热力学gydF4y2Ba
NCERT解决方案gydF4y2Ba
第2部分gydF4y2Ba
问题12:有限公司(g)的生成焓变,有限公司gydF4y2Ba2gydF4y2Ba(g), NgydF4y2Ba2gydF4y2BaO (g)和NgydF4y2Ba2gydF4y2BaOgydF4y2Ba4gydF4y2Ba(g)是-110、-393、81和9.7 kJ摩尔gydF4y2Ba1gydF4y2Ba分别。找到Δ的价值gydF4y2BargydF4y2BaH反应:gydF4y2Ba
NgydF4y2Ba2gydF4y2BaOgydF4y2Ba4gydF4y2Ba(g) + 3 (g)→NgydF4y2Ba2gydF4y2BaO (g) + 3有限公司gydF4y2Ba2gydF4y2Ba(g)gydF4y2Ba
答:gydF4y2BaΔgydF4y2BargydF4y2BaHgydF4y2Ba
= 9.7 + 3×(-110)- (81 - 3×393)gydF4y2Ba
= 9.7 - 330 - 81 + 1179 = 777.7 = 778 kJ摩尔gydF4y2Ba1gydF4y2Ba
问题13:鉴于:NgydF4y2Ba2gydF4y2Ba(g) + 3 hgydF4y2Ba2gydF4y2Banh (g)→2gydF4y2Ba3gydF4y2Ba
ΔgydF4y2BargydF4y2BaHgydF4y2BaΘgydF4y2Ba= -92.4 kJ摩尔gydF4y2Ba1gydF4y2Ba
NH的标准生成焓是什么gydF4y2Ba3gydF4y2Ba气体?gydF4y2Ba
答:gydF4y2BaNH的标准生成焓gydF4y2Ba3gydF4y2Ba= ' (-92.4)/ 2 = -46.2 ' kJ摩尔gydF4y2Ba1gydF4y2Ba
问题14:计算的标准生成焓CHgydF4y2Ba3gydF4y2Ba哦(l)从以下数据:gydF4y2Ba
CHgydF4y2Ba3gydF4y2Ba哦(l) + ' 3/2'OgydF4y2Ba2gydF4y2Ba(g)→有限公司gydF4y2BaggydF4y2Ba+ 2 hgydF4y2Ba2gydF4y2BaO(左)gydF4y2Ba
ΔgydF4y2BargydF4y2BaHgydF4y2BaθgydF4y2Ba= -726 kJ摩尔gydF4y2Ba1gydF4y2Ba
C(石墨)+ OgydF4y2Ba2gydF4y2Ba(g)→有限公司gydF4y2Ba2gydF4y2Ba(g)gydF4y2Ba
ΔgydF4y2BargydF4y2BaHgydF4y2BaθgydF4y2Ba= -393 kJ摩尔gydF4y2Ba1gydF4y2Ba
HgydF4y2Ba2gydF4y2Ba(g) + 1/2'OgydF4y2Ba2gydF4y2Ba(g)→HgydF4y2Ba2gydF4y2BaO(左)gydF4y2Ba
ΔgydF4y2BargydF4y2BaHgydF4y2BaθgydF4y2Ba= -286 kJ摩尔gydF4y2Ba1gydF4y2Ba
答:gydF4y2Ba的方程CH的形成gydF4y2Ba3gydF4y2Ba哦,如下:gydF4y2Ba
C (s) + 2 hgydF4y2Ba2gydF4y2Ba(g) + 1/2'OgydF4y2Ba2gydF4y2Ba(g)→CHgydF4y2Ba3gydF4y2Ba哦(左)gydF4y2Ba
这个方程可以实现如下:gydF4y2Ba
方程(2)+ 2×方程(3),方程(1)gydF4y2Ba
添加:方程(2)+ 2×方程(3)gydF4y2Ba
C + OgydF4y2Ba2gydF4y2Ba+ 2 hgydF4y2Ba2gydF4y2Ba+ 2啊gydF4y2Ba2gydF4y2Ba→公司gydF4y2Ba2gydF4y2Ba+ 2 hgydF4y2Ba2gydF4y2Ba
↠C + 2 hgydF4y2Ba2gydF4y2Ba+ 2啊gydF4y2Ba2gydF4y2Ba→公司gydF4y2Ba2gydF4y2Ba+ 2 hgydF4y2Ba2gydF4y2BaOgydF4y2Ba
减去:方程(1)从上面方程gydF4y2Ba
C + 2 hgydF4y2Ba2gydF4y2Ba+ 2啊gydF4y2Ba2gydF4y2Ba- - - - - - CHgydF4y2Ba3gydF4y2Ba哦- 3/2 OgydF4y2Ba2gydF4y2Ba→公司gydF4y2Ba2gydF4y2Ba+ 2 hgydF4y2Ba2gydF4y2Ba- - - - - -有限公司gydF4y2Ba2gydF4y2Bah - 2gydF4y2Ba2gydF4y2BaOgydF4y2Ba
