cnverted t aqueus ins which have much higher entrpy. This is mre than enugh t cmpensate fr the negative entrpy change in the surrundings caused by absrptin f heat as the slid disslves. 7. Calculate the entrpy change when 1 ml f water is heated frm 250 K t 300 K. The enthalpy change f fusin fr water is at 0 C is +6.01 kj ml -1 and the heat capacities are Cp = 75.3 J K -1 ml -1 and 37.2 J K -1 ml -1 fr water and ice respectively. There are three stages, each with its wn entrpy change. Stage 1: heating H2O(s) frm 250 K t its melting pint at 273 K (equatin 14.7). Stage 2: melting H2O(s) t H2O(l) (equatin 14.6) Stage 3: heating H2O(l) frm 273 K t 300 K (equatin 14.7) The ttal entrpy change is given by adding tgether the entrpy change fr each f these three stages.
C (slid) H Cp(liquid) S T dt T T T 273 p 300 fus ttal d 250 273 m 273 fus 300 Sttal Cp(slid) ln Cp(liquid) ln 250 Tm 273 H 273K H 300K 7.2 ln 75.3 ln 250 273K -1-1 fus -1-1 Sttal 3 J K ml J K ml K Tm Sttal = 3.27 J K -1 ml -1 + 22.01 J K -1 ml -1 + 7.10 J K -1 ml -1 Sttal = 32.38 J K -1 ml -1 8. 100.0 g f water at 30 C were placed in a refrigeratr at 4 C. When the water cls dwn, what is the entrpy change f (a) the water and (b) the refrigeratr? What is the verall entrpy change? Cmment n the results. (a) Use equatin 14.7 t find the entrpy change when water is cled frm 30 C t 4 C. (b) Use equatin 13.2 t calculate the quantity f heat, q, released by the water when it is cled frm 30 C t 4 C. Then use equatin 14. 4 t find the entrpy change in the refrigeratr (the surrundings). Add the tw entrpy changes in (a) and (b) and see whether the verall entrpy change is psitive as the Secnd Law f Thermdynamics wuld predict. 100 g (a). The amunt f water in 100.0 g n = -1 18.02 g ml = 5.55 ml S nc T 277 K 3 K f -1-1 p ln 5.55 ml 75.3 J K ml ln Ti 30
S (water) = - 37.5 J K -1 (b) The heat released by the water as it cls, and absrbed by the refrigeratr, q = n Cp T q = 5.55 ml 75.3 J K -1 ml -1 ( 303 K 277 K) = + 10870 J The refrigeratr is maintained at 277 K s S = q T = 10870 J 277 K S (refrigeratr) = + 39.2 J K -1 The verall entrpy change is therefre (- 37.5 J K -1 + 39.2 J K -1 ) = + 1.7 J K -1. There is an verall increase in entrpy as expected frm the Secnd Law f Thermdynamics. 9. Given that the nrmal melting and biling pints f CO are 74 K and 82 K respectively, sketch a plt t shw hw the entrpy f CO varies with temperature between 0 K and 273 K. Think abut S fr each f the fllwing stages: Stage 1: heating CO(s) frm 0 t 74 K Stage 2: melting CO(s) t CO(l) at 74 K Stage 3: heating CO(l) frm 74 K t 82 K Stage 4: vaprising CO(l) t CO(g) at 82 K (this will be larger than S fr Stage 2) Stage 5: heating CO(g) frm 82 K t 273 K
