北京奥运会“祥云”火炬燃料是丙烷(C3H8),亚特兰大奥运会火炬燃料是丙烯(C3H6).(1)丙烷脱氢可得

2025-05-17 02:33:21
推荐回答(1个)
回答(1):

(1)已知:①C3H8(g)═CH4(g)+HC≡CH(g)+H2(g)△H1=156.6kJ?mol-1
②CH3CH=CH2(g)═CH4(g)+HC≡CH(g )△H2=32.4kJ?mol-1
根据盖斯定律,①-②得C3H8(g)═CH3CH=CH2(g)+H2(g),所以△H=△H1-△H2=156.6kJ?mol-1-32.4kJ?mol-1=124.2kJ?mol-1
故答案为:124.2;
(2)①CH4(g)+H2O(g)═CO (g)+3H2(g)△H=+206.2kJ?mol-1
②CH4(g)+CO2(g)═2CO (g)+2H2(g)△H=+247.4kJ?mol-1
由盖斯定律可知,①×2-②可得CH4(g)+2H2O(g)═CO2(g)+4H2(g),
其反应热△H=(+206.2kJ?mol-1)×2-(+247.4kJ?mol-1)=+165.0 kJ?mol-1
即热化学方程式为CH4(g)+2H2O(g)═CO2(g)+4H2(g)△H=+165.0 kJ?mol-1
故答案为:CH4(g)+2H2O(g)═CO2(g)+4H2(g)△H=+165.0 kJ?mol-1
(3)①Al2O3(s)+AlCl3(g)+3C(s)═3AlCl(g)+3CO(g)△H=+a kJ?mol-1
②3AlCl(g)═3Al(l)+AlCl3(g)△H=+b kJ?mol-1
依据盖斯定律计算①+②得到:Al2O3(s)+3C(s)═2Al(l)+3CO(g)△H=a+bKJ/mol;
故答案为:a+b.