基礎(chǔ)化學(xué)第六章后習(xí)題解答

1、第六章后習(xí)題解答 習(xí)題1.略2. 計(jì)算以下系統(tǒng)內(nèi)能的變化：(1)系統(tǒng)放出2.5kJ的熱量，并且對(duì)環(huán)境作功500J。(2)系統(tǒng)放出650J的熱量，環(huán)境對(duì)系統(tǒng)作功350J。解 (1) U = Q+W = -2.5kJ+(-50010-3kJ)=-3.0kJ(2) U=-650J+350J=-300J3.略4. 已知反應(yīng)：A+B=C+D =-40.0kJmol-1C+D=E =60.0kJmol-1求以下各反應(yīng)的。(1)C+D=A+B(2)2C+2D=2A+2B(3)A+B=E解 (1)C+D=A+B =- =40.0kJmol-1(2)2C+2D=2A+2B =2 (-)=80.0kJmol-1(

2、3)A+B=E =+=-40.0 kJmol-1+60.0 kJmol-1=20.0 kJmol-15.在一定溫度下，4.0molH2(g)與2.0molO2(g)混合，經(jīng)一定時(shí)間反應(yīng)后，生成了0.6molH2O(g)，請(qǐng)按以下兩個(gè)不同反應(yīng)式計(jì)算反應(yīng)進(jìn)度。(1)2H2(g)+O2(g)=2H2O(g)(2)H2(g)+O2(g)=H2O(g)解 t=0 =0 n(H2)=4.0mol n(O2)=2.0mol n(H2O)=0 t=t =t n(H2)=3.4mol n(O2)=1.7mol n(H2O)=0.6mol按(1)式： 2H2(g)+O2(g)=2H2O(g)= = =0.30mo

3、l= = =0.30mol= = =0.30mol按(2)式： H2(g)+O2(g)= H2O(g)= = =0.60mol= = =0.60mol= =0.60mol6.略7. 已知以下反應(yīng)的標(biāo)準(zhǔn)反應(yīng)熱：(1)C6H6(l)+7O2(g)=6CO2(g)+3H2O(l) =-3267.6kJmol-1(2)C(gra)+O2(g)=CO2(g) =-393.5kJmol-1(3)H2(g)+O2(g)=H2O(l) =-285.8kJmol-1求下述不直接發(fā)生反應(yīng)的標(biāo)準(zhǔn)反應(yīng)熱：解 6C(gra)+3H2(g)=C6H6(l) =？由62+331得所求的反應(yīng) 6C(g)+3H2 (g)=C6

4、H6(l)=63=6(-393.5) kJmol-13(-285.8) kJmol-1(-3267.6)kJmol-1= 49.2kJmol-18. 肼N2H4(l)是火箭的燃料，N2O4作氧化劑，其燃燒反應(yīng)的產(chǎn)物為N2(g)和H2O(l)，假設(shè)(N2H4，l)=50.63kJmol-1，(N2O4，g)=9.16kJmol-1寫出燃燒反應(yīng)，并計(jì)算此反應(yīng)的反應(yīng)熱。 解 2N2H4(l)+N2O4(g)=3N2(g)+4H2O(l)=0+4(-285.8 kJmol-1)-2(50.63 kJmol-1)+9.16 kJmol-1-1254 kJmol-19已知以下反應(yīng)在298.15K,標(biāo)準(zhǔn)狀態(tài)

5、下：(1) Fe2O3(s)+ 3CO(g) 2Fe(s) + 3CO2(g)； rH=-24.8kJmol-1，rG=- 29.4kJmol-1(2) 3 Fe2O3(s)+CO(g) 2Fe3O4(s) + CO2(g)；rH=-47.2kJmol-1，rG=-61.41kJmol-1(3)Fe3O4(s) + CO(g)3FeO(s)+CO2(g)；rH=19.4kJmol-1， rG=5.21kJmol-1試求(4) FeO(s) + CO(g) Fe(s)+ CO2(g)的r H、 rG和rS。解 反應(yīng)式4可由反應(yīng)式1、2和3組合求出：-在298.15K,標(biāo)準(zhǔn)狀態(tài)下，反應(yīng)： FeO(

6、s) + CO(g) Fe(s)+CO2(g)的r H、 rG和rS分別為 r H=-rH+- =-19.4kJmol-1 + =-11.0 kJmol-1 =- =-6.20 kJmol-1 =-6.20JK-1 mol-110. 甲醇的分解反應(yīng)為：CH3OH(l)CH4(g)+O2(g)(1)在298.15K的標(biāo)準(zhǔn)狀態(tài)下此反應(yīng)能否自發(fā)進(jìn)行？(2)在標(biāo)準(zhǔn)態(tài)下此反應(yīng)的溫度應(yīng)高于多少才能自發(fā)進(jìn)行？解 (1) =-74.6 kJmol-1+0- (-239.2 kJmol-1)=164.6kJmol-1=186.3JK-1mol-1+205.2 JK-1mol-1-126.8 JK-1mol-1=

7、162.1JK-1mol-1=-T=164.6kJmol-1- 298.15K162.1JK-1mol-1=116.3 kJmol-10或=-50.5kJmol-1+0- (-166.6 kJmol-1)=116.3kJmol-1025和標(biāo)準(zhǔn)狀態(tài)下反應(yīng)不能自發(fā)進(jìn)行。2T =1015.42K(742.37)11. 試計(jì)算298.15K，標(biāo)準(zhǔn)態(tài)下的反應(yīng)：H2O(g)+CO(g)=H2(g)+CO2(g)的和，并計(jì)算298.15K時(shí)H2O(g)的。解 = 0+(-393.5kJmol-1)+(-241.8kJmol-1)-(-110.5 kJmol-1)=-41.2kJmol-1= 0+(-394.

8、4kJmol-1)+(-228.6kJmol-1)-(-137.2 kJmol-1)=-28.6kJmol-1=-42.26 JK-1mol-1=+-=130.7 JK-1mol-1213.8 JK-1mol-1197.77 JK-1mol-1-42.26 JK-1mol-1=189.1 JK-1mol-112. 計(jì)算以下反應(yīng)在298.15K標(biāo)準(zhǔn)態(tài)下的，判斷自發(fā)進(jìn)行的方向，求出標(biāo)準(zhǔn)平衡常數(shù)。(可用于汽車尾氣的無(wú)害化)解 1= -228.6 kJmol-100=-228.6 kJmol-10正向反應(yīng)可以自發(fā)進(jìn)行。=1.110402= 287.6kJmol-100=175.2kJmol-1=2.0

9、410-30(3) = 124.5 kJmol-13209.9 kJmol-1=-505.2kJmol-1=3.131088(4) = (-394.4 kJmol-1+0)(-137.2 kJmol-1+87.6 kJmol-1)=-344.8 kJmol-10=2.51060(5) = 2(-394.4kJmol-1)+2(-174.8 kJmol-1)(-910.6 kJmol-1)=-227.8 kJmol-1=8.01103913. 某病人平均每天需要6300kJ能量以維持生命。假設(shè)每天只能吃250g牛奶(燃燒值為3.0kJ.g-1)和50g面包(燃燒值為12kJ.g-1)，問(wèn)每天還需

10、給他輸入多少升質(zhì)量濃度為50.0gL-1的葡萄糖(燃燒值為15.6kJg-1)溶液？解 設(shè)每天需輸入50.0gL-1葡萄糖的體積為V3.0 kJg-1250g+12 kJg-150g+50gL-1V15.6 kJg-1=6300kJV=6.31L14. 糖代謝的總反應(yīng)為：C12H22O11(s) + 12O2(g) = 12CO2(g)+ 11H2O(l)(1)從附表的熱力學(xué)數(shù)據(jù)求298.15 K，標(biāo)準(zhǔn)態(tài)下的和。(2)如果在體內(nèi)只有30%的自由能變轉(zhuǎn)化為非體積功，求在37下，1.00molC12H22O11(s)進(jìn)行代謝時(shí)可以得到多少非體積功。解 (1) =-393.5 kJmol-112+

11、(-285.8 kJmol-1)11-012-(-2226.1 kJmol-1)-5640 kJmol-1= 213.8JK-1mol-112+70.0 JK-1mol-111-205.2 JK-1mol-112-360.2 JK-1mol-1= 513JK-1mol-1=-T =-5640kJmol-1-298.15K51310-3kJK-1mol-1=-5793 kJ mol-1或 = =-394.4 kJ mol-112+(-237.1 kJ mol-1)11-012-(-1544.6 kJmol-1)=-5796 kJ mol-1(2) ，310.5 =-T =-5640kJmol-1

12、-310.5K51310-3kJK-1mol-1=-5799.11kJ mol-1Wf =(310.5K)30%=-5799.11kJ mol-130%=-1740 kJ mol-115在823K，標(biāo)準(zhǔn)態(tài)下以下反應(yīng)的：(1)CO2(g)+H2(g)=CO(g)+H2O(g) (2)CoO(s)+H2(g)=Co(s)+H2O(g) 試求823K，標(biāo)準(zhǔn)態(tài)下反應(yīng)(3)的：(3)CoO(s)+CO(g)=Co(s)+CO2(g)并求反應(yīng)(2)和(3)的，比較CO(g)和H2(g)對(duì)CoO(s)的復(fù)原能力誰(shuí)更強(qiáng)些。解 (2)- (1)為(3) 反應(yīng)(2) =-28.78 kJ mol-1反應(yīng)(3) =