10. What are the signs f H, S and G fr the prcess f a vapur cndensing t frm a liquid? Cnsider whether heat is released by r transferred t the vapur n cndensatin and hence determine the sign f the enthalpy change. Als cnsider hw the degree f disrder changes and thus the effect n the sign f the entrpy change. Finally, use Equatin 14.16 t predict the sign f the Gibbs energy change based n yur previus answers. Heat is released during cndensatin s vaph is negative. In cndensing, a disrdered vapur becmes a mre rdered liquid, s that vaps is negative. Since, frm Equatin 14.16 Δ vap G = Δ vap H TΔ vap S then if Δ vap H < 0 and if Δ vap S > 0 then TΔ vap S > 0
The enthalpic and entrpic terms are thus in cmpetitin. The balance between the tw, and thus the sign f the change in the Gibbs energy will depend upn the temperature. Cndensatin will becme spntaneus at a temperature when and which is given by Δ vap G = Δ vap H TΔ vap S = 0 T = Δ vaph Δ vap S 11. An ice cube f mass 18 g is added t a large glass f water just abve 0 C. Calculate the change f entrpy fr the ice and fr the water, (withut the ice). The enthalpy change f fusin fr water is +6.01 kj ml 1. Determine the amunt f ice that crrespnds t a mass f 18 g frm the mlar mass f water. Hence determine the enthalpy change and the heat transferred n melting. Use Equatin 14.4 t determine the entrpy change. The mlar mass f water is M = {16.00 + (2 1.01)} g ml 1 = 18.02 g ml 1 Thus, 18 g is equivalent t 1 ml. The heat transferred frm the water t the ice as it melts reversibly is q rev = q = ΔH = Δ fus H = +6.01 10 3 J ml 1 Treating the ice as the system and the water as the surrundings, then frm Equatin 14.4, the entrpy change is ΔS ice = q rev T = +6.01 103 J ml 1 273 K assuming that the temperature remains cnstant. = +22.0 J K 1 ml 1 Nw cnsidering the water, the heat transferred frm the ice t the water is thus q rev = 6.01 10 3 J ml 1
and s ΔS water = q rev T = 6.01 103 J ml 1 273 K = 22.0 J K 1 ml 1 12. Calculate the change in entrpy when 100 g f water at 90 C is added t an insulated flask cntaining 100 g f water at 10 C. Deduce the final temperature f the water. Calculate the mlar mass f water and hence determine the amunt f water that crrespnds t 100 g. Then use Equatin 14.7, using the value fr the heat capacity f water, t determine the entrpy change fr this amunt. Since 100 g f ht water is added t 100 g f cld water, the final temperature will be the mean f the tw starting temperatures, i.e. 50 C. The mlar mass f water is M = {16.00 + (2 1.01)} g ml 1 = 18.02 g ml 1 s that 100 g is equivalent t n = m/m = 100 g /18.02 g ml 1 = 5.55 ml The heat transferred frm the water t the ice Frm Equatin 14.7 S Tf = S Ti + C p ln(t f /T i ) then the mlar entrpy change ΔS m = S Tf S Ti = C p,m ln(t f /T i ) Remembering that the temperatures must be in units f kelvin, s that fr the cld water, T i = (273 + 10) K = 283 K T f = (273 + 50) K = 323 K then the entrpy change fr an amunt n is
ΔS = nδs m = nc p,m ln(t f /T i ) = 5.55 ml 75.2 J K 1 ml 1 ln(323 K/283 K) = +55.2 J K 1 ml 1 Fr the ht water, T i = (273 + 90) K = 363 K T f = (273 + 50) K = 323 K then the entrpy change fr an amunt n is ΔS = nδs m = nc p,m ln(t f /T i ) = 5.55 ml 75.2 J K 1 ml 1 ln(323 K/363 K) = 48.7 J K 1 ml 1 The ttal entrpy change is thus ΔS ttal = +55.2 J K 1 ml 1 48.7 J K 1 ml 1 = +6.5 J K 1 ml 1 13. Calculate the standard entrpy change at 298 K fr the reactins: (a) (b) (c) N2 (g) + 3 H2 (g) 2 NH3 (g) Hg (l) + Cl2 (g) HgCl2 (s) C6H12O6 (s) + 6 O2 (g) 6 CO2 (g) + 6 H2O (g) Use Equatin 14.11, and the data fr the standard entrpies f the varius reactants and prducts in Appendix 7. (a) Using Equatin 14.11 Δ r S 298 = ν i S 298 (prducts) ν i S 298 (reactants) where the terms in i represent the stichimetric cefficients f the prducts and reactants. Using data frm Appendix 7, then Δ r S 298 = 2 S 298 (NH 3 (g)) {S 298 (N 2 (s)) + 3 S 298 (H 2 (g))} = 2 192.5 J K 1 ml 1 {191.6 J K 1 ml 1 + 3 130.7 J K 1 ml 1 }
(b) In the same way, = 198.7 J K 1 ml 1 Δ r S 298 = S 298 (HgCl 2 (s)) {S 298 (Hg(l)) + S 298 (Cl 2 (g))} = 146.0 J K 1 ml 1 {76.0 J K 1 ml 1 + 223.1 J K 1 ml 1 } = 153.1 J K 1 ml 1 (c) Δ r S 298 = {6 S 298 (CO 2 (g)) + 6 S 298 (H 2 O(g))} {S 298 (C 6 H 12 O 6 (s)) + 6 S 298 (O 2 (g))} = {6 213.7 J K 1 ml 1 + 6 188.8 J K 1 ml 1 } = +972.4 J K 1 ml 1 {212.0 J K 1 ml 1 + 6 205.1 J K 1 ml 1 } 14. Calculate the entrpy changes fr the system and surrundings when 1.00 ml f NaCl melts at 1100 K. Calculate fusg and estimate the melting pint f NaCl. (Fr the melting f sdium chlride (NaCl) fush = +30.2 kj ml 1 and fuss = +28.1 J K 1 ml 1.) Use the value fr the standard entrpy f fusin t determine the entrpy change fr the system and the surrundings.. Use Equatin 14.16 t calculate the change in Gibbs energy. Rearrange Equatin 14.16 and use it t determine the melting temperature at which the slid and liquid are in equilibrium and the Gibbs energy change is 0. Standard mlar entrpy changes refer t the system rather than the surrundings. Thus, ΔS system = Δ fus S = +28.1 J K 1 ml 1 Fr the surrundings, it is necessary t determine the entrpy change frm the Secnd Law f Thermdynamics, ΔS surrundings = q rev T = ΔH system T = 30.2 103 J ml 1 1100 K = 27.5 J K 1 ml 1
where qrev refers t the heat transferred frm the surrundings t the system. The ttal entrpy change at this temperature is therefre ΔS ttal = ΔS system + ΔS surrundings = +28.1 J K 1 ml 1 27.5 J K 1 ml 1 = +0.6 J K 1 ml 1 The ttal entrpy change f system and surrundings is therefre psitive, and s melting must be spntaneus at 1100 K. Using Equatin 14.16, ΔG = ΔH TΔS = +30.2 10 3 J ml 1 1100 K +28.1J K 1 ml 1 = 710 J ml 1 The Gibbs energy change is therefre negative, which is cnsistent with the bservatin that melting must be spntaneus at this temperature. At the melting pint, and s, using Equatin 14.16 then ΔG = 0 Δ fus G = Δ fus H T m Δ fus S = 0 T m = Δ fush Δ fus S = +30.2 103 J ml 1 +28.1 J K 1 = 1075 K ml 1 15. Calcium carbnate (CaCO3) decmpses t frm CaO and CO2 with rh298 = +178 kj ml 1 and rs298 = +161 J K 1 ml 1. Estimate the temperature at which the decmpsitin becmes spntaneus. Rearrange Equatin 14.16 and use it t determine the temperature at which the Gibbs energy f reactin is 0.
