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

ÓÖÖªÑõÈ²ÑæµÄζȱÈÒÒϩȼÉÕʱ»ðÑæµÄζȸߡ£¾Ý´Ë£¬ÏÂÁÐ˵·¨´íÎóµÄÊÇ£¨    £©

A.ÎïÖʵÄȼÉÕÈÈÔ½´ó£¬»ðÑæÎ¶ÈÔ½¸ß

B.ÌþÍêȫȼÉÕʱ£¬»ðÑæÎ¶ȵĸߵͲ»½ö½öÈ¡¾öÓÚÆäȼÉÕÈȵĸߵÍ

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

D.1 molÒÒÏ©ÍêȫȼÉÕÉú³ÉÆøÌ¬²úÎïʱ£¬·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¿ÉÍÆ¶ÏDÕýÈ·¡£

´ð°¸£ºA

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø