A.CH
4(g)+2O
2(g)=CO
2(g)+2H
2O(l)△H
3=-890.3kJ/mol①,C
2H
2(g)+
O
2(g)═2CO
2(g)+H
2O(l)△H=-1299.6 kJ/mol②,H
2(g)+
O
2(g)=H
2O(l)△H=-285.8kJ/mol③,依据②+③×3-①×2可得:C
2H
2(g)+3H
2(g)═2CH
4(g)△H=-376.4 kJ/mol,△H=反应物总键能-生成物总键能=C≡C键的键能+2×413.4kJ/mol+3×436.0kJ/mol-8×413.4kJ/mol=-376.4KJ/mol,则C≡C键的键能=796.0 kJ/mol,故A正确;
B.形成共价键的两原子半径之和越小,共用电子对数越多,共价键越牢固,则键长越长,作用力越小,键能越小,C原子半径比H大,所以C-H键键长大于H-H键,故B错误;
C.氢气燃烧热为285.8kJ/mol,热化学方程式为:H
2(g)+
O
2(g)=H
2O(l)△H=-285.8kJ/mol,则2H
2(g)+O
2(g)═2H
2O(l);△H=-571.6 kJ/mol,2H
2(g)+O
2(g)═2H
2O(l)△H=-571.6 kJ?mol
-1,故C错误;
D.由A可知:C
2H
2(g)+3H
2(g)═2CH
4(g)△H=-376.4 kJ/mol,则 2CH
4(g)═C
2H
2(g)+3H
2(g)△H=+376.4 kJ?mol
-1,故D错误;
故选A.