Assuming that rh and rs remain cnstant, the reactin becmes spntaneus at the temperature when the value f rg changes frm psitive t negative i.e. when rg = 0. Thus, using Equatin 14.16 then Δ r G = Δ r H TΔ r S = 0 T = Δ rh Δ r S = +178 103 J ml 1 +161 J K 1 = 1106 K ml 1 16. Yu expend abut 100 kj a day keeping yur heart beating. What is the minimum mass f glucse yu must xidise per day in rder t prduce this much energy? C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6 H2O (l) G = 2872 kj ml 1 Determine the amunt f glucse that is required t prduce 100 kj f energy as a prprtin f the mlar change in Gibbs energy. Calculate the mlar mass f glucse and hence determine the mass t which this amunt crrespnds. If the mlar change in Gibbs energy f reactin is 2872 kj ml 1, then 100 kj must be prvided by The mlar mass f glucse is Thus, 100 kj = 0.035 ml 2872 kj ml 1 M = {(6 12.01) + (12 1.01) + (6 16.00)} g ml 1 = 180.18 g ml 1 m = n M = 0.035 ml 180.18 g ml 1 = 6.27 g
17. Fr the reactin: CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ch 298 = 890 kj ml 1. (a) Calculate cg 298 fr the cmbustin f methane. (b) Hw much f the heat prduced by burning 1.00 ml f methane cannt be used t d wrk? (c) A heat engine uses methane as a fuel. What height culd a 1.00 kg mass be raised t by burning 1.00 dm 3 f methane? (a) Use Equatin 14.16 t determine the Gibbs energy change frm the enthalpy and entrpy f cmbustin. The value fr the enthalpy f cmbustin is given, but it is necessary t use Equatin 14.11 and the data in Appendix 7 t determine the entrpy change. (b) Calculate the energy that is nt available fr wrk as the difference between the Gibbs energy and enthalpy changes. (c) The Gibbs energy change is equivalent t the amunt f energy that is available fr nn-expansin wrk. Determine the amunt f methane that crrespnds t a vlume f 1 dm 3 frm the ideal gas equatin, and hence calculate the Gibbs energy change fr this vlume. Use Equatin 1.11 t determine the height thrugh which a mass may be lifted by ding this amunt f wrk. (a) Using Equatin 14.16, Δ c G = Δ c H TΔ c S The value fr the standard enthalpy change is given, but in rder t determine the entrpy change, then it is necessary t use Equatin 14.11 Δ r S 298 = ν i S 298 (prducts) ν i S 298 (reactants) where the terms in i represent the stichimetric cefficients f the prducts and reactants. Using data frm Appendix 7, then Δ r S 298 = {S 298 (CO 2 (s)) + 2 S 298 (H 2 O(l))} {S 298 (CH 4 (g)) + 2 S 298 (O 2 (g))}
Thus, = {213.7 J K 1 ml 1 + 2 69.9 J K 1 ml 1 } = 243.0 J K 1 ml 1 {186.3J K 1 ml 1 + 2 205.1 J K 1 ml 1 } Δ c G = Δ c H TΔ c S = 890 10 3 J ml 1 298 K 243.0 J K 1 ml 1 = 818 10 3 J ml 1 = 818 kj ml 1 (b) The Gibbs energy change represents the maximum amunt f nn-expansin wrk that can be dne. The amunt f energy that cannt be used t d wrk is therefre the difference between the enthalpy and Gibbs energy changes. = 890kJ ml 1 ( 818 kj ml 1 ) = 72 kj ml 1 (c) Rearranging the ideal gas equatin, Equatin 8.5, then a vlume f 1 dm 3 10 3 m 3 is equivalent