BEIJING UNIVERSITY OF CHEMICAL TECHNOLOGY 第，、章镜计热力学和步美文司数及参考答素 this temperature and 1 atm pressure). m=1007852x10kgmo1=1673559×10-”ke 6.022045×1025mo -y7 -[2红L67359x109g038062x10J-KX30K™x02461705m (6.626176×10Js)7 =7.69115×100 Note that the molecular partition function is a dimensionless quantity and may be interpreted approximately as the number of energy levels accessible to the single hydrogen atom in this volume 唱cw2” For H(g) J厂20.67359x10-”kg1.38062x10-JKX300K)] (6.626176×10-4Js)2 ×1380662×10-KX300K=1636274 101325Nm2 =&31K.mr得+16362g=1368W.Kmr G=-(8.314J.K-.mol-)3000K16.36274)=-408.160 kJ.mol- For Ha(g) s=831Knm得+174024时-16s4Ka TEL010-64434903 1
BEIJING UNIVERSITY OF CHEMICAL TECHNOLOGY TEL:010-64434903 1 第八章 统计热力学初步英文习题及参考答案 1.Calculate the molecular partition function for translational motion for a hydrogen atom at 3000 K in a volume of 0.2461705m3 (This is the volume that would contain 1 mol of perfect gas at this temperature and 1 atm pressure). 1.673559 10 kg 6.022045 10 mol 1.007852 10 kg mol 27 23 1 3 1 - - - - = ´ ´ ´ × m = V h mkT qt 3 / 2 2 2 ÷ ø ö ç è æ = p (0.2461705 ) (6.626176 10 Js) 2 (1.673559 10 kg)(1.380662 10 J K )(300K) 3 3/ 2 34 2 27 23 1 ú ´ m û ù ê ë é ´ ´ ´ × = - - - - p =7.69115×1030 Note that the molecular partition function is a dimensionless quantity and may be interpreted approximately as the number of energy levels accessible to the single hydrogen atom in this volume. 2.Calculate the translational contributions to S°and G°for H(g) and H2(g) at 3000K. ïþ ï ý ü ïî ï í ì ú û ù ê ë é = + θ 3 / 2 2 0 2 ln 2 5 p kT h mkT St R p ， ú ú û ù ê ê ë é ÷ ø ö ç è æ = - θ 3 / 2 2 0 2 ln p kT h mkT Gt RT p For H(g) ú ú û ù ê ê ë é ÷ ø ö ç è æ θ 3 / 2 2 2 ln p kT h pmkT 16.36274 101325Nm (1.380662 10 JK )(3000K) (6.626176 10 Js) 2 (1.673559 10 kg)(1.380662 10 JK )(3000K) ln -2 23 -1 3 / 2 34 2 27 23 -1 = þ ý ´ ü ´ ïî ï í ì ú û ù ê ë é ´ ´ ´ = - - - - p θ 1 1 1 1 16.36274 156.84 K mol 2 5 (8.314J K mol ) - - - - ÷ = × × ø ö ç è æ S = × × + J t ( ) θ 1 1 1 (8.314J K mol ) 3000 (16.36274) 408.160kJ mol - - - G = - × × K = - × t For H2(g) ú ú û ù ê ê ë é ÷ ø ö ç è æ θ 3/ 2 2 2 ln p kT h pmkT =17.40246 θ 1 1 1 1 17.40246 165.485J K mol 2 5 (8.314J K mol ) - - - - ÷ = × × ø ö ç è æ St = × × +
BEIJING UNIVERSITY OF CHEMICAL TECHNOLOGY G=-8.314J.K-4.mol-)3000K17.40246)=-434.073kJ-mol 3.Calculate the Characteristic rotational temperatureand the rotational partition q.for H:(g)at 3000K (The moment of inertia is 4.6033 10kgm) 2 (6.62618×10J-S2 0,8rm8x4603x100gmX138066×104-k可87494K 3000K o6208749417144 4.Calculate the characteristic vibrational temperatureand the vibrational partition function for H(g)at 3000K.The fundamental vibrational frequency is4405.3cm 0.=hcm -6.626176x10JsX2.97925x10msX4053cmX102cmm 1.380662×10-2JK =6338.3K 5.Calculate the molar S'and G'at I atm and 3000K for Ha using the translational conributions calculated inquestion16and the rotational,vibrational,and electrncnbuions shown in Table. Table:Rotational,Vibrational,and Electronic Contributions to Molar Thermodynamic Quantities for Perfect Diatomic Gases Rotation Vibration Electronic U RT -N,D9 Ho RT T -N Do R-xer (e-) 0 货 0 So g 风e-e可] RIngo -m到 RTIn(1-e-) -N,D°-RTIngo TEL:010-64434903 2
