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(1)氢气、甲醇是优质的清洁燃料,可制作燃料电池.已知:①2CH3OH(l)+3O2(g)═2CO2(g)+4H2O(g)△H1=-1275.6kJ•mol-1②2CO(g)+O2(g)═2CO2(g)△H2=-566.0kJ•mol-1③H2O(g)═H2O(l)△H3=-
题目详情
(1)氢气、甲醇是优质的清洁燃料,可制作燃料电池.
已知:①2CH3OH(l)+3O2(g)═2CO2(g)+4H2O(g)△H1=-1275.6kJ•mol-1
②2CO(g)+O2(g)═2CO2(g)△H2=-566.0kJ•mol-1
③H2O(g)═H2O(l)△H3=-44.0kJ•mol-1写出甲醇不完全燃烧生成一氧化碳和液态水的热化学方程式______.
(2)处理NOx的一种方法是利用甲烷催化还原NOx.
CH4(g)+4NO2(g)═4NO(g)+CO2(g)+2H2O(g)△H1=-574kJ•mol-1
CH4(g)+4NO(g)═2N2(g)+CO2(g)+2H2O(g)△H2
CH4(g)+2NO2(g)═N2(g)+CO2(g)+2H2O(g)△H3=-867kJ•mol-1则△H2=______.
32221-1
222-1
223-1
xx
42221-1
42222
422223-12
已知:①2CH3OH(l)+3O2(g)═2CO2(g)+4H2O(g)△H1=-1275.6kJ•mol-1
②2CO(g)+O2(g)═2CO2(g)△H2=-566.0kJ•mol-1
③H2O(g)═H2O(l)△H3=-44.0kJ•mol-1写出甲醇不完全燃烧生成一氧化碳和液态水的热化学方程式______.
(2)处理NOx的一种方法是利用甲烷催化还原NOx.
CH4(g)+4NO2(g)═4NO(g)+CO2(g)+2H2O(g)△H1=-574kJ•mol-1
CH4(g)+4NO(g)═2N2(g)+CO2(g)+2H2O(g)△H2
CH4(g)+2NO2(g)═N2(g)+CO2(g)+2H2O(g)△H3=-867kJ•mol-1则△H2=______.
32221-1
222-1
223-1
xx
42221-1
42222
422223-12
▼优质解答
答案和解析
(1)已知:①2CH33OH(1)+3O22(g)=2CO22(g)+4H22O(g)△H11=-1275.6kJ/mol
②2CO(g)+O22(g)=2CO22(g)△H22=-566.0kJ/mol
③H22O(g)=H22O(1)△H33=-44.0kJ/mol
根据盖斯定律,①×
-②×
+③×2得:CH3OH(1)+O2(g)=CO(g)+2H2O(1)△H=-442.8 kJ•mol-1,
故答案为:CH3OH(1)+O2(g)=CO(g)+2H2O(1)△H=-442.8 kJ•mol-1;
(2)①CH4(g)+4NO2(g)═4NO(g)+CO2(g)+2H2O(g)△H1=-574kJ•mol-1
②CH4(g)+4NO(g)═2N2(g)+CO2(g)+2H2O(g)△H2
③CH4(g)+2NO2(g)═N2(g)+CO2(g)+2H2O(g)△H3=-867kJ•mol-1
依据盖斯定律计算,③×2-①得到CH4(g)+4NO(g)═2N2(g)+CO2(g)+2H2O(g)△H2=-1160kJ•mol-1;
故答案为:-1160kJ•mol-1.
