题目内容
已知下列热化学方程式:
2C(s)+O2(g)═2CO(g)△H=-221kJ/mol
2H2(g)+O2(g)═2H2O(g)△H=-483.6kJ/mol
由此可知C(s)+H2O(g)═CO(g)+H2(g)△H等于( )
2C(s)+O2(g)═2CO(g)△H=-221kJ/mol
2H2(g)+O2(g)═2H2O(g)△H=-483.6kJ/mol
由此可知C(s)+H2O(g)═CO(g)+H2(g)△H等于( )
| A.-131.3 kJ/mol | B.131.3 kJ/mol |
| C.373.1 kJ/mol | D.-373.1 kJ/mol |
①2C(S)+O2(g)=2CO(g)△H=-221kJ/mol
②2H2(g)+O2(g)=2H2O(g)△H=-483.6kJ/mol
根据盖斯定律分析计算得到:①-②得到热化学方程式为:2C(S)+2H2O(g)=2CO(g)+2H2(g)△H=+262.6kJ/mol;
即C(S)+H2O(g)=CO(g)+H2(g)△H=+131.3kJ/mol;
所以△H=+131.3kJ/mol,
故选B.
②2H2(g)+O2(g)=2H2O(g)△H=-483.6kJ/mol
根据盖斯定律分析计算得到:①-②得到热化学方程式为:2C(S)+2H2O(g)=2CO(g)+2H2(g)△H=+262.6kJ/mol;
即C(S)+H2O(g)=CO(g)+H2(g)△H=+131.3kJ/mol;
所以△H=+131.3kJ/mol,
故选B.
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