发布时间 : 星期一 文章工程化学基础(第二版)练习题参考答案更新完毕开始阅读
8.玻璃钢是一种复合材料,它由合成树脂,如酚醛树脂、环氧树脂及玻璃纤维等组成,将玻璃纤维(增强相)浸渍在树脂(粘结相,基体)中,再加以固化剂、稀释剂、填充剂、增塑剂等辅助材料制成。它的主要优点是质轻,电绝缘性好,不受电磁作用,不反射无线电波,微波透过性能好,耐磨,耐腐蚀,成型简便。可用作汽车、轮船外壳、室内器具等。
第四章 化学反应与能源
§4.1热化学与能量转化 练习题(p.106)
1.(1) a、b;(2) b、d;(3) c;(4) b
2. C2H2(g) + 5/2O2 (g) = 2CO2 (g) + H2O(g)
?fHm? (298.15)/kJ.mol
-1
227.4 0 -393.5 -241.8
-1
?rHm?(298.15)=[2×(-393.5) -241.8-227.4] kJ?mol
=-1256.2 kJ?mol1
--1
CH4(g) + 2O2 (g) = CO2 (g) + 2H2O(g)
?fHm? (298.15)/kJ.mol
-74.6 0 -393.5 -241.8
-1
?rHm?(298.15)=[ (-393.5) +2×(-241.8)-(-74.6)] kJ?mol
=-802.5 kJ?mol1
--1
C2H4(g) + 3O2 (g) = 2CO2 (g) + 2H2O(g)
?fHm? (298.15)/kJ.mol
52.4 0 -393.5 -241.8
-1
?rHm?(298.15)=[2×(-393.5) +2×(-241.8)-52.4] kJ?mol
=-1323 kJ?mol1
--1
C2H6(g) + 7/2O2 (g) = 2CO2 (g) + 3H2O(g)
?fHm? (298.15)/kJ.mol
-84.0 0 -393.5 -241.8
-1
?rHm?(298.15)=[2×(-393.5) +3×(-241.8)-(-84.0)] kJ?mol
=-1428.4 kJ?mol1
-
可见,燃烧1molC2H4或C2H6放出的热量大于C2H2,因此可以代替,而CH4不行。 3. Na2S(s) + 9H2O(g) = Na2S?9H2O(s)
?fHm?rHm
/kJ.mol
-1
-372.86 -241.8 -3079.41
-1
(298.15)={(-3079.41)-[(-372.86)+(-241.8)×9]}kJ?mol=-530.35 kJ?mol1
-
-
1kg Na2S的物质的量:n=1000g/(22.99×2+32.07)g?mol1=12.81mol Q= Qp=?H= (-530.35kJ?mol1 )×12.81mol=-6794 kJ
-
17
4. 2N2H4(l) + N2O4(g) = 3N2(g) + 4H2O(l)
?fHm ?rHm
/kJ.mol1 50.63 9.66 0 -285.8
-
(298.15)=[(-285.8)×4-(50.63×2+9.66)] kJ?mol1
-
-
=-1254.12 kJ?mol1
32 g N2H4的物质的量为:n=32g/(14×2+1×4)g?mol1=1.0 mol
-
1.0molN2H4完全反应,其反应进度为2mol,所以: Q= Qp=?H=-1254.12 kJ?mol1×2mol=-627.06 kJ
-
115. CaO(s) + H2O(l) = Ca2+(aq) + 2OH(aq)
-
?fHm ?rHm
(298.15)/kJ?mol
-1
-634.9 -285.8 -542.8 -230.0
-
(298.15)=[(-543.20)-2× (-230.0)]-[(-634.9)+(-285.8)] kJ?mol1
-
=-82.1 kJ?mol1 罐头从25℃?80℃需吸收的热量: Q= Qp=?H=400 J?K1?(80-25)K=22000 J
-
设需CaO为W克,则其物质的量 n=W/[(40.08+16.00) g?mol1]=Q/[-?rHm
-
(298.15) ×80%]
-
? W=[22000/(82.1×103×80%)] mol ×56.08 g?mol1 =18.78 g 6. C6H6(l) + 15/2O2 (g) = 6CO2 (g) + 3H2O(l)
?U?QV??209.2kJ
?H??U??ngRT?[(?209.2)?(5/78)?(6?7.5)?8.314?298.15?10?3]kJ??209.4kJ1mol液态苯在弹式量热计中完全燃烧放热:
Q?QV?(?209.2)?78/5kJmol?1??3263.5kJmol?1
7.恒容反应热QV??U,恒压反应热Qp??H,当液体、固体相对于气体体积可以忽略且气体可以看作理想气体时有:?H??U??ngRT。所以:
(1) H2(g)十(2) H2(g)十
10.5O2(g)==H2O(g) ?ng??,?H??U,Qp?QV;
211.5O2(g)==H2O(l) ?ng??,?H??U,Qp?QV。
2-1
8. Fe2O3(s) + 3CO(g) = 2Fe(s) + 3CO2(g) ?fHm ?rHm
(298.15)/kJ?mol
-824.2 -110.5 0 -393.5
-
(298.15)=[3× (-393.5)-(-824.2)+3×(-110.5)] kJ?mol1
