题目内容

【题目】用化学知识填空:

1)丙烷通过脱氢反应可得丙烯。

已知:①C3H8(g)= CH4(g)C2H2(g)H2(g) ΔH1=+156.6 kJ·mol

C3H6(g)= CH4(g)C2H2(g) ΔH2=+32.4 kJ·mol

则相同条件下,反应C3H8(g)=== C3H6(g)H2(g)ΔH__________kJ·mol

2)将煤转化为水煤气是通过化学方法将煤转化为洁净燃料的方法之一。煤转化为水煤气的主要化学反应为C(s)H2O(g) CO(g)H2(g)C(s)CO(g)H2(g)完全燃烧的热化学方程式分别为

C(s)O2(g)===CO2(g) ΔH1=-393.5 kJ·mol

H2(g)O2(g)===H2O(g) ΔH2=-242.0 kJ·mol

CO(g)O2(g)===CO2(g) ΔH3=-283.0 kJ·mol

写出C(s)与水蒸气反应生成COH2的热化学方程式: ______________________

33mol甲烷燃烧时,生成液态水和二氧化碳,同时放出2 670.9kJ的热量,写出甲烷燃烧热的热化学方程式_________________________________________________

4)如图是298 K101 kPa时,N2H2反应过程中能量变化的曲线图。该反应的热化学方程式为____________

5)已知:H2(g)Cl2(g)===2HCl(g) ΔH=-185 kJ·mol Cl—Cl 的键能为247 kJ·molH—H的键能为436kJ·molH—Cl的键能为______________kJ·mol

【答案】+124.2 C(s)H2O(g) === CO(g) H2(g) ΔH+131.5 kJ·mol1 CH4(g)2O2(g) === CO2(g) 2 H2O (l) ΔH-890.3 kJ·mol1 N2(g)+3H2(g) 2NH3(g) H=-92 kJ·mol1 434

【解析】

从盖斯定律、燃烧热的定义、反应热与焓变的关系角度进行分析;

1)利用盖斯定律,根据目标反应方程式,因此有①-②得出△H=H1-△H2=156.6kJ·mol1(32. 4kJ·mol1)=124.2kJ·mol1

故答案为+124.2kJ·mol1

2C(s)H2O(g) CO(g)H2(g) ①,H2(g)O2(g)===H2O(g) ②,CO(g)O2(g)===CO2(g) ③,利用盖斯定律,得出①-②-③得出:△H=H1-△H2-△H3=(393.5kJ·mol1)(242.0kJ·mol1283.0kJ·mol1)=131.5kJ·mol1,则热化学反应方程式为C(s)H2O(g) = CO(g) H2(g) ΔH+131.5 kJ·mol1

故答案为C(s)H2O(g) = CO(g) H2(g) ΔH+131.5 kJ·mol1

3)根据燃烧热的定义,得出1mol甲烷完全燃烧放出的热量为==890.3kJ·mol1,因此甲烷燃烧热的热化学方程式为CH4(g)2O2(g) = CO2(g) 2 H2O (l) ΔH-890.3 kJ·mol1

故答案为CH4(g)2O2(g) = CO2(g) 2 H2O (l) ΔH-890.3 kJ·mol1

4)根据图像,反应物总能量大于生成物的总能量,即该反应为放热反应,因此△H=(508kJ·mol1600kJ·mol1)=92kJ·mol1,热化学反应方程式为N2(g)+3H2(g) 2NH3(g) H=-92 kJ·mol1

故答案为N2(g)+3H2(g) 2NH3(g) H=-92 kJ·mol1

5)利用△H=反应物键能总和-生成物键能总和,令HCl的键能为akJ·mol1,则有:436kJ·mol1247kJ·mol12akJ·mol1=185kJ·mol1,解得a=434kJ·mol1

故答案为434kJ·mol1

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