↠C + 2 hgydF4y2Ba2gydF4y2Ba+½OgydF4y2Ba2gydF4y2Ba- - - - - - CHgydF4y2Ba3gydF4y2Ba哦= 0gydF4y2Ba
↠C + 2 hgydF4y2Ba2gydF4y2Ba+½OgydF4y2Ba2gydF4y2Ba→CHgydF4y2Ba3gydF4y2Ba哦gydF4y2Ba
所以,CH的生成焓gydF4y2Ba3gydF4y2Ba哦,可以计算如下:gydF4y2Ba
方程(2)+ 2×方程(3),方程(1)gydF4y2Ba
= -393 - 2×286 + 726 = -239 kJ摩尔gydF4y2Ba1gydF4y2Ba
问题15:计算焓的变化过程gydF4y2Ba
创新领导力gydF4y2Ba4gydF4y2Ba(g)→C (g) + 4 cl (g)gydF4y2Ba
并计算债券CCl C-Cl的焓gydF4y2Ba4gydF4y2Ba(g)gydF4y2Ba
ΔgydF4y2Ba特许经销商gydF4y2BaHgydF4y2BaΘgydF4y2Ba(同gydF4y2Ba4gydF4y2Ba)= 30.5 kJ摩尔gydF4y2Ba1gydF4y2Ba
ΔgydF4y2BafgydF4y2BaHgydF4y2BaΘgydF4y2Ba(同gydF4y2Ba4gydF4y2Ba)= -135.5 kJ摩尔gydF4y2Ba1gydF4y2Ba
ΔgydF4y2Ba一个gydF4y2BaHgydF4y2BaΘgydF4y2Ba[C] = 715 kJ摩尔gydF4y2Ba1gydF4y2Ba
ΔgydF4y2Ba一个gydF4y2BaHgydF4y2BaΘgydF4y2Ba(ClgydF4y2Ba2gydF4y2Ba)= 242 kJ摩尔gydF4y2Ba1gydF4y2Ba
在这里,ΔgydF4y2Ba一个gydF4y2BaHgydF4y2BaΘgydF4y2Ba是焓gydF4y2Ba
答:gydF4y2Ba以下方程可以写为各种焓变的问题:gydF4y2Ba
创新领导力gydF4y2Ba4gydF4y2Ba(左)→创新领导力gydF4y2Ba4gydF4y2Ba(g) [30.5 kJ摩尔gydF4y2Ba1gydF4y2Ba
C (s) + 2 clgydF4y2Ba2gydF4y2BaCCl→gydF4y2Ba4gydF4y2Ba(左)(- 135.5 kJ摩尔gydF4y2Ba1gydF4y2Ba]gydF4y2Ba
C (s)→C (g) [715 kJ摩尔gydF4y2Ba1gydF4y2Ba]gydF4y2Ba
ClgydF4y2Ba2gydF4y2Ba(g)→2 cl (g) [242 kJ摩尔gydF4y2Ba1gydF4y2Ba]gydF4y2Ba
目标方程是:三地gydF4y2Ba4gydF4y2Ba(g)→C (g) + 4 cl (g)gydF4y2Ba
这可以实现如下:gydF4y2Ba
方程(3)+ 2×方程(1)-(4)-方程方程(2)gydF4y2Ba
增加方程3和4,如下所示:gydF4y2Ba
C + 2氯gydF4y2Ba2gydF4y2Ba→C + 4 clgydF4y2Ba
减去方程(1)和(2)从上面gydF4y2Ba
C + 2氯gydF4y2Ba2gydF4y2BaCCl -gydF4y2Ba4gydF4y2Bacl - C - 2gydF4y2Ba2gydF4y2BaCCl→C + 4 cl -gydF4y2Ba4gydF4y2BaCCl -gydF4y2Ba4gydF4y2Ba
CCl↠gydF4y2Ba4gydF4y2Ba→C + 4 clgydF4y2Ba
所以,需要焓可以添加相同的顺序的计算:gydF4y2Ba
方程(3)+ 2×方程(1)-(4)-方程方程(2)gydF4y2Ba
= 715 + 2×242 - 30.5 + 135.5 = 1304 kJgydF4y2Ba
所以,键焓在每个C-Cl键' = (1304)/ 4 = 326 kJ摩尔gydF4y2Ba1gydF4y2Ba
问题16:对于一个孤立系统,ΔU = 0,ΔS是什么呢?gydF4y2Ba
答:gydF4y2Ba对孤立系统,ΔS > 0,因为熵会增加。gydF4y2Ba
问题17:在298 K的反应gydF4y2Ba
2 a + B→CgydF4y2Ba
ΔH = 400 kJ摩尔gydF4y2Ba1gydF4y2Ba和ΔS K = 0.2 kJgydF4y2Ba1gydF4y2Ba摩尔gydF4y2Ba1gydF4y2Ba
在什么温度下反应会自发考虑ΔH和ΔS ranage恒定温度。gydF4y2Ba
答:gydF4y2Ba根据吉布斯方程,gydF4y2Ba
ΔG =ΔH - TΔSgydF4y2Ba
