ÌâÄ¿ÄÚÈÝ


ÔÚijһÈÝ»ýΪ2 LµÄÃܱÕÈÝÆ÷ÄÚ£¬¼ÓÈë0.8 molµÄH2ºÍ0.6 molµÄI2£¬ÔÚÒ»¶¨Ìõ¼þÏ·¢ÉúÈçÏ·´Ó¦£ºH2(g)+I2(g)2HI(g)

¦¤H<0£¬·´Ó¦Öи÷ÎïÖʵÄŨ¶ÈËæÊ±¼ä±ä»¯Çé¿öÈçͼ1£º

(1)¸Ã·´Ó¦µÄ»¯Ñ§Æ½ºâ³£Êý±í´ïʽΪ__________¡£

(2)¸ù¾Ýͼ1Êý¾Ý£¬·´Ó¦¿ªÊ¼ÖÁ´ïµ½Æ½ºâʱ£¬Æ½¾ù·´Ó¦ËÙÂÊv(HI)Ϊ_______¡£

(3)·´Ó¦´ïµ½Æ½ºâºó£¬µÚ8·ÖÖÓʱ£º

¢ÙÈôÉý¸ßζȣ¬»¯Ñ§Æ½ºâ³£ÊýK_________(Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±)£¬HIŨ¶ÈµÄ±ä»¯ÕýÈ·µÄÊÇ___________(ÓÃͼ2ÖÐa¡«cµÄ±àºÅ»Ø´ð)¡£

¢ÚÈô¼ÓÈëI2£¬H2Ũ¶ÈµÄ±ä»¯ÕýÈ·µÄÊÇ_______(ÓÃͼ2ÖÐd¡«fµÄ±àºÅ»Ø´ð)¡£


(1)ÓÉ»¯Ñ§Æ½ºâ³£Êý¸ÅÄîÖ±½Óд³öÆä±í´ïʽ

(2)ÓÉͼ1ÖªÔÚ3 minʱ·´Ó¦´ïƽºâ״̬£¬¹Êv(HI)=¦¤c(HI)/¦¤t=

0.5 mol¡¤L-1/3 min¡Ö0.167 mol¡¤L-1¡¤min-1¡£

(3)¢ÙÒòÕý·´Ó¦·ÅÈÈ£¬¹ÊÉýÎÂÆ½ºâ×óÒÆ£¬KÖµ±äС£¬c(HI)¼õС£¬Í¼2ÖÐcÇúÏß·ûºÏÕâÒ»ÊÂʵ£»¢Úµ±¼ÓÈëI2ʱƽºâÓÒÒÆ£¬c(H2)¼õС£¬Í¼2ÖÐfÇúÏß·ûºÏÕâÒ»ÊÂʵ¡£

´ð°¸£º(1)

(2)0.167 mol¡¤L-1¡¤min-1

(3)¢Ù¼õС       c       ¢Úf


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

½«Ãº×ª»¯ÎªË®ÃºÆøµÄÖ÷Òª»¯Ñ§·´Ó¦ÎªC(s)£«H2O(g)CO(g)£«H2(g)£»C(s)¡¢CO(g)ºÍH2(g)ÍêȫȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ£º

C(s)£«O2(g)===CO2(g)    ¦¤H£½£­393.5 kJ·mol£­1

H2(g)£«O2(g)===H2O(g)  ¦¤H£½£­242.0 kJ·mol£­1

CO(g)£«O2(g)===CO2(g)  ¦¤H£½£­283.0 kJ·mol£­1

Çë»Ø´ð£º

(1)¸ù¾ÝÒÔÉÏÊý¾Ý£¬Ð´³öC(s)ÓëË®ÕôÆø·´Ó¦µÄÈÈ»¯Ñ§·´Ó¦·½³Ìʽ£º________________________________________________________________________¡£

(2)±È½Ï·´Ó¦ÈÈÊý¾Ý¿ÉÖª£¬1 mol CO(g)ºÍ1 mol H2(g)ÍêȫȼÉշųöµÄÈÈÁ¿Ö®ºÍ±È1 mol C(s)ÍêȫȼÉշųöµÄÈÈÁ¿¶à¡£¼×ͬѧ¾Ý´ËÈÏΪ¡°Ãº×ª»¯ÎªË®ÃºÆø¿ÉÒÔʹúȼÉշųö¸ü¶àµÄÈÈÁ¿¡±£»ÒÒͬѧ¸ù¾Ý¸Ç˹¶¨ÂÉ×ö³öÏÂÁÐÑ­»·Í¼£º

²¢¾Ý´ËÈÏΪ¡°Ãº×ª»¯ÎªË®ÃºÆøÔÙȼÉշųöµÄÈÈÁ¿Óëúֱ½ÓȼÉշųöµÄÈÈÁ¿ÏàµÈ¡±¡£

Çë·ÖÎö£º¼×¡¢ÒÒÁ½Í¬Ñ§¹ÛµãÕýÈ·µÄÊÇ__________(Ìî¡°¼×¡±»ò¡°ÒÒ¡±)£»ÅжϵÄÀíÓÉÊÇ________________________________________________________________________

________________________________________________________________________¡£

(3)½«Ãº×ª»¯ÎªË®ÃºÆø×÷ΪȼÁϺÍúֱ½ÓȼÉÕÏà±ÈÓкܶàÓŵ㣬ÇëÁÐ¾ÙÆäÖеÄÁ½¸öÓŵ㣺________________________________________________________________________

________________________________________________________________________¡£

(4)Ë®ÃºÆø²»½öÊÇÓÅÁ¼µÄÆøÌåȼÁÏ£¬Ò²ÊÇÖØÒªµÄÓлú»¯¹¤Ô­ÁÏ¡£COºÍH2ÔÚÒ»¶¨Ìõ¼þÏ¿ÉÒԺϳɣº¢Ù¼×´¼¡¡¢Ú¼×È©¡¡¢Û¼×Ëá¡¡¢ÜÒÒËá¡£ÊÔ·ÖÎöµ±COºÍH2°´1¡Ã1µÄÌå»ý±È»ìºÏ·´Ó¦£¬ºÏ³ÉÉÏÊö________(ÌîÐòºÅ)ÎïÖÊʱ£¬¿ÉÒÔÂú×ã¡°ÂÌÉ«»¯Ñ§¡±µÄÒªÇó£¬ÍêÈ«ÀûÓÃÔ­ÁÏÖеÄÔ­×Ó£¬ÊµÏÖÁãÅÅ·Å¡£

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

¾«Ó¢¼Ò½ÌÍø