-
=-24.8 kJ?mol1
18
9. C5H12(l) + 8O2 (g) = 5CO2 (g) + 6H2O(g)
?fHm? (298.15)/kJ.mol
-1
-149.9 0 -393.5 -285.8
-1
?rHm?(298.15)=[5×(-393.5) +6×(-285.8)-(-149.9)] kJ?mol
=-3532.4kJ?mol1
-
燃烧1克汽油所放出的热量:Q?3532.4kJmol?72gmol
§4.2 化学反应的方向和限度
练习题(p.114)
1.(1)X;(2)√;(3)X;(4)X;(5)X;(6)√。 2.(1)Smθ[H2O(s)]<Smθ[H2O(l)] <Smθ[H2O(g)] (2)Smθ(298.15K) <Smθ(398.15K)<Smθ(498.15K)
(3)同一温度下:Smθ(Fe)<Smθ(FeO)<Smθ(Fe2O3)。 3. C(s)+CO2(g)==2CO(g) ?fGm ?rGm
(298.15)/kJ?mol
-1
?1?1?49.06kJg?1
0 -394.4 -137.2
-
(298.15)=[ 2× (-137.2) -(-394.4)] kJ?mol1
-
=120.0 kJ?mol1
4. CaCO3(s) == CaO(s)+ CO2(g) ?fHm
Sm
?rHm
(298.15)/kJ?mol
--1
-
-1207.6 -634.9 -393.5
(298.15)/J?mol1?K1 91.7 38.1 213.8 (298.15)=[(-634.9)- (-393.5)-(-1207.6)] kJ?mol1
-
-
=179.2 kJ?mol1 ?rSm
(298.15)=(38.1+213.8-91.7) J?mol1?K1
-
-
-
-
=160.2 J?mol1?K1
ΔrGmθ(1222K)? ?rHm
(298.15) -T?rSm
-
(298.15)
-
-
= 179.2 kJ?mol1-1222K×160.2 J?mol1?K1
= -16.56 kJ?mol1
-
ΔrGmθ(1222K) <0,能自发进行。
5. SiO2(s) + 2C(s) = Si(s) + 2CO(g)
0?fHm(298.15K)/kJ·mol-1 -910.7 0 0 -110.5 0Sm(298.15K)/J·mol-1·K-1 41.5 5.7 18.8 -197.7 0?fGm(298.15K)/kJ mol-1 -856.3 0 0 -137.2
0?1?1(1) ?rHm(298.15K)?[2?(?110.5)?(?910.7)]kJmol?689.7kJmol
19
0?rSm(298.15K)?[2?197.7?18.8?(41.5?2?5.7)]Jmol?1K?1?361.3Jmol?1K?1
0?1?1(2) ?rGm(298.15K)?[2?(?137.2)?(?856.3)]kJmol?581.9kJmol
或
000?rGm(298.15K)??rHm?T?rSm?[689.7?298.15?361.3?10?3]kJmol?1?582.0kJmol?10?3?1?1 (3) ?rGm(1000K)?[689.7?1000?361.3?10]kJmol?328.4kJmol?0
不能自发。
000(4) ?rGm??rHm?T?rSm?0,自发,所以:
00T??rHm/?rSm?(689.7?103/361.3)K?1909K
6.(1)大于零;(2)大于零;(3)小于零;(4)小于零。 7. C(s) + H2O (g) = CO(s) + H2(g)
0?fHm(298.15K)/kJ·mol-1 0 -241.8 -110.5 0 0Sm(298.15K)/J·mol-1·K-1 5.7 188.8 197.7 130.7 0?fGm(298.15K)/kJ mol-1 0 -228.6 -137.2 0
0?1?1(1) ?rGm(298.15K)?[(?137.2)?(?228.6)]kJmol?91.4kJmol?0
不能向正方向进行。
0?1?1(2) ?rHm?[(?110.5)?(?241.8)]kJmol?131.3kJmol?0 0?rSm?[197.7?130.7?(188.8?5.7)]Jmol?1K?1?133.9Jmol?1K?1?0 000?rGm??rHm?T?rSm
因此,升高温度能向正方向进行。
000(3) ?rGm??rHm?T?rSm?0
00T??rHm/?rSm?(131.3?103/133.9)K?980.6K
8.已知?rHm?rSm
(298.15)=-402.0kJ?mol
-1
,?rGm
=-345.7kJ.mol
-1
,则 298.15K时的
(298.15)值可以从下式求出:
(298.15)-298.15K×?rSm(298.15)=[?rHm
(298.15)=?rGm
(298.15)
?rHm ?rSm
(298.15)-?rGm
(298.15)]/298.15K
-
=[-402.0-(-345.7)]kJ?mol1/298.15K =0.1888 kJ.mol
当?rGm ?rHm
-1.
K
-1
(T)=0时的温度可用下式表示: (298.15)-T??rSm
(298.15)]/?rSm
(298.15)?0 (298.15)
20
算得:T?[?rHm