t an amunt n = pv RT = 101325 Pa 1.00 10 3 m 3 8.3145 J K 1 ml 1 = 0.041 ml 298 K Burning this amunt f methane wuld result in a Gibbs energy change, ΔG = n Δ c G = 0.041 ml 818 kj ml 1 = 33.5 kj = 33.5 10 3 J This energy is available as wrk t lift the mass. w = ΔG The ptential energy f an bject and thus the wrk that must be dne t raise the bject t that psitin, is given by Equatin 1.16, Rearranging, w = ΔG = E PE = mgh h = ΔG/mg = 33.5 10 3 J/1.00 kg 9.81 m s 2 = 3.42 10 3 m = 3.42 km 18. Ethanic acid can be prduced by a number f methds, including (a) the reactin f methanl with carbn mnxide, CH3OH (l) + CO (g) CH3COOH (l)
(b) the xidatin f ethanl with xygen gas, CH3CH2OH (l) + O2 (g) CH3COOH (l) + H2O (l) (c) the reactin f carbn dixide with methane, CO2 (g) + CH4 (g) CH3COOH (l) Calculate the standard Gibbs energy change fr each f these reactins at 298 K and 773 K. Use Equatin 13.6 and 14.11 t determine the standard enthalpy and entrpy changes fr the reactins at 298 K using the data given in Appendix 7. Als calculate the change in the heat capacity fr the reactins using Equatin 13.11 using the data in the Appendix and use Equatins 13.10 and 14.12 t determine the standard enthalpy and entrpy changes fr the reactins at 773 K. Determine the Gibbs energy change at bth temperatures using Equatin 14.16. (a) Using Equatin 13.6 Δ r H 298 = ν i Δ f H 298 (prducts) ν i Δ f H 298 (reactants) = Δ f H 298 (CH 3 COOH(l)) {Δ f H 298 (CH 3 OH(g)) + Δ f H 298 (CO(g))} = 484.5 kj { 238.4 kj ml 1 + ( 110.5 kj ml 1 )} = 135.6 kj ml 1 and using Equatin 14.11 Δ r S 298 = ν i S 298 (prducts) ν i S 298 (reactants) Δ r S 298 = S 298 (CH 3 COOH(l)) {S 298 (CH 3 OH(g)) + S 298 (CO(g))} = 158 J K 1 ml 1 {126.8 J K 1 ml 1 197.7 J K 1 ml 1 } = 166.5 J K 1 ml 1 Cmbining the tw values using Equatin 14.16 t calculate rg 298, Δ r G 298 = Δ r H 298 TΔ r S 298 = 135.6 10 3 J ml 1 298 K 166.5 J K 1 ml 1 = 86.0 10 3 J ml 1 = 86.0 kj ml 1
Calculating Cp frm Equatin 13.11 t cnvert H 298 and S 298 t the higher temperature, Δ r C p = ν i C p (prducts) ν i C p (reactants) Δ r C p = C p (CH 3 COOH(l)) {C p (CH 3 OH(g)) + C p (CO(g))} = 123.1 J K 1 ml 1 {81.6 J K 1 ml 1 + 29.1 J K 1 ml 1 } = +12.4 J K 1 ml 1 Using Equatin 13.10 t cnvert rh 298 t the higher temperature Δ r H T2 = Δ r H T1 + ΔC p (T 2 T 1 ) s that Using Equatin 14.12 Δ r H 773 = Δ r H 298 + ΔC p (773 K 298 K) = 135.6 10 3 J ml 1 + 12.4 J K 1 ml 1 475 K = 130 10 3 J ml 1 = 130 kj ml 1 ΔS T2 = ΔS T1 + ΔC p ln(t 2 /T 1 ) then ΔS 650 = ΔS 298 + ΔC p ln(773 K/298 K) = 166.5 J K 1 ml 1 + 12.4 J K 1 ml 1 ln(773 K/298 K) = 154.7 J K 1 ml 1 s that Δ r G 773 = Δ r H 773 TΔ r S 773 = 129.7 10 3 J ml 1 773 K 154.7 J K 1 ml 1 = 10.1 10 3 J ml 1 = 10.1 kj ml 1 (b) In the same way fr the reactin using Equatin 13.6 CH3CH2OH (l) + O2 (g) CH3COOH (l) + H2O (l) Δ r H 298 = ν i Δ f H 298 (prducts) ν i Δ f H 298 (reactants)