BEIJING UNIVERSITY OF CHEMICAL TECHNOLOGY TEL:010-64434903 2 ( ) θ 1 1 1 (8.314J K mol ) 3000 (17.40246) 434.073kJ mol - - - G = - × × K = - × t 3.Calculate the Characteristic rotational temperature Θr and the rotational partition fumcfion qr for H2(g) at 3000K (The moment of inertia is 4.6033×10-48kgm2 ) 87.494K 8 (4.6033 10 kg m )(1.38066 10 J ) (6.62618 10 J S) 8 2 48 2 23 1 34 2 2 2 = ´ × ´ × ´ × Q = = - - - - Ik K h r p p 17.144 2(87.494 ) 3000 = = Q = K T K q r r s 4.Calculate the characteristic vibrational temperatureΘv and the vibrational partition function qv for H2(g) at 3000K. The fundamental vibrational frequency is 4405.3cm -1 . k hc v v Q = = 23 -1 34 8 -1 -1 2 -1 1.380662 10 JK (6.626176 10 Js)(2.997925 10 ms )(4405.3cm )(10 cm m ) - - ´ ´ ´ × =6338.3K 1.13753 1 1 1 1 / 6338.3/ 3000 = - = - = -Q - e e qv v T 5.Calculate the molar S°and G°at 1 atm and 3000K for H2 using the translational contributions calculated in question 16 and the rotational, vibrational, and electronic contributions shown in Table. Since the characteristic rotational temperature Θr calculated in question 17 is 87.494K. Table： Rotational, Vibrational, and Electronic Contributions to Molar Thermodynamic Quantities for Perfect Diatomic Gases a Rotation Vibration Electronicb 0 U RT -1 x e x RT 0 - NAD 0 H RT -1 x e x RT 0 - NAD 0 CV R 2 2 ( -1) x x e x e R 0 0 CP R 2 2 ( -1) x x e x e R 0 0 S ÷ ÷ ø ö ç ç è æ Qr eT R s ln ú û ù ê ë é - - - - ln(1 ) 1 x x e e x R 0 Rln g 0 A ÷ ÷ ø ö ç ç è æ Q - r T RT s ln ln(1 ) x RT e - - 0 0 N D RT ln g - A -
BEIJING UNIVERSITY OF CHEMICAL TECHNOLOGY G RTIn(1-e-) -N,D°-RTIngo =/andx=K=/T.D Is the spectroscopic dissociation energy? The ND terms are included only if the objective of the calculation is to compare H,orwith the values for the constituent atoms 828300 K=31.942.K1.mol 2(87.4943K c-kT石-31 u-K-.X0K)恤72-nur Since the characteristic vibrational temperature calculated in Example 13.is6338.3K s=lb0-e-8314.K.moX0291+0129=348K-mo G°=RT1n1-e)=(8.314J.K-1mol-'3000K)n1-e2)=-3.214kJ-mol1 The electronic ground state for H(g)is nondegenerate and the spectroscopic dissociation is 4.47797eV.Therefore,there is no electronic contribution tos and the contribution to Gis simply. G=-N,D°=-432.077kmol- Thus the total molar entropy of H(g)at 3000K and 1atm is S=S9+S+S9+S8=200.92J-K1mol1 This may be compared with the value 202.782J.Kmol in Table A.2.The difference between these values arises from the fact that we have not included anharmonicity of vibrational The total molar is G°=G+G+G8+G8=-940.2699·mol 6.Using the results of questions 16andI7 calculate△s°，△G°，Kand degree of dissociation for Ha(g)=2H(g)at 3000K △°=2S(H-S(H,)=2162.59)-200.92-K-1.mol-=124.280J.K-1.mol- △G°=2G(H0+G(H2)=[2(-425.439)+940.269]W-mol-=89.368 J.mol- This is twice the standard Gibbs energy of formation of H(g)so we have calculated AG=44.686 kJ.mol for H(g)at 3000K.According to Table A.2.AG for H(g)is 46.171kJmol.Approximations in our calculation of S for Ha(g)have been pointed out in TEL010-64434903 3
BEIJING UNIVERSITY OF CHEMICAL TECHNOLOGY TEL:010-64434903 3 0 G ÷ ÷ ø ö ç ç è æ Q - r T RT s ln ln(1 ) x RT e - - 0 0 N D RT ln g - A - a h Ik r / 2 2 Q = and x = hv / KT = Qv /T . 0 D Is the spectroscopic dissociation energy? b The 0 NAD terms are included only if the objective of the calculation is to compare 0 U ， 0 H ， 0 A or 0 G with the values for the constituent atoms. 0 1 1 1 1 31.942J K mol 2(87.4943K) (2.71828)(3000K) ln (8.314J K mol )ln - - - - = × × = × × Q = r r eT S R s 0 1 1 1 70.885kJ mol 2(87.4943K) (3000K) ln (8.314J K mol )(3000K)ln - - - = × × = - × Q = - r r T G RT s Since the characteristic vibrational temperature Qv calculated in Example 13.5 is 6338.3K 0 1 1 1 1 ln(1 ) (8.314J K mol )(0.291 0.129) 3.489J K mol 1 - - - - - = × × + = × × ú û ù ê ë é - - - = x v x e e x S R 0 1 1 2.113 1 ln(1 ) (8.314J K mol )(3000K)ln(1 ) 3.214kJ mol - - - - - G = RT - e = × × - e = - × x v The electronic ground state for H2(g) is nondegenerate and the spectroscopic dissociation is 4.47797 eV. Therefore, there is no electronic contribution to 0 S and the contribution to 0 G is simply. 