ÌâÄ¿ÄÚÈÝ

ÒÑÖªÒÒÏ©ºÍÒÒȲȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ·Ö±ðΪ£º

2C2H2(g)+5O2(g)====4CO2(g)+2H2O(l);¦¤H=-2 600 kJ¡¤mol-1

C2H4(g)+3O2(g) ====2CO2(g)+2H2O(l);¦¤H=-1 411 kJ¡¤mol-1

ÓÖÖªÑõȲÑæµÄζȱÈÒÒϩȼÉÕʱ»ðÑæµÄζȸߡ£¾Ý´Ë£¬ÏÂÁÐ˵·¨´íÎóµÄÊÇ(    )

A.ÎïÖʵÄȼÉÕÈÈÔ½´ó£¬»ðÑæζÈÔ½¸ß

B.ÌþÍêȫȼÉÕʱ£¬»ðÑæζȵĸߵͲ»½ö½öÈ¡¾öÓÚÆäȼÉÕÈȵĸߵÍ

C.ÏàͬÌõ¼þϵÈÌå»ýÒÒÏ©ºÍÒÒȲÍêȫȼÉÕʱ£¬ÒÒȲ·ÅÈȽÏÉÙ

D.1 molÒÒÏ©ÍêȫȼÉÕÉú³ÉÆø̬²úÎïʱ£¬·Å³öµÄÈÈÁ¿Ð¡ÓÚ1 411 kJ

½âÎö£ºC2H2(g)+O2(g)====2CO2(g)+H2O(l);¦¤H=-1 300 kJ¡¤mol-1,½«ÒÔÉÏÈÈ»¯Ñ§·½³ÌʽÓëC2H4ȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ±È½Ï£¬¿ÉÖªB¡¢CÕýÈ·£¬A´íÎó£»ÓÖH2O(g)====H2O(l);¦¤H,·´Ó¦ÎªÒ»·ÅÈÈ·´Ó¦£¬¦¤H£¼0,¸ù¾Ý¸Ç˹¶¨ÂÉ¿ÉÍƶÏDÕýÈ·¡£

´ð°¸£ºA

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø