反应热的计算与热化学方程式

1、突破反应热的计算与热化学方程式的书写一、根据盖斯定律计算H的步骤和方法1.计算步骤2.计算方法二、热化学方程式的书写与反应热大小的比较1.热化学方程式书写的“六个注意”三、相关题练习考点一:利用化学键计算反应热,热化学方程式的书写1.已知:C(s)H2O(g)=CO(g)H2(g)Ha kJmol12C(s)O2(g)=2CO(g)H220 kJmol1 ,HH键、O=O键和OH键的键能分别为436 kJmol1、496 kJmol1和462 kJmol1,则a为()A332 B. 118C350 D. 130解析 将两个热化学方程式分别标为、,则根据盖斯定律可知2即得到热化学方程式O2(g)

2、+2H2(g)=2H2O(g)H(2202a) kJ/mol.由于反应热等于断键吸收的能量与形成新化学键所放出的能量的差值,则496 kJ/mol2436 kJ/mol22462 kJ/mol(2202a) kJ/mol,解得a130,答案选D.2.(1)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原理是:CH3OH(g)H2O(g)=CO2(g)3H2(g) H49.0 kJmol1CH3OH(g)O2(g)=CO2(g)2H2(g) H192.9 kJmol1又知:H2O(g)=H2O(l)H44 kJmol1则甲醇燃烧生成液态水的热化学方程式为_化学键PPPOO=OP=O键能/

3、kJmol1abcx已知1 mol白磷(P4)完全燃烧放热为d kJ,白磷及其完全燃烧的产物结构如图所示,则上表中x_解析(1)322得CH3OH(g)O2(g)=CO2(g)2H2O(l)H3(192.9 kJmol1)249.0 kJmol1(44 kJmol1)2764.7 kJmol1.(2)反应热反应物键能总和生成物键能总和,即6a5c(4x12b)d,解得x(12bd6a5c)或(d6a5c12b).答案(1)CH3OH(g)O2(g)=CO2(g)2H2O(l)H764.7 kJmol1(2)(12bd6a5c)考点二:盖斯定律及其应用3.(1)已知:甲醇脱水反应:2CH3OH(

4、g)=CH3OCH3(g)H2O(g) H123.9 kJmol1甲醇制烯烃反应:2CH3OH(g)=C2H4(g)2H2O(g) H229.1 kJmol1乙醇异构化反应:C2H5OH(g)=CH3OCH3(g) H350.7 kJmol1则乙烯气相直接水合反应C2H4(g)H2O(g)=C2H5OH(g)的H_kJmol1.(2)二甲醚(CH3OCH3)是无色气体,可作为一种新型能源.由合成气(组成为H2、CO和少量的CO2)直接制备二甲醚,其中的主要过程包括以下四个反应:甲醇合成反应:()CO(g)2H2(g)=CH3OH(g) H190.1 kJmol1()CO2(g)3H2(g)=C

5、H3OH(g)H2O(g) H249.0 kJmol1水煤气变换反应:()CO(g)H2O(g)=CO2(g)H2(g) H341.1 kJmol1二甲醚合成反应:()2CH3OH(g)=CH3OCH3(g)H2O(g) H424.5 kJmol1由H2和CO直接制备二甲醚(另一产物为水蒸气)的热化学方程式为_.解析(1)2CH3OH(g)=CH3OCH3(g)H2O(g) H123.9 kJmol12CH3OH(g)=C2H4(g)2H2O(g) H229.1 kJmol1C2H5OH(g)=CH3OCH3(g) H350.7 kJmol1根据盖斯定律,由得:C2H4(g)H2O(g)=C2H5OH(g)H45.5 kJmol1.(2)写出由H2和CO直接制备二甲醚的化学方程式:4H22CO= CH3OCH3H2O.由()得:2CO(g)4H2(g)=2CH3OH(g)H1(90.1)2 kJmol1；由()得:2CH3OH(g)=CH3OCH3(g)H2O(g)H424.5 kJmol1；以上两式相加得所求热化学方程式的H(-90.12)kJmol1+(-24.5) kJmol1204.7 kJmol1.故热化学方程式为4H2(g)2C