and using Equatin 14.11 = {Δ f H 298 (CH 3 COOH(l)) + Δ f H 298 (H 2 O(l))} {Δ f H 298 (CH 3 CH 2 OH(g)) + Δ f H 298 (O 2 (g))} = { 484.5 kj ml 1 285.8 kj ml 1 } { 277.64 kj ml 1 0} = 492.7 kj ml 1 Δ r S 298 = ν i S 298 (prducts) ν i S 298 = {S 298 (CH 3 COOH(l)) + S 298 (H 2 O(l))} {S 298 (CH 3 CH 2 OH(g)) + S 298 (O 2 (g))} = {158 J K 1 ml 1 + 69.9J K 1 ml 1 } {159.9 J K 1 ml 1 + 205.1 J K 1 ml 1 } = 137.1 J K 1 ml 1 (reactants) Cmbining the tw values using Equatin 14.16 t calculate rg 298, Δ r G 298 = Δ r H 298 TΔ r S 298 = 492.7 10 3 J ml 1 298 K 137.1 J K 1 ml 1 = 452 10 3 J ml 1 = 452.0 kj ml 1 Calculating Cp frm Equatin 13.11 t cnvert H 298 and S 298 t the higher temperature, Δ r C p = ν i C p (prducts) ν i C p (reactants) = {C p (CH 3 COOH(l)) + C p (H 2 O(l))} {C p (CH 3 CH 2 OH(g)) + C p (O 2 (g))} = {123.1 J K 1 ml 1 + 75.3 J K 1 ml 1 } {111.5 J K 1 ml 1 + 29.4 J K 1 ml 1 } = +57.5 J K 1 ml 1 Using Equatin 13.10 t cnvert rh 298 t the higher temperature Δ r H T2 = Δ r H T1 + ΔC p (T 2 T 1 ) s that
Using Equatin 14.12 Δ r H 773 = Δ r H 298 + ΔC p (773 K 298 K) = 492.7 10 3 J ml 1 + 57.1 J K 1 ml 1 475 K = 465 10 3 J ml 1 = 465 kj ml 1 ΔS T2 = ΔS T1 + ΔC p ln(t 2 /T 1 ) then ΔS 650 = ΔS 298 + ΔC p ln(773 K/298 K) = 137.1 J K 1 ml 1 + 57.1 J K 1 ml 1 ln(773 K/298 K) = 82.7 J K 1 ml 1 s that Δ r G 773 = Δ r H 773 TΔ r S 773 (c) And fr the reactin using Equatin 13.6 = 465 10 3 J ml 1 773 K 82.7 J K 1 ml 1 = 401 10 3 J ml 1 = 401 kj ml 1 CO2 (g) + CH4 (g) CH3COOH (l) Δ r H 298 = ν i Δ f H 298 (prducts) ν i Δ f H 298 (reactants) = Δ f H 298 (CH 3 COOH(l)) {Δ f H 298 (CO 2 (g)) + Δ f H 298 (CH 4 (g))} = 484.5 kj ml 1 { 74.8 kj ml 1 393.5 kj ml 1 } = 16.2 kj ml 1 and using Equatin 14.11 Δ r S 298 = ν i S 298 (prducts) ν i S 298 (reactants) = S 298 (CH 3 COOH(l)) {S 298 (CO 2 (g)) + S 298 (CH 4 (g))} = 158.0 J K 1 ml 1 {186.3 J K 1 ml 1 + 213.7 J K 1 ml 1 } = 242.0 J K 1 ml 1 Cmbining the tw values using Equatin 14.16 t calculate rg 298,
Δ r G 298 = Δ r H 298 TΔ r S 298 = 16.2 10 3 J ml 1 298 K 242.0 J K 1 ml 1 = +55.9 10 3 J ml 1 = +55.9 kj ml 1 Calculating Cp frm Equatin 13.11 t cnvert H 298 and S 298 t the higher temperature, Δ r C p = ν i C p (prducts) ν i C p (reactants) = S 298 (CH 3 COOH(l)) {S 298 (CO 2 (g)) + S 298 (CH 4 (g))} = 123.1 J K 1 ml 1 {35.7 J K 1 ml 1 + 37.1 J K 1 ml 1 } = +50.3 J K 1 ml 1 Using Equatin 13.10 t cnvert rh 298 t the higher temperature Δ r H T2 = Δ r H T1 + ΔC p (T 2 T 1 ) s that Using Equatin 14.12 Δ r H 773 = Δ r H 298 + ΔC p (773 K 298 K) = 16.2 10 3 J ml 1 + 50.3 J K 1 ml 1 475 K = +7.7 10 3 J ml 1 = +7.7 kj ml 1 ΔS T2 = ΔS T1 + ΔC p ln(t 2 /T 1 ) then ΔS 650 = ΔS 298 + ΔC p ln(773 K/298 K) = 242.0 J K 1 ml 1 + 50.3 J K 1 ml 1 ln(773 K/298 K) = 194.1 J K 1 ml 1 s that Δ r G 773 = Δ r H 773 TΔ r S 773 = 7.7 10 3 J ml 1 773 K 194.1 J K 1 ml 1 = +158 10 3 J ml 1 = +158 kj ml 1
