高中化学复习压轴题热点练习:盖斯定律的应用
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热点2 盖斯定律的应用
1.联氨(又称联肼,N 2H 4,无色液体)是一种应用广泛的化工原料,可用作火箭燃料,回答下列问题: ①2O 2(g)+N 2(g)===N 2O 4(l) ΔH 1
②N 2(g)+2H 2(g)===N 2H 4(l) ΔH 2
③O 2(g)+2H 2(g)===2H 2O(g) ΔH 3
④2N 2H 4(l)+N 2O 4(l)===3N 2(g)+4H 2O (g)ΔH 4=-1048.9 kJ/mol
上述反应热效应之间的关系式为ΔH 4=__________________________,联氨和N 2O 4可作为火箭推进剂的主要原因为_________________.
答案 2ΔH 3-2ΔH 2-ΔH 1 反应放热量大、产生大量气体
解析 根据盖斯定律,2×③-2×②-①即得2N 2H 4(l)+N 2O 4(l)===3N 2(g)+4H 2O(g)的ΔH 4,所以反应热效应之间的关系式为ΔH 4=2ΔH 3-2ΔH 2-ΔH 1.联氨有强还原性,N 2O 4有强氧化性,两者在一起易发生氧化还原反应,反应放热量大、产生大量气体,所以联氨和N 2O 4可作为火箭推进剂.
2.已知:C(s,石墨)+O 2(g)===CO 2(g)ΔH 1=-393.5 kJ ·mol -1
2H 2(g)+O 2(g)===2H 2O (l)ΔH 2=-571.6 kJ·mol -1
2C 2H 2(g)+5O 2(g)===4CO 2(g)+2H 2O (l)ΔH 3=-2599 kJ·mol -1
根据盖斯定律,计算反应2C(s,石墨)+H 2(g)===C 2H 2(g)的ΔH =_________.
答案 +226.7 kJ·mol -1
解析 ①C(s ,石墨)+O 2(g)===CO 2(g)ΔH 1=-393.5 kJ·mol -1,
②2H 2(g)+O 2(g)===2H 2O (l)ΔH 2=-571.6 kJ·mo l -1,
③2C 2H 2(g)+5O 2(g)===4CO 2(g)+2H 2O (l)ΔH 3=-2599 kJ·mol -1;
根据盖斯定律计算(①×2+②×12-③×12
)得2C(s,石墨)+H 2(g)===C 2H 2(g) ΔH =(-393.5 kJ·mol -1)×2+12×(-571.6 kJ·mol -1)-12
×(-2599 kJ ·mol -1)=+226.7 kJ·mol -1. 3.由金红石(TiO 2)制取单质Ti,涉及的步骤为:
TiO 2―→TiCl 4――→镁/800 ℃/Ar
Ti
已知:①C(s)+O 2(g)===CO 2(g)ΔH 1=-393.5 kJ·mol -1
②2CO(g)+O 2(g)===2CO 2(g)ΔH 2=-566 kJ·mol -1
③TiO 2(s)+2Cl 2(g)===TiCl 4(s)+O 2(g)ΔH 3=+141 kJ·mol -1
则TiO 2(s)+2Cl 2(g)+2C(s)===TiCl 4(s)+2CO(g)的ΔH =_________________.
答案 -80 kJ·mo l -1
解析 ③+①×2-②就可得TiO 2(s)+2Cl 2(g)+2C(s)===TiCl 4(s)+2CO(g),则ΔH =ΔH 3+ΔH 1×2-ΔH 2=-80 kJ·mol -1
.
4.甲醇质子交换膜燃料电池中将甲醇蒸气转化为氢气的两种反应的热化学方程式如下:
①CH 3OH(g)+H 2O(g)===CO 2(g)+3H 2(g)ΔH =+49.0 kJ·mol -1
②CH 3OH(g)+12
O 2(g)===CO 2(g)+2H 2(g)ΔH =-192.9 kJ·mol -1 又知③H 2O(g)===H 2O(l) ΔH =-44 kJ·mol -1,
则甲醇蒸气燃烧生成液态水的热化学方程式为___________.
答案 CH 3OH(l)+32
O 2(g)===CO 2(g)+2H 2O(l) ΔH =-764.7 kJ·mol -1
解析 根据盖斯定律计算(②×3-①×2+③×2)得:CH 3OH(l)+32
O 2(g)===CO 2(g)+2H 2O(l) ΔH =3×(-192.9 kJ·mol -1)-2×49.0 kJ·mol -1+(-44 kJ·mol -1)×2=-764.7 kJ·mol -1;则甲醇蒸气燃烧
为液态水的热化学方程式为:CH 3OH(l)+32
O 2(g)===CO 2(g)+2H 2O (l)ΔH =-764.7 kJ·mol -1. 5.已知:①H 2的热值为142.9 kJ·g -1(热值是表示单位质量的燃料完全燃烧生成稳定的化合物时所放出的热量);
②N 2(g)+2O 2(g)===2NO 2(g)ΔH =+133 kJ·mol -1
③H 2O(g)===H 2O(l) ΔH =-44 kJ·mol -1
催化剂存在下,H 2还原NO 2生成水蒸气和其他无毒物质的热化学方程式:_________________________________________________.