0 0 1 432.077kJ mol - G = -N D = - × e A Thus the total molar entropy of H2(g) at 3000K and 1 atm is 0 0 0 0 0 1 1 200.92J K mol - - = + + + = × × t r v e S S S S S This may be compared with the value 202.782J·K -1·mol-1 in Table A.2. The difference between these values arises from the fact that we have not included anharmonicity of vibrational motion and vibration-rotation interaction. The total molar Gibbs energy is 0 0 0 0 0 1 940.2699 J mol - G = G + G + G + G = - k × t r v e 6.Using the results of questions 16 and 17 calculate ΔS°, ΔG°,Kp, and degree of dissociation for H2(g) == 2H(g) at 3000K. 1 1 2 0 0 0 2 (H) (H ) [2(162.59) 200.92]J K mol - - DS = S - S = - × × 1 1 124.280J K mol - - = × × 1 2 0 0 0 2 (H) (H ) [2( 425.439) 940.269] J mol - DG = G + G = - + k × 1 89.368 J mol- = k × This is twice the standard Gibbs energy of formation of H(g), so we have calculated 0 DGf =44.686 kJ·mol-1 for H(g) at 3000K. According to Table A.2, q DGf for H(g) is 46.171kJ·mol-1 . Approximations in our calculation of 0 S for H2(g) have been pointed out in
BEIJING UNIVERSITY OF CHEMICAL TECHNOLOGY Example 13.6. The value of AG that we have calculated in the series of examples yieldsK=2.79x10 This yields a degree of dissociation of 0.084 at I atm,compared with 0783 calculated from Table A.2 in Example 4.4. 7.Calculate the molar heat capacity C of water vapor at 1000K,given the frequencies of the three normal modes of vibration,1595.3652,and 3756cm Translation comtributesR To Cand assuming ideal gas behavior C=C.+R. Since three angles are required to describe the orientation of a water molecule,there are three degesof ion rdmhatc(Sti1.)Theofthefirst normal mode is given by the following expression from Table 13.3. R31aKn22g R-xer =5.460J-K-1.mol (e224-1)2 where =2=(6.626x10"Js2998x10'msX1595cm100cm:m-22 (1.381x10J-K-X1000K The other two normal modes contribute 1.213 and 1.100 J.Kmol The sum of these contributions is41.03J·K·molr',compared with41l.218J·K·mofr'from Table A.2.In the statistical mechanical calculation we have assumed that water vapor is a perfect gas,and this is not quite true. TEL:010-64434903
BEIJING UNIVERSITY OF CHEMICAL TECHNOLOGY TEL:010-64434903 4 Example 13.6. The value of 0 DG that we have calculated in the series of examples yields K p =2.79×10-2 . This yields a degree of dissociation of 0.084 at 1 atm, compared with 0.0783 calculated from Table A.2 in Example 4.4. 7.Calculate the molar heat capacity Cp of water vapor at 1000K, given the frequencies of the three normal modes of vibration, 1595,3652, and 3756cm -1 . Translation comtributes R 2 5 To Cp , and assuming ideal gas behavior Cp = Cv + R . Since three angles are required to describe the orientation of a water molecule, there are three degrees of rotational freedom that contribute R 2 3 (Section 1.12). The contribution of the first normal mode is given by the following expression from Table 13.3. 1 1 2.294 2 1 1 2 2.294 2 2 5.460J K mol ( 1) (8.314J K mol )(2.294) ( 1) - - - - = × × - × × = - e e e x e R x x where (1.381 10 J K )(1000 ) (6.626 10 J s)(2.998 10 m s )(1595cm )(100cm m ) 23 1 34 8 -1 1 1 kT K hc x - - - - - ´ × ´ × ´ × × = = v =2.295 The other two normal modes contribute 1.213 and 1.100 J·K -1·mol-1 . The sum of these contributions is 41.03 J·K -1·mol-1 , compared with 41.218J·K -1·mol-1 from Table A.2. In the statistical mechanical calculation we have assumed that water vapor is a perfect gas, and this is not quite true