19. The reactin f methanl, (CH3OH (l)), with xygen can be used in a fuel cell. Using data frm Appendix 7 (p.1350), calculate the enthalpy change fr the cmbustin reactin at 298 K and the maximum wrk that can be prduced by the xidatin f 1.00 ml f methanl. Write a balanced chemical equatin fr the cmbustin f methanl. Use it, alng with Equatins 13.6, 14.11 and 14.16 t determine the standard enthalpy, entrpy and hence Gibbs energy change fr cmbustin. The maximum nnexpansin wrk that may be perfrmed is equivalent t the Gibbs energy change. The chemical equatin fr the cmbustin f methanl is CH3OH (l) + 1½ O2 (g) CO2 (g) + H2O (l) Using Equatin 13.6 t calculate rh 298 Δ r H 298 = ν i Δ f H 298 (prducts) ν i Δ f H 298 (reactants) = {Δ f H 298 (CO 2 (g)) + Δ f H 298 (H 2 O(g))} {Δ f H 298 (CH 3 OH(g)) + 1.5 Δ f H 298 (O 2 (g))} = { 393.5 kj ml 1 + ( 285.8 kj ml 1 )} { 238.4 kj ml 1 0} = 440.9 kj ml 1 and using Equatin 14.11 Δ r S 298 = ν i S 298 (prducts) ν i S 298 Δ r S 298 = {S 298 (CO 2 (g)) + S 298 (H 2 O(g))} (reactants) {S 298 (CH 3 OH(g)) + 1.5 S 298 (O 2 (g))} = {213.7 J K 1 ml 1 + 69.9 J K 1 ml 1 } {126.8 J K 1 ml 1 + (1.5 205.1 J K 1 ml 1 )} = 150.9 J K 1 ml 1 Cmbining the tw values using Equatin 14.16 t calculate rg 298, Δ r G 298 = Δ r H 298 TΔ r S 298 = 440.9 10 3 J ml 1 298 K 150.9 J K 1 ml 1
= 396 10 3 J ml 1 = 396 kj ml 1 The maximum wrk that can be dne by cnsuming 1 ml f methanl is thus w = ΔG = 396 kj ml 1 20. Calculate the nrmal biling pint f ethanl given that vaph = +42.6 kj ml -1 and vaps = +122.0 J K -1 ml -1. Use the fact that, at the nrmal biling pint, the liquid and vapur are in equilibrium s that vapg = 0. G = H - T S vapg = vaph - Tb vaps vapg = 0 = vaph - Tb vaps T b vap vap H S 3 42.6 10 J ml -1-1 122 J K ml -1 = 349.2 K 21. Synthesis gas is a mixture f hydrgen and carbn mnxide, prepared by reacting water r steam with a surce f carbn such as cal. The reactin can be represented as: C(s) + H2O(g) CO(g) + H2(g) Estimate the temperature at which the reactin becme thermdynamically spntaneus.
The reactin just becmes spntaneus when rg = 0. T find rg, we need rh and rs. These can be calculated frm tabulated values fr standard enthalpy change f frmatin, fh 298, and standard entrpy, S 298. H = H (prducts) H (reactants) r 298 i f 298 i f 298 rh 298 = [0 + (-110.5 kj ml -1 )]-[(-241.8 kj ml -1 ) + (0)] = +131.3 kj ml -1 S = S (prducts) S (reactants) r 298 i 298 i 298 rs 298 = [197.7 J K -1 ml -1 + 130.7 J K -1 ml -1 ] - [5.7 J K -1 ml -1 + 188.8 J K -1 ml -1 ] = +133.9 J K -1 ml -1. rg rh 298 - T rs 298 0 = + 131.3 kj ml -1 - T (+133.9 J K -1 ml -1 ) T = 3-1 131.3 10 J ml 133. 9 J K ml -1-1 = 980.6 K This assumes that rh and rs d nt change with temperature and s is nly an estimate. 22. Use the mean heat capacities and standard entrpies listed in Appendix 7 t estimate the Gibbs energy change fr the decmpsitin f water vapur int hydrgen and xygen at 2000 C and 1 bar pressure. (The standard enthalpy f frmatin f water vapur is 241.8 kj ml 1.) Deduce the enthalpy change fr the decmpsitin reactin at 298 K frm the standard enthalpy f frmatin. Calculate the entrpy change at this temperature using Equatin 14.12. Use Equatin 13.11 t calculate the change in the heat capacity and use this value, alng with Equatin 13.10 and Equatin 14.11 t determine the enthalpy and entrpy changes at 2000 C. Cmbine the