ÌâÄ¿ÄÚÈÝ

ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£º
¢ÙH2£¨g£©+
1
2
O2£¨g£©=H2O£¨l£©¡÷H=-285kJ?mol-1
¢ÚH2£¨g£©+
1
2
O2£¨g£©=H2O£¨g£©¡÷H=-241.8kJ?mol-1
¢Û2C£¨s£©+O2£¨g£©=2CO£¨g£©¡÷H=-110.5kJ?mol-1
¢ÜC£¨s£©+O2£¨g£©=CO2£¨g£©¡÷H=-393.5kJ?mol-1
¢ÝCH4£¨g£©+2O2£¨g£©¨TCO2£¨g£©+2H2O£¨l£©¡÷H=-890kJ?mol-1
»Ø´ðÏÂÁÐÎÊ£º
£¨1£©H2µÄȼÉÕÈÈΪ
 
kJ?mol-1£»CµÄȼÉÕÈÈΪ
 
kJ?mol-1
£¨2£©È¼ÉÕ1gH2Éú³ÉҺ̬ˮ£¬·Å³öµÄÈÈÁ¿Îª
 
kJ
£¨3£©·´Ó¦2CO£¨g£©+O2£¨g£©=2CO2£¨g£©¡÷H=
 
kJ?mol-1
£¨4£©½«2mol µÄH2ºÍCH4»ìºÏÆøÌå³ä·ÖȼÉÕÉú³ÉCO2ºÍҺ̬H2O¹²·Å³ö1381.8KJµÄÈÈÁ¿£¬Ôò¸Ã»ìºÏÆøÌåÖÐH2ºÍCH4µÄÎïÖʵÄÁ¿Ö®±ÈÊÇ
 
£®
¿¼µã£ºÈÈ»¯Ñ§·½³Ìʽ,Óйط´Ó¦ÈȵļÆËã
רÌ⣺»¯Ñ§·´Ó¦ÖеÄÄÜÁ¿±ä»¯
·ÖÎö£º£¨1£©È¼ÉÕÈÈÊÇ1mol¿ÉȼÎïÍêȫȼÉÕÉú³ÉÎȶ¨Ñõ»¯Îï·Å³öµÄÈÈÁ¿£¬½áºÏÈÈ»¯Ñ§·½³Ìʽ·ÖÎöµÃµ½£»
£¨2£©È¼ÉÕ1gH2Éú³ÉҺ̬ˮ£¬·Å³öµÄÈÈÁ¿ÒÀ¾ÝÈÈ»¯Ñ§·½³Ìʽ¢Ù¼ÆËãµÃµ½£»
£¨3£©ÒÀ¾Ý¢Û¢Ü½áºÏ¸Ç˹¶¨ÂɼÆËãµÃµ½ËùÐèÈÈ»¯Ñ§·½³Ìʽ£»
£¨4£©ÒÀ¾ÝÈÈ»¯Ñ§·½³Ìʽ¢Ù¢ÝµÄ¶¨Á¿¹ØÏµ¼ÆËãµÃµ½£»
½â´ð£º ½â£º£¨1£©È¼ÉÕÈÈÊÇ1mol¿ÉȼÎïÍêȫȼÉÕÉú³ÉÎȶ¨Ñõ»¯Îï·Å³öµÄÈÈÁ¿£¬ÈÈ»¯Ñ§·½³Ìʽ¢ÙºÍ¢Ü·ûºÏȼÉÕÈȵĸÅÄ¹ÊÇâÆøÈ¼ÉÕÈÈΪ285kJ?mol-1£¬Ì¼µÄȼÉÕÈÈΪ£»393.5kJ?mol-1£»
¹Ê´ð°¸Îª£º285£¬393.5£»
£¨2£©È¼ÉÕ1gH2ÎïÖʵÄÁ¿Îª0.5mol£¬Éú³ÉҺ̬ˮ£¬ÒÀ¾ÝÈÈ»¯Ñ§·½³Ìʽ£¬H2£¨g£©+
1
2
O2£¨g£©=H2O£¨l£©¡÷H=-285kJ?mol-1£¬È¼ÉÕ1gH2Éú³ÉҺ̬ˮ£¬·Å³öµÄÈÈÁ¿Îª142.5KJ£»
¹Ê´ð°¸Îª£º142.5£»
£¨3£©¢Û2C£¨s£©+O2£¨g£©=2CO£¨g£©¡÷H=-110.5kJ?mol-1
¢ÜC£¨s£©+O2£¨g£©=CO2£¨g£©¡÷H=-393.5kJ?mol-1
ÒÀ¾Ý¸Ç˹¶¨ÂÉ£¬¢Ü¡Á2-¢ÛµÃµ½2CO£¨g£©+O2£¨g£©=2CO2£¨g£©¡÷H=-676.5KJ/mol£»
¹Ê´ð°¸Îª£º-676.5£»
£¨4£©2mol µÄH2ºÍCH4»ìºÏÆøÌå³ä·ÖȼÉÕÉú³ÉCO2ºÍҺ̬H2O¹²·Å³ö1381.8KJµÄÈÈÁ¿£¬ÉèÇâÆøÈ¼ÉÕÎïÖʵÄÁ¿Îªx£¬¼×ÍéȼÉÕÎïÖʵÄÁ¿Îªy£¬ÒÀ¾ÝÈÈ»¯Ñ§·½³ÌʽÁÐʽ¼ÆË㣬
¢ÙH2£¨g£©+
1
2
O2£¨g£©=H2O£¨l£©¡÷H=-285kJ?mol-1
  1                             285KJ
  x                             285xKJ
CH4£¨g£©+2O2£¨g£©¨TCO2£¨g£©+2H2O£¨l£©¡÷H=-890kJ?mol-1
1                                        890
y                                        890y
285x+890y=1381.8
x+y=2
¼ÆËãµÃµ½x=0.658mol
y=1.342£¬
x£ºy¡Ö1£º2
¹Ê´ð°¸Îª£º1£º2£»
µãÆÀ£º±¾Ì⿼²éÁËÈÈ»¯Ñ§·½³ÌʽµÄÊéд·½·¨ºÍ¸Ç˹¶¨ÂɼÆËãÓ¦Óã¬ÈÈ»¯Ñ§·½³ÌʽµÄ¼ÆËãÊǹÌÌ壬կÎÕ»ù´¡Êǹؼü£¬ÌâÄ¿ÄѶÈÖеȣ®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
¼×¡¢ÒÒ¡¢±û¡¢¶¡ËÄÖÖµ¥ÖÊÔÚµãȼÌõ¼þÏÂÁ½Á½»¯ºÏÉú³ÉX¡¢Y¡¢Z¡¢WËÄÖÖ»¯ºÏÎת»¯¹ØÏµÈçͼËùʾ£®ÓÖÖª£º¢Ù¼×¡¢ÒÒ¡¢±û¾ùΪǰÈýÖÜÆÚÔªËØµÄµ¥ÖÊ£¬³£ÎÂϾùÎªÆøÌ¬£¬¶¡ÊÇÈÕ³£Éú»îÖеÄÒ»ÖÖ³£¼û½ðÊô£»¢Ú³£ÎÂÏ£¬XÊÇÎÞɫҺÌ壬YÊǺÚÉ«¹ÌÌ壻¢Û±ûÔÚÒÒÖÐȼÉÕ·¢³ö²Ô°×É«»ðÑæ£¬¶¡ÔÚÒÒÖÐȼÉÕÉú³Éר»ÆÉ«µÄÑÌ£¬WµÄË®ÈÜÒº³Ê»ÆÉ«£®Çë»Ø´ð£º
£¨1£©Ð´³öʵÑéÊÒÖÆÈ¡ÒÒµÄÀë×Ó·½³Ìʽ
 