1 1 12 2 2-②×
+③×2得:CH3OH(1)+O2(g)=CO(g)+2H2O(1)△H=-442.8 kJ•mol-1,
故答案为:CH3OH(1)+O2(g)=CO(g)+2H2O(1)△H=-442.8 kJ•mol-1;
(2)①CH4(g)+4NO2(g)═4NO(g)+CO2(g)+2H2O(g)△H1=-574kJ•mol-1
②CH4(g)+4NO(g)═2N2(g)+CO2(g)+2H2O(g)△H2
③CH4(g)+2NO2(g)═N2(g)+CO2(g)+2H2O(g)△H3=-867kJ•mol-1
依据盖斯定律计算,③×2-①得到CH4(g)+4NO(g)═2N2(g)+CO2(g)+2H2O(g)△H2=-1160kJ•mol-1;
故答案为:-1160kJ•mol-1.
1 1 12 2 2+③×2得:CH33OH(1)+O22(g)=CO(g)+2H22O(1)△H=-442.8 kJ•mol-1-1,
故答案为:CH33OH(1)+O22(g)=CO(g)+2H22O(1)△H=-442.8 kJ•mol-1-1;
(2)①CH44(g)+4NO22(g)═4NO(g)+CO22(g)+2H22O(g)△H11=-574kJ•mol-1-1
②CH44(g)+4NO(g)═2N22(g)+CO22(g)+2H22O(g)△H22
③CH44(g)+2NO22(g)═N22(g)+CO22(g)+2H22O(g)△H33=-867kJ•mol-1-1
依据盖斯定律计算,③×2-①得到CH44(g)+4NO(g)═2N22(g)+CO22(g)+2H22O(g)△H22=-1160kJ•mol-1-1;
故答案为:-1160kJ•mol-1-1.
②2CO(g)+O22(g)=2CO22(g)△H22=-566.0kJ/mol
③H22O(g)=H22O(1)△H33=-44.0kJ/mol
根据盖斯定律,①×
1 |
2 |
1 |
2 |
故答案为:CH3OH(1)+O2(g)=CO(g)+2H2O(1)△H=-442.8 kJ•mol-1;
(2)①CH4(g)+4NO2(g)═4NO(g)+CO2(g)+2H2O(g)△H1=-574kJ•mol-1
②CH4(g)+4NO(g)═2N2(g)+CO2(g)+2H2O(g)△H2
③CH4(g)+2NO2(g)═N2(g)+CO2(g)+2H2O(g)△H3=-867kJ•mol-1
依据盖斯定律计算,③×2-①得到CH4(g)+4NO(g)═2N2(g)+CO2(g)+2H2O(g)△H2=-1160kJ•mol-1;
故答案为:-1160kJ•mol-1.
1 |
2 |
1 |
2 |
故答案为:CH3OH(1)+O2(g)=CO(g)+2H2O(1)△H=-442.8 kJ•mol-1;
(2)①CH4(g)+4NO2(g)═4NO(g)+CO2(g)+2H2O(g)△H1=-574kJ•mol-1
②CH4(g)+4NO(g)═2N2(g)+CO2(g)+2H2O(g)△H2
③CH4(g)+2NO2(g)═N2(g)+CO2(g)+2H2O(g)△H3=-867kJ•mol-1
依据盖斯定律计算,③×2-①得到CH4(g)+4NO(g)═2N2(g)+CO2(g)+2H2O(g)△H2=-1160kJ•mol-1;
故答案为:-1160kJ•mol-1.
1 |
2 |
故答案为:CH33OH(1)+O22(g)=CO(g)+2H22O(1)△H=-442.8 kJ•mol-1-1;
(2)①CH44(g)+4NO22(g)═4NO(g)+CO22(g)+2H22O(g)△H11=-574kJ•mol-1-1
②CH44(g)+4NO(g)═2N22(g)+CO22(g)+2H22O(g)△H22
③CH44(g)+2NO22(g)═N22(g)+CO22(g)+2H22O(g)△H33=-867kJ•mol-1-1
依据盖斯定律计算,③×2-①得到CH44(g)+4NO(g)═2N22(g)+CO22(g)+2H22O(g)△H22=-1160kJ•mol-1-1;
故答案为:-1160kJ•mol-1-1.
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