13、-42.25 kJ mol-1反應(yīng)3的自由能比2更負(fù)，所以CO(g)對(duì)CoO的復(fù)原能力大于H2(g)對(duì)CoO的復(fù)原能力。16. 在某細(xì)胞內(nèi)ADP和H3PO4濃度分別為3.0mmolL-1和1.0mmolL-1。ATP的水解反應(yīng)為：在310.15時(shí)，=-31.05kJmol-1,試求ATP在細(xì)胞內(nèi)的平衡濃度；如實(shí)際上ATP的濃度是10mmolL-1,求反應(yīng)的。 解 ADP與H3PO4濃度即可以看成是平衡濃度，也可以看成任意時(shí)刻濃度。 =-51.97 kJ mol-117. 欲用MnO2與HCl溶液反應(yīng)制備Cl2(g)，已知該反應(yīng)的方程式為：(1)寫出此反應(yīng)的標(biāo)準(zhǔn)平衡常數(shù)的表達(dá)式。(2)根據(jù)附表中

14、的熱力學(xué)數(shù)據(jù)，求出298.15K標(biāo)準(zhǔn)狀態(tài)下此反應(yīng)的及值，并指出此反應(yīng)能否自發(fā)進(jìn)行？(3)假設(shè)HCl溶液濃度為12.0molL-1,其它物質(zhì)仍 為標(biāo)準(zhǔn)態(tài)，反應(yīng)在298K時(shí)能否自發(fā)進(jìn)行？解 (1) (2) = (-237.1 kJmol-1)2+0+(-228.11)-(-465.1 kJmol-1)+40+(-131.2 kJmol-1)2= 25.2 kJmol-10在298.15K時(shí)標(biāo)態(tài)下不能自發(fā)進(jìn)行。(3) = 25.2 kJmol-1+ 8.314JK-1mol-1298K=-11.77 kJmol-1 0 此反應(yīng)在298.15K、標(biāo)準(zhǔn)狀態(tài)下不能自發(fā)進(jìn)行。 (2) 由于此過(guò)程的 0，0，

15、升高溫度有利于反應(yīng)自發(fā)進(jìn)行，T即T1015K時(shí)，此過(guò)程自發(fā)進(jìn)行。Exercises1. (omit)2. (omit)3. Phosgene, COCl2, was used as a war gas during World War 1. It reacts with the moisture in the lungs to produce HCl, which causes the lungs to fill fluid, and CO2. Write an equation of the reaction and compute . For COCl2(g), (g)=-210 kJmo

16、l-1Solution COCl2(g) + H2O(l) 2HCl(g) + CO2(g)=2(-95.3kJmol-1) + (-394.4kJmol-1)- -210kJmol-1+(-237.1 kJmol-1)=-137.9kJmol-14. Given the following,4NO(g) 2 N2O(g) + O2(g) = -139.56 kJmol-12 NO(g) + O2(g) 2 NO2(g) = -69.70kJmol-1calculate for the reaction 2 N2O(g) + 3O2(g) 4 NO2(g).Solution =0.16 kJm

17、ol-15. Chloroform, formerly used as aneshetic and now belived to be a carcinogen (cancer- causing agent), has a heat of vaporization =31.4 kJmol-1 . The change, CHCl3(l) CHCl3(g), has =94.2 JK-1mol-1. At what temperature do we expect CHCl3 to boil? .Solution 6. A reaction that can convert coal to me

18、thane (the cihef component of natural gas) is C(s) +2 H2(g) = CH4(g)for which = -50.79 kJmol-1 . What is the value of K for the reaction at 25 0C ? Does this value of K suggest that studying this reaction as a mean of methane production is worth pursuing? Solution According to the result, this react

19、ion as a mean of methane production is worth pursuing .7Triglyceride is one of typical fatty acids, its metabolic reaction is :C57H104O6(s) + 80O2(g) = 57CO2(g) +52H2O(l), =-3.3510-4kJ/mol.Calculate the of Triglyceride.Solution The necessary data can be found from table in the addenda of the text bo

20、ok: (CO2 g)=-393.5 kJ/mol (H2O,l)=-285.8 kJ/mol (O2 g)=0So =57 (CO2 g)+52 (H2O,l)- (C57H104O6)-80 (O2 g) (C57H104O6)=57(-393.5)+52(-285.8)-(-3.3510-4)=-3.79103 kJ/mol8predict the sign of the entropy change for each of the following reactions:(a) O2(g)(100kPa,298K)O2(g) (10kPa,298K)(b) NH4Cl(s)NH3(g)+HCl(g)(c)CO(g)+H2O(g)CO2(g)+H2(g)Solution (a) S0; (b) S0; (c) For no change take place in the number of moles of gas, the entropy change