£®ÇëÓõç×Óʽ±íʾZµÄÐγɹý³Ì
 
£®
£¨2£©Í¨³£¶¡µÄÖÆÆ·Ôڼ׺ÍXͬʱ´æÔÚµÄÌõ¼þÏ¿ÉÒÔ±»¸¯Ê´£¬·¢Éú¸¯Ê´Ê±µÄÕý¼«·´Ó¦Ê½Îª
 
£®
£¨3£©¹¤ÒµÉÏÒ±Á¶¶¡³£²ÉÓÃ
 
·¨£»½«ÊÊÁ¿µÄ¶¡¼ÓÈëWµÄÈÜÒºÖУ¬Ó¦¸Ã¹Û²ìµ½µÄÏÖÏóÊÇ
 
£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ
 
£»ÈôÒ»¶¨Ìõ¼þ϶¡·Ö±ðÓëX¡¢Z·¢Éú·´Ó¦Éú³ÉµÈÎïÖʵÄÁ¿µÄÆøÌ壬ÔòÏûºÄ¶¡µÄÎïÖʵÄÁ¿Ö®±ÈΪ
 
£®
£¨4£©ÔÚÊ¢ÓÐZµÄË®ÈÜÒºµÄÊÔ¹ÜÀï¼ÓÈëÊÊÁ¿µÄ¶¡³ä·Ö·´Ó¦£®Çëд³öÏò·´Ó¦ºóµÄÈÜÒºÖеμÓNaOHÈÜÒº£¬²¢·ÅÖÃÒ»¶Îʱ¼äËù·¢Éú·´Ó¦µÄ·½³Ìʽ£¨ÊôÓÚÀë×Ó·´Ó¦µÄÓÃÀë×Ó·½³Ìʽ±íʾ£©
 
£®

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

¾«Ó¢¼Ò½ÌÍø