.-635 -393-|||-9 .5 .5-|||-({S)_(r)(G)_(m)/(kJ)_(1)cdot (mol)^-1}-1128.8-|||--604 -394-|||-.2 .4-|||-._(12)^0(1)(200)^circ (1)(1000)^circ (1)=1-604.2-394.4+1128.81kJcdot (mol)^-1=1.3021.5cdot (mol)^-1-|||-^theta (298K)=1.52times (10)^-23-|||-由于 Delta (C)_(p)=0 ,故 ({Delta )_(1)}^Delta (H)_(m)^theta 不随温度变化。设分解温-|||-度为T2,则 ^theta ((T)_(2))=10 .-|||-ln dfrac ({K)^theta ((I)_(2))}({K)^0(298K)}=dfrac ({A)_(1)(H)_(m)^theta }(R)(dfrac (1)(298)-dfrac (1)({T)_(2)})-|||-解得:-|||-_(2)=1112K-|||-36.试根据下列数据求算反应-|||-._(2)(H)_(4)(g)+(H)_(2)(g)=(C)_(2)(H)_(6)(g)-|||-在1000K时的标准平衡常数k°。已知:-|||-(1)298K时,乙烯和乙烷的标准燃烧热分别-|||-为 -1411 和 -1560kJ(mol)^-1 ,液态水的标准生成热为-286 kJ·mol-1;(2) 298 K时,C2H4(g)、C2H6(g)和H2(g)的标准熵分别为219.5、229.5和130.6 J·K-1·mol-1;(3) 在298 ~1000 K范围内,反应的平均热容差Cp = 10.8 J·K-1·mol-1。

石灰窖中烧石灰的反应为

CaCO3(s) CaO(s) + CO2(g)

欲使石灰石能以一定速率分解为石灰,分解压最

为-286 kJ·mol-1;

(2) 298 K时,C2H4(g)、C2H6(g)和H2(g)的标准熵分别为219.5、229.5和130.6 J·K-1·mol-1;

(3) 在298 ~1000 K范围内,反应的平均热容差Cp = 10.8 J·K-1·mol-1。

参考答案与解析:

相关试题

1.设有关系R和S:-|||-R A B C-|||-6 4 2 2-|||-4-|||-6 5 3 3-|||-5-|||-5 6 8-|||-S B C D-|||-4 4 9-|||-4 9-|

1.设有关系R和S:-|||-R A B C-|||-6 4 2 2-|||-4-|||-6 5 3 3-|||-5-|||-5 6 8-|||-S B C D

  • 查看答案
  • 已知1 2 -2 4-|||-2 2 2 2-|||-1 4 -3 5-|||--1 4 2 7,1 2 -2 4-|||-2 2 2 2-|||-1 4 -3 5-|||--1 4 2 7为元素1

    已知1 2 -2 4-|||-2 2 2 2-|||-1 4 -3 5-|||--1 4 2 7,1 2 -2 4-|||-2 2 2 2-|||-1 4 -3

  • 查看答案
  • ,P点是明纹B 相位差为R 4-|||-1 P-|||-5-|||-3-|||-5-|||-2. ,P点是暗纹C 相位差R 4-|||-1 P-|||-5-|||-3-|||-5-|||-2.,P点是

    ,P点是明纹B 相位差为R 4-|||-1 P-|||-5-|||-3-|||-5-|||-2. ,P点是暗纹C 相位差R 4-|||-1 P-|||-5-||

  • 查看答案
  • =1:4:9-|||-(2) _(1):(x)_(square ):(x)_(square ):... =1:3:5-|||-(3) Delta S=(S)_(2)-(S)_(1)=(S)_(3)-(

    =1:4:9-|||-(2) _(1):(x)_(square ):(x)_(square ):... =1:3:5-|||-(3) Delta S=(S)_(

  • 查看答案
  • S2 S.-|||--4/4- 一 3/4-|||-r2 r1-|||-习题 9-18 图

    S2 S.-|||--4/4- 一 3/4-|||-r2 r1-|||-习题 9-18 图

  • 查看答案
  • 计算5阶行列式-|||-0 2 3 4 5-|||-1 1 3 4 5-|||-1 2 2 4 5 = __-|||-1 2 3 3 5-|||-1 2 3 4 4

    计算5阶行列式-|||-0 2 3 4 5-|||-1 1 3 4 5-|||-1 2 2 4 5 = __-|||-1 2 3 3 5-|||-1 2 3 4

  • 查看答案
  • _(2)(g)+dfrac (1)(2)(O)_(2)(g)=!=!= (H)_(2)O(1) Delta (H)_(2)=-285.5kJ-|||-cdot (mol)^-1;-|||-Ⅲ. _(

    _(2)(g)+dfrac (1)(2)(O)_(2)(g)=!=!= (H)_(2)O(1) Delta (H)_(2)=-285.5kJ-|||-cdot

  • 查看答案
  • 设2 -2 -4 2 -3 -5-|||-A= -1 3 4 B= -1 4 5-|||-1 -2 -3 1 -3 -4(1)验证2 -2 -4 2 -3 -5-|||-A= -1 3 4 B= -1

    设2 -2 -4 2 -3 -5-|||-A= -1 3 4 B= -1 4 5-|||-1 -2 -3 1 -3 -4(1)验证2 -2 -4 2 -3 -5

  • 查看答案
  • 设3 2 -1 4-|||-1 5 0 -1-|||-2 -1 3 2-|||-4 9 2 1,3 2 -1 4-|||-1 5 0 -1-|||-2 -1 3 2-|||-4 9 2 1,3 2 -

    设3 2 -1 4-|||-1 5 0 -1-|||-2 -1 3 2-|||-4 9 2 1,3 2 -1 4-|||-1 5 0 -1-|||-2 -1 3

  • 查看答案
  • 计算:(1)/(3)×(3)/(5)+1(5)/(7)-(5)/(9)×(5)/(7)1-(5)/(7)×(21)/(25)(1)/(2)+(5)/(4)×(4)/(5)(1)/(6)×(5-(2)/

    计算:(1)/(3)×(3)/(5)+1(5)/(7)-(5)/(9)×(5)/(7)1-(5)/(7)×(21)/(25)(1)/(2)+(5)/(4)×(4

  • 查看答案