答案 4H 2(g)+2NO 2(g)===N 2(g)+4H 2O (g)ΔH =-1100.2 kJ·mol -1
解析 已知:①H 2的热值为142.9 kJ·g -1,则H 2(g)+12
O 2(g)===H 2O(l) ΔH =-285.8 kJ·mol -1; ②N 2(g)+2O 2(g)===2NO 2(g)ΔH =+133 kJ·mol -1;
③H 2O(g)===H 2O(l) ΔH =-44 kJ·mol -1;
根据盖斯定律由①×4-②-③×4可得4H 2(g)+2NO 2(g)===4H 2O(g)+N 2(g) ΔH =(-285.8 kJ·mol -1)×4-(+133 kJ·mol -1)-(-44 kJ·mol -1)×4=-1100.2 kJ·mol -1,故此反应的热化学方程式为4H 2(g)
+2NO 2(g)===N 2(g)+4H 2O (g)ΔH =-1100.2 kJ·mol -1.
6.能源问题是人类社会面临的重大课题,H 2、CO 、CH 3OH 都是重要的能源物质,它们的燃烧热依次为-285.8 k J·mol -1、-282.5 kJ·mol -1、-726.7 kJ·mol -1.已知CO 和H 2在一定条件下可以合成甲醇CO(g)+2H 2(g)===CH 3OH(l).则CO 与H 2反应合成甲醇的热化学方程式为______________________________.
答案 CO(g)+2H 2(g)===CH 3OH (l)ΔH =-127.4 kJ·mol -1
解析 根据目标反应与三种反应热的关系,利用盖斯定律,计算出目标反应的反应热ΔH =2×(-285.8 kJ·mol -1)+(-282.5 kJ·mol -1)-(-726.7 kJ ·mol -1)=-127.4 kJ·mol -1.
7.已知:25 ℃、101 kPa 时,Mn(s)+O 2(g)===MnO 2(s)ΔH =-520 kJ·mol -1
S(s)+O 2(g)===SO 2(g) ΔH =-297 kJ·mol -1
Mn(s)+S(s)+2O 2(g)===MnSO 4(s)ΔH =-1065 kJ·mol -1
则SO 2与MnO 2反应生成无水MnSO 4的热化学方程式是___________________.
答案 MnO 2(s)+SO 2(g)===MnSO 4(s)ΔH =-248 kJ·mol -1
解析 将题给三个热化学方程式依次编号为①②③,根据盖斯定律,由③-①-②可得SO 2(g)+MnO 2(s)===MnSO 4(s) ΔH =(-1065 kJ ·mol -1)-(-520 kJ·mol -1)-(-297 kJ·mol -1)=-248 kJ·mol -1.
8.已知下列热化学方程式:
①Fe 2O 3(s)+3CO(g)===2Fe(s)+3CO 2(g)ΔH =-25 kJ·mol -1
②3Fe 2O 3(s)+CO(g)===2Fe 3O 4(s)+CO 2(g)ΔH =-47 kJ·mol -1
③Fe 3O 4(s)+CO(g)===3FeO(s)+CO 2(g)ΔH =+19 kJ·mol -1
写出FeO(s)被CO 还原成Fe 和CO 2的热化学方程式______________.
答案 FeO(s)+CO(g)===Fe(s)+CO 2(g)ΔH =-11 kJ·mol -1
解析 ①×3-②-③×2就可得6FeO(s)+6CO(g)===6Fe(s)+6CO 2(g) ΔH =-66 kJ·mol -1, 即FeO(s)+CO(g)===Fe(s)+CO 2(g)ΔH =-11 kJ·mol -1.
9.已知:①2Cu 2S(s)+3O 2(g)===2Cu 2O(s)+2SO 2(g)ΔH =-768.2 kJ·mol -1
②2Cu2O(s)+Cu2S(s)===6Cu(s)+SO2(g)ΔH=+116.0 kJ·mol-1
则Cu2S(s)+O2(g)===2Cu(s)+SO2(g) ΔH=____________________. 答案-217.4 kJ·mol-1
解析根据盖斯定律,将方程式1
3
×(①+②)得Cu2S(s)+O2(g)===2Cu(s)+SO2(g) ΔH=
1
3
×(-768.2
+116.0) kJ·mol-1=-217.4 kJ·mol-1.