ΔG = 0gydF4y2Ba
ΔH = TΔSgydF4y2Ba
或者,' T =(ΔH) /(ΔS) 'gydF4y2Ba
' = (400)/ (0.2)= 2000 KgydF4y2Ba
问题18:反应,2 Cl (g)→ClgydF4y2Ba2gydF4y2Ba(g)ΔH和ΔS的迹象是什么?gydF4y2Ba
答:gydF4y2Ba债券形成的能量释放,所以ΔH是负的。在形成分子熵减少,所以ΔS将是负的。gydF4y2Ba
问题19:反应:gydF4y2Ba
2 (g) + B (g)→2 d (g)gydF4y2Ba
ΔUgydF4y2BaΘgydF4y2Ba= -10.5 kJ,ΔSgydF4y2BaΘgydF4y2Ba= -44.1 JKgydF4y2Ba1gydF4y2Ba
计算ΔGgydF4y2BaΘgydF4y2Ba反应和预测是否可能发生自发的反应。gydF4y2Ba
答:gydF4y2BaΔH =ΔU +ΔngydF4y2BaggydF4y2BaRTgydF4y2Ba
考虑到;ΔU = -10.5 kJgydF4y2Ba
ΔngydF4y2BaggydF4y2Ba= 2 - 3 = - 1摩尔gydF4y2Ba
R = 8.3145×10gydF4y2Ba3gydF4y2BakJ摩尔gydF4y2Ba1gydF4y2Ba
T = 298 KgydF4y2Ba
所以,ΔH = -10.5 - 1×8.314×10gydF4y2Ba3gydF4y2Ba×298gydF4y2Ba
= - 10.5 - 8.314×0.298 = 12.978 kJgydF4y2Ba
根据吉布斯方程:gydF4y2Ba
ΔG =ΔH - TΔSgydF4y2Ba
= - 12.978 - 298×- 0.0441gydF4y2Ba
= - 12.978 + 13.142 = 0.164 kJgydF4y2Ba
ΔG积极,反应会发酵gydF4y2Ba
问题20:反应的平衡常数是10。ΔG的价值gydF4y2BaΘgydF4y2Ba吗?R = 8.314 JKgydF4y2Ba1gydF4y2Ba摩尔gydF4y2Ba1gydF4y2BaT = 300 K。gydF4y2Ba
答:gydF4y2BaΔG RT日志= - RT ln K = 2.303 KgydF4y2Ba
鉴于:JK R = 8.314gydF4y2Ba1gydF4y2Ba摩尔gydF4y2Ba1gydF4y2BaT = 300 K, K = 10gydF4y2Ba
所以,ΔG = - 2.303×8.314×300×10日志gydF4y2Ba
= - 5527 J摩尔gydF4y2Ba1gydF4y2Ba= - 5.527 kJ摩尔gydF4y2Ba1gydF4y2Ba
问题21:评论的热力学稳定性没有(g),gydF4y2Ba
“1/2'NgydF4y2Ba2gydF4y2Ba(g) + 1/2'OgydF4y2Ba2gydF4y2Ba(g)→没有(g)gydF4y2Ba
ΔgydF4y2BargydF4y2BaHgydF4y2BaΘgydF4y2Ba= 90 kJ摩尔gydF4y2Ba1gydF4y2Ba
没有(g) + ' 1/2'OgydF4y2Ba2gydF4y2Ba(g)→不gydF4y2Ba2gydF4y2Ba(g)gydF4y2Ba
ΔgydF4y2BargydF4y2BaHgydF4y2BaΘgydF4y2Ba= -74 kJ摩尔gydF4y2Ba1gydF4y2Ba
答:gydF4y2Ba没有(g):ΔgydF4y2BargydF4y2BaHgydF4y2BaΘgydF4y2Ba= + ve:本质上不稳定gydF4y2Ba
没有gydF4y2Ba2gydF4y2Ba(g):ΔgydF4y2BargydF4y2BaHgydF4y2BaΘgydF4y2Ba=负:稳定gydF4y2Ba
问题22:计算熵变在周围1.00摩尔的HgydF4y2Ba2gydF4y2BaO (l)是在标准条件下形成的。ΔgydF4y2BafgydF4y2BaHgydF4y2BaΘgydF4y2Ba= -286 kJ摩尔gydF4y2Ba1gydF4y2Ba
答:gydF4y2Ba问gydF4y2Ba牧师gydF4y2Ba= -ΔgydF4y2BafgydF4y2BaHgydF4y2BaΘgydF4y2Ba
= - 286 kJ摩尔gydF4y2Ba1gydF4y2Ba= 286000 J摩尔gydF4y2Ba1gydF4y2Ba
ΔSgydF4y2BasurrgydF4y2Ba' = (q_ (re \ v)) / T 'gydF4y2Ba
' = (286000)/ (298)= 959 J KgydF4y2Ba1gydF4y2Ba摩尔gydF4y2Ba1gydF4y2Ba