已知下列熱化學(xué)方程式:
(1)CH
3COOH(l)+2O
2(g)═2CO
2(g)+2H
2O(l)△H
1=-870.3kJ/mol
(2)C(s)+O
2(g)═CO
2(g)△H
2=-393.5kJ/mol
(3)H
2(g)+
O
2(g)═H
2O(l)△H
3=-285.8kJ/mol
則反應(yīng)2C(s)+2H
2(g)+O
2(g)═CH
3COOH(l)的焓變?yōu)椋ā 。?/div>
分析:依據(jù)熱化學(xué)方程式和蓋斯定律計(jì)算分析,反應(yīng)的焓變與反應(yīng)過程無關(guān),只與起始狀態(tài)和終了狀態(tài)有關(guān).
解答:解:(1)CH
3COOH(l)+2O
2(g)═2CO
2(g)+2H
2O(l)△H
1=-870.3kJ/mol
(2)C(s)+O
2(g)═CO
2(g)△H
2=-393.5kJ/mol
(3)H
2(g)+
O
2(g)═H
2O(l)△H
3=-285.8kJ/mol
依據(jù)蓋斯定律(2)×2-(1)+(3)×2得到
2C(s)+2H
2(g)+O
2(g)═CH
3COOH(l)△H=-488.3 kJ/mol,
故選:A.
點(diǎn)評(píng):本題考查熱化學(xué)方程式的書寫和蓋斯定律的計(jì)算應(yīng)用,題目較簡單.