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写出下列燃烧热的热化学方程式.(1)0.5molCH4完全燃烧生成CO2和液态水时,放出445kJ的热量.写出CH4的燃烧热化学方程式.(2)在标准状况下,44.8LH2完全燃烧生成液态水放出571.6kJ的热

题目详情
写出下列燃烧热的热化学方程式.
(1)0.5molCH4完全燃烧生成CO2和液态水时,放出445kJ的热量.写出CH4的燃烧热化学方程式______.
(2)在标准状况下,44.8LH2完全燃烧生成液态水放出571.6kJ的热量,求在此条件下H2的燃烧热化学方程式
H2(g)+
1
2
O2(g)=H2O(l)△=-285.8KJ/mol
H2(g)+
1
2
O2(g)=H2O(l)△=-285.8KJ/mol

(3)已知碳的燃烧热为393.5kJ/mol,写出碳的燃烧热化学方程式
C(s)+
1
2
O2(g)=CO2(g)△H=-393.5kJ/mol;
C(s)+
1
2
O2(g)=CO2(g)△H=-393.5kJ/mol;

424
22
H2(g)+
1
2
O2(g)=H2O(l)△=-285.8KJ/mol
2
1
2
O2(g)=H2O(l)△=-285.8KJ/mol
1
2
112222
H2(g)+
1
2
O2(g)=H2O(l)△=-285.8KJ/mol
2
1
2
O2(g)=H2O(l)△=-285.8KJ/mol
1
2
112222
C(s)+
1
2
O2(g)=CO2(g)△H=-393.5kJ/mol;
1
2
O2(g)=CO2(g)△H=-393.5kJ/mol;
1
2
112222
C(s)+
1
2
O2(g)=CO2(g)△H=-393.5kJ/mol;
1
2
O2(g)=CO2(g)△H=-393.5kJ/mol;
1
2
112222
▼优质解答
答案和解析
(1)0.5mol CH44完全燃烧生成CO22和液态H22O时,放出445kJ热量,1mol甲烷燃烧生成二氧化碳和液态水放热890kJ,反应的热化学方程式为:
CH44(g)+2O22(g)→CO22(g)+2H22O(l)△H=-890kJ/mol;
故答案为:CH44(g)+2O22(g)→CO22(g)+2H22O(l)△H=-890kJ/mol;
(2)标准状况下44.8L氢气物质的量为2mol,燃烧生成液态水时放出571.6KJ的热量,反应的热化学方程式为:2H22(g)+O22(g)=2H22O(l)△=-571.6KJ/mol,则氢气的燃烧热为:H22(g)+
1
2
O2(g)=H2O(l)△=-285.8KJ/mol;
故答案为:H2(g)+
1
2
O2(g)=H2O(l)△=-285.8KJ/mol;
(3)已知碳的燃烧热为393.5kJ/mol,碳的燃烧热化学方程式为:C(s)+
1
2
O2(g)=CO2(g)△H=-393.5kJ/mol;
故答案为:C(s)+
1
2
O2(g)=CO2(g)△H=-393.5kJ/mol;
1
2
111222O22(g)=H22O(l)△=-285.8KJ/mol;
故答案为:H22(g)+
1
2
O2(g)=H2O(l)△=-285.8KJ/mol;
(3)已知碳的燃烧热为393.5kJ/mol,碳的燃烧热化学方程式为:C(s)+
1
2
O2(g)=CO2(g)△H=-393.5kJ/mol;
故答案为:C(s)+
1
2
O2(g)=CO2(g)△H=-393.5kJ/mol;
1
2
111222O22(g)=H22O(l)△=-285.8KJ/mol;
(3)已知碳的燃烧热为393.5kJ/mol,碳的燃烧热化学方程式为:C(s)+
1
2
O2(g)=CO2(g)△H=-393.5kJ/mol;
故答案为:C(s)+
1
2
O2(g)=CO2(g)△H=-393.5kJ/mol;
1
2
111222O22(g)=CO22(g)△H=-393.5kJ/mol;
故答案为:C(s)+
1
2
O2(g)=CO2(g)△H=-393.5kJ/mol;
1
2
111222O22(g)=CO22(g)△H=-393